SOME BASIC CONCEPTS OF CHEMISTRY
Chapter Outline
1.1 Importance of Chemistry
1.2 Nature of Matter
1.3 Properties of Matter and Their Measurement
1.4 Uncertainty in Measurement
1.5 Laws of Chemical Combinations
1.6 Dalton’s Atomic Theory
1.7 Atomic and Molecular Mass
1.8 Mole Concept and Molar Masses
1.9 Percentage Composition
1.10 Stoichiometry and Stoichiometric Calculations
This chapter deals with quantitative aspects of chemical entities, namely atoms, molecules, ions etc.,
1.1 IMPORTANCE OF CHEMISTRY
Chemistry deals with the study of materials, their preparations, their properties, their mutual interactions and their uses in all possible walks of life. Chemistry plays an important role in daily life.
Chemistry is responsible for designing safer alternatives to environmentally hazardous refrigerates, like CFCs, responsible for ozone depletion in the stratosphere, have been successfully synthesized.
1.2 NATURE OF MATTER
Anything that has mass and occupies space is called matter. Matter can exist in three physical states, namely, solid, liquid, and gas. Solids have definite volume and definite shape liquids have definite volume but not definite shape. Liquids take the shape of the container in which they are placed. Gases have neither definite volume nor definite shape. They completely occupy the container in which they are placed.
The constituent particles of solids are held very close to each other in an orderly fashion. In liquids, the particles are close to each other but they can move around. The particles are far apart in gases, as compared to those present in solid and liquid states, and so, their movement is easy and fast.
By changing temperature and pressure, these three states of matter are interconvertible.
Matter can be classified as mixtures or pure substances. These can be further sub-divided as shown below:
The components in a homogeneous mixture completely mix with each other and its composition is uniform throughout. The composition is not uniform throughout in a heterogeneous mixture and, sometimes, the different components can also be observed. The components of a mixture can be separated by physical methods, such as hand picking, filtration, crystallisation, distillation, etc. Pure substances have fixed composition. The constituents of pure substances cannot be separated by simple physical methods.
Pure substances can be further classified into elements and compounds. An element consists of only one type of particle. These particles may be atoms or molecules. In some others, two or more atoms combine to give molecules of the element. When two or more atoms of different elements combine, the molecules of a compound are formed.
The atoms of different elements are present in a compound in a fixed and definite ratio and this ratio is characteristic of a particular compound. The properties of a compound are different from those of its constituent elements.
The smallest particle of an element or compound which is capable of independent existence is called ‘molecule.’
Two or more atoms of the same or different elements combine to give a molecule of compound. Atoms may lose or gain electrons to form ions.
Pure substances are of two types. The substance that gives similar atoms on fine division is called element.
Example: H2, Cl2, O2, O3, P4, S8, Cn, etc.
The substance that gives different atoms on division is called a compound.
Example: HCl, H2O, NH3, NaOH, H2SO4, C6H12O6, etc.
Compounds may be ionic or covalent. Compounds have definite composition of the combining elements.
Number of atoms constituting a molecule is called atomicity.
Mass of one H-atom = 1.6736×10 –24 g = 1.66×10–24 g.
C-12 scale is also called unified scale and, on this scale, the symbol ‘u’ is used in place of amu.
Based on C-12 scale, atomic masses of hydrogen and oxygen are calculated as 1.007825 amu and 15.995 amu, respectively.
Now, atomic mass of an element may be defined as “the number of times an atom of an element is heavier than 1 amu” or 1 12 th mass of C-12 isotope. Atomic weight is also called atomic mass.
1. Noble gases are monoatomic, i.e., each molecule has only one atom.
Example: He, Ne, Ar, Kr, Xe, and Rn
2. Elementary gases are diatomic, i.e, each molecule has two atoms.
Example: H2, N2, O2, and Cl2
3. Ozone (O3) is triatomic. White phosphorus (P4) is tetraatomic.
4. Rhombic sulphur (S 8 ) is octaatomic. Atomicity of some familiar substances is given in Table 1.1
Table 1.1 Atomicity of familiar substances
1.3 PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substance in nature shows both physical and chemical properties. Colour, odour, melting point, boiling point, density, etc. are examples of physical properties.
Two different systems of measurement, namely English system and metric system, are used in different parts of the world. The metric system is more convenient as it is based on the decimal system.
1.3.1 The International System of Units (SI)
The international system of units was established by the 11 th General Conference on Weights and Measures (CGPM). The SI system has seven base units and they are listed in Table 1.2.
Table 1.2 Physical quantities and their units
Base
Length metre m
Mass m kilogram kg
Time t second s
Electric current I ampere A
Thermodynamic temperature T kelvin K
Amount of substance n mole mol
Luminous intensity I u candela cd
The definitions of the SI base units are given in Table 1.3. The SI system allows the use of prefixes to indicate the multiples or submultiples of a unit. These prefixes are listed in Table 1.4.
1.3.2 Mass
Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object. The mass of a substance remains a constant but its weight may vary from one place to another due to change in gravity.
The SI unit of mass is kilogram (1 kg = 1000 g)
1.3.3 Volume
The SI units of volume is m 3 . 1 m3 = 1000 dm3; 1 dm3 = 1000 cm3
1 L = 1000 mL
In the laboratory, volume of liquids or solutions can be measured by graduated cylinder, burette, pipette, etc.
1.3.4 Density
Density of a substance is its amount of mass per unit volume. SI unit of density is kg. m –3 .
1.3.5 Temperature
There are three common scales to measure temperature: (1) degree Celsius (°C), (2) degree Fahrenheit (°F), (3) Kelvin (K). Kelvin scale is SI unit of temperature.
The Fahrenheit and Celsius scales are related by () 9 FC32 5 °=°+
The Kelvin scale is related to Celsius scale as follows; K = °C + 273.15
Many a times, in the study of chemistry, one has to deal with experimental data as well as theoretical calculations.
There are meaningful ways to handle the numbers conveniently and present the data realistically with certainty to the extent possible. These ideas are discussed below in detail.
Table 1.3 Definitions of SI base units
Properties
Length metre
Mass kilogram
Time second
Electric current ampere
The metre is the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second.
The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram.
The second is the duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.
The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length and negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to 2 × 10–7 newton per metre of length.
Temperature kelvin
Amount of substance mole
Luminous intensity candela
The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.
The mole is the amount of substance of a contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12; its symbol is ‘mol’.
When the mole is used, the elementary entities must be specified; they may be atoms, molecules, ions, electrons, and other particles or specific groups of such particles.
The candela is the luminous intensity in a given direction, of a source that emits monochromatic radiation of frequency 540 × 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.
Table 1.4 Prefixes used in the SI system
1.4 UNCERTAINTY IN MEASUREMENT
As chemistry is the study of atoms and molecules, which have extremely low masses and are present in extremely large numbers, a chemist has to deal with numbers as large as 602, 200, 000, 000, 000, 000, 000, 000 for the molecules of 2 g of hydrogen gas or as small as 0.00000000000000000000000166 g mass of a hydrogen atom.
Similarly, other constants, such as Plank’s constant, speed of light, charges on particles, etc., involve numbers of the above magnitude.
It may look funny for a moment to write or count numbers involving so many zeros but it offers a real challenge to do simple mathematical operations of addition, subtraction, multiplication, or division with such numbers.
You can write any two numbers of the above type and try any one of the operations you like to accept the challenge and, then, you will really appreciate the difficulty in handling such numbers.
This problem is solved by using scientific notation for such numbers, i.e., exponential notation, in which any number can be represented in the form N × 10n, where n is an exponent having positive or negative values and N can vary between 1 to 10. Thus, we can write 462.207 as 4.62207 × 10 2 in scientific notation.
Note that while writing it, the decimal had to be moved to the left by two places and same is the exponent (2) of 10 in the scientific notation.
Similarly, 0.00032 can be written as 3.2 × 10–4. Here the decimal has to be moved four places to the right and (–4) is the exponent in the scientific notation.
Now, for performing mathematical operations on numbers expressed in scientific notations, the following points are to be kept in mind.
1.4.1 Significant Figures
Every experimental measurement has some amount of uncertainty associated with it. However, one would always like the results to be precise and accurate.
Precision and accuracy are often referred to while we talk about the measurement. Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value of the result.
For example, if the true value for a result is 2.00 g and a student ‘A’ takes two measurements and reports the results as 1.95 g and 1.93 g, these values are precise as they are close to each other but are not accurate. Another student repeats the experiment and obtains 1.94 g and 2.05 g as the results for the measurements. These observations are neither precise nor accurate. When a third student repeats these measurements and reports 2.01 g and 1.99 g as the result, these values are both precise and accurate.
The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures.
Significant figures are meaningful digits which are known with certainty. The uncertainty is indicated by writing the certain digits and the last uncertain digit.
Thus, if we write a result as 11.2 mL, we say that 11 is certain and 2 is uncertain, and the uncertainty would be 1 in the last digit, which is always understood. There are certain rules for determining the number of significant figures. These are stated one by one.
1. All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.
2. Zeros preceding the first non-zero digit are not significant. Such a zero indicates the position of decimal point.
Thus, 0.03 has one significant figure and 0.0052 has two significant figures.
3. Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.
4. Zeros at the end or right of a number are significant, provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures.
But, if otherwise, the terminal zeros are not significant if there is no decimal point. For example, 100 has only one significant figure, but 10.0. has three significant figures and 100.0 has four significant figures. Such numbers are better represented in scientific notation.
We can express the number 100 as 1 × 102 for one significant figure, 1.0 × 102 for two significant figures, and 1.00 × 102 for three significant figures.
5. Numbers of objects, for example, 2 balls or 20 eggs, have infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal, i.e., 2=2.000000.... or 20=20.000000....
In numbers written in scientific notation, all digits are significant, e.g., 4.01 × 102 has three significant figures and 8.256 × 10 –3 has four significant figures.
Addition and Subtraction
In addition and subtraction, the final result should be reported to the same number of decimal places as present in the quantity with the minimum number of decimal places. For that, first, the numbers are written in such a way that they have the same exponent.
Example:
1) 2.6 kg (one decimal place) + 3.44 kg (two decimal places) 6.04 kg
The final result is corrected to one decimal place as 6.0 kg.
2) 8.62 + 11.354 + 14.4 052 34.3 792
The number should be reported up to two decimal places, and hence, is rounded off to 34.38.
3) 6.65×104 + 8.95×103 = 6.65×104 + 0.895×104 = (6.65 + 0.895)104 = 7.545×104
Answer is 7.54 × 104
4) 35.648 – 22.12 13.528
On rounding off, it is reported as 13.53.
5) 2.5×10–2 – 4.8×10–3 = (2.5×10–2) – (0.48×10–2) = (2.5 – 0.48)10–2 = 2.02×10–2
Multiplication and Division
In multiplication and division, the final result should be reported to the same number of significant figures as present in the quantity with minimum number of significant figures.
1) 2.5 × 1.25 = 3.125 It is 3.1
2) 46.512 × 1.67 = 77.67504 It is 77.7
3) 5.4221 × 15.6 = 84.58476 It is 84.6
4) 312.6 ÷ 14.68 = 21.244277 It is 21.24
5) 515.69 ÷ 2.6101 = 197.57480 It is 197.57
While limiting the result to the required number of significant figures, as done in the above mathematical operation, one has to keep in mind the following points for rounding off the numbers.
■ If the right-most digit to be removed is more than 5, the preceding number is increased by one, e.g., in 2.487, if we have to remove 7, we have to round it to 2.49.
■ If the right-most digit to be removed is less than 5, the preceding number is not changed, e.g., in 7.923, if we have to remove 3, we have to round it off 7.92.
■ If the right most digit to be removed is 5, then
i) the preceding number is not changed if it is an even number e.g., 7.45 should be rounded off to 7.4.
ii) the preceding number is increased by one, if it is an odd number.
Example: 6.35 should be rounded off to 6.34.
1.4.2 Dimensional Analysis
Often, while calculat ing, there is a need to convert units from one system to another. The method used to accomplish this is called factor label method or unit factor or dimensional analysis.
1. A jug contains 2 L of milk. Calculate the volume of the milk in m3 .
Sol. Since 1 L = 1000 cm3 and 1 m = 100 cm, 1m100cm =1= 100cm1 m
To get m3 from the above unit factors, the first unit factor is taken and it is cubed.
Sol. Find the exact mass and volume of the ring and calculate its density. By comparing the density obtained with that of pure gold, one can get the idea of its purity.
Try yourself:
1. How many significant figures are there in ‘ p ’? Ans: This is a non-terminating and nonrecurring value. Hence, the number of significant figures in ‘ p ’ is infinity.
TEST YOURSELF
1. How many seconds are there in 2 days?
(1) 172800 (2) 15820 (3) 16466 (4) 28800
2. Add 6.65 × 104 and 8.95 × 103 . (1) 7.545 × 104 (2) 7.54 × 104 (3) 7 × 104 (4) 7 × 103
3. Give the answer for 2.5 × 1.25 in significant figures.
(1) 3.1 (2) 3.12 (3) 3.13 (4) 3.125
4. Which one of the following has four significant figures?
(1) 0.00050 (2) 6.300 (3) 0.052 (4) 6.02 × 1023
5. 0.00025 has how many significant figures? (1) 5 (2) 3 (3) –4 (4) 2
6. The number of significant figures in electronic charge 1.602 × 10 –19 C is (1) 1 (2) 2 (3) 3 (4) 4
Now, 2L = 2 × 1000 cm
The above is multiplied by the unit factor. 33 3 33 633 1m2m 21000cm 210m 10cm10 ××==×
2. Density of pure gold is 19.3 g/cm3. When a gold ring is purchased from a jewellery shop, how will you check its purity?
7. The correctly reported answer of the addition of 4.523, 2.3, and 6.24 will have how many significant figures?
(1) Two (2) Three (3) Four (4) Five
8. After rounding off 1.235 and 1.225 to three significant figures, we will have their answers, respectively, as (1) 1.23, 1.22 (2) 1.24, 1.123 (3) 1.23, 1.23 (4) 1.24, 1.22
9. 8.281 has how many significant figures? (1) 1 (2) 2 (3) 3 (4) 4
10 Planck’s constant has the dimensions of (1) force (2) work (3) angular momentum (4) torque
11. Which of the following units represents the largest amount of energy? (1) calorie (2) erg (3) joule (4) electron-volt
Answer Key
(1) 1 (2) 2 (3) 1 (4) 2 (5) 4 (6) 4 (7) 1 (8) 4 (9) 4 (10) 3 (11) 1
1.5 LAWS OF CHEMICAL COMBINATIONS
There are five imp ortant laws of chemical combination:
1. The law of conservation of mass
2. The law of definite proportions
3. The law of multiple proportions
4. The law of combining volumes
5. The law of reciprocal proportions
1.5.1 The Law of Conservation of Mass
This law was proposed by Antoine Lavoisier. He performed careful experimental studies for combustion reactions.
Statement : According to the law of conservation of mass, “the matter can neither be created nor destroyed during a chemical change.”
This law can also be stated as: the total mass of the products formed during a chemical change is exactly equal to the total mass of the reactants.
Example 1: (NH4)2SO4(aq)+ BaCl2(aq)→BaSO4(s)+ 2NH4Cl(aq)
Total mass of (NH4)2SO4 + BaCl2 is equal to the total mass of BaSO4 + NH4Cl. Thus, the law of conservation of mass is verified.
Example 2: AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)
Law of conservation of mass is not applicable to unbalanced equation.
3. When 8.4 g of NaHCO3 is added to 3.65 g of HCl, then 5.85 g NaCl and 1.8 g water are formed. How many grams of CO2 gas escaped from the container?
Sol. Equation for the reaction: NaHCO3 + HCl → NaCl + H2O + CO2
8.4 g 3.65 g 5.85 g 1.8 g x g
According to the law of conservation of mass, 8.4 + 3.65 = 5.85 + 1.8 + x grams of CO2
From this equation, x = 4.4 g.
Try yourself:
2. For a hypothetical equation, 2A(g) + 3B(g) → A2B3(g), 25 g of reactants are allowed to react. In the end, sum of the masses of the products formed and reactants unreacted is equal to 25 g. Why?
Ans: This is according to the law of conservation of mass.
1.5.2 The Law of Definite Proportions
The law of definite proportions was proposed by the French chemist Joseph Proust. This law is also called the law of constant or the law of fixed proportions.
Statement: A given chemical substance always contains the same elements combined in a fixed proportion by weight.
Proust worked with two samples of cupric carbonate.
Irrespective of the source, cupric carbonate always contains Cu, C, and O combined in the same proportion by weight.
Example: A sample of carbon dioxide may be prepared by chemical methods or by fermentation.
5. 60 g of compound (x) contains 25 g of element ( y ) and 35 g of element (z ). What mass of compound (x) is formed when 5 g of ( x) is combined with 9 g of (z), if the given data obeys the law of definite proportions?
Sol. Ratio of masses ( y ) and ( z ) in 60 g of compound (x) = 25 : 35 = 5 : 7
5 g of (y) reacts with 7 g of (z) to form 12 g of compound ( x ) and 9 – 7 = 2 g of ( z ) is left unreacted.
Try yourself:
3. Why is the law of definite proportions not applicable to 12CO2 and CO18 2 , though both oxides are made up of the same elements C and O?
It is found that carbon dioxide is made up of the same elements, carbon and oxygen, irrespective of the method of preparation. The elements carbon and oxygen are combined together in the same fixed ratio of 3 : 8 by weight.
The importance of the law of definite proportions is that it is useful in deriving the chemical formulae of compounds.
Percent weight of a constituent element = 100 Atomic weightofelementNumber of atoms Grammolecularweightofcompound × ×
4. Two oxides of metal M have 27.6% and 30% oxygen by weight. If the formula of the first oxide is M3O4, what is the formula of the second oxide?
Sol. Oxide (A) : Oxygen 27.6%; 4’O’ atoms
Metal 72.4%; 3 ‘M’ atoms
Oxide (B) : Oxygen 30%; Metal 70% Number of ‘O’ atoms
Ratio of M and O atoms in oxide B = 2.9 : 4.35 = 2 : 3
Formula of the second oxide is M 2O3
Combining ratio is different; thus, it is not applicable.
Combining ratio of masses of C and O in 18CO 2 is 12 : 36 = 3:9
Ans: Combining ratio of masses of C and O in 2CO12 is 12 : 32 = 3.8.
1.5.3 The Law of Multiple Proportions
The law of multiple proportions was proposed by John Dalton.
Statement: If two elements chemically combine to give two or more compounds, then the weights of one element that combined with a fixed weight of the other element, in those compounds are in the ratio of small whole numbers.
Based on the law of definite proportions and the law of conservation of mass, law of multiple proportions was proposed.
Example : Hydrogen combines with oxygen to form two compounds, water and hydrogen peroxide.
Hydrogen(2 g) + Oxygen(16 g) → Water (18 g)
Hydrogen (2 g) + Oxygen (32 g) →H2O2(34 g)
The masses of oxygen which combines with a fixed mass of hydrogen bears a simple ratio of 1:2.
The ratio of different weights of oxygen that combine with 14 g of N2 to form oxides, i.e, N2O, NO, N2O3, NO2, and N2O5, is 1 : 2 : 3 : 4 : 5.
Some other examples illustra ting this law are given below:
1. S2Cl2, SCl2, and SCl4
2. N2O, NO, N2O3, NO2, and N2O5
3. CrO, Cr2O3, and CrO3
4. Cl2O, ClO2, Cl2O6, and Cl2O7
5. H2S, H2S2, and H2S5
6. In two oxides of nitrogen, oxide-1 has 63.64% N and oxide-2 has 46.67% N. Prove that these figures illustrate the law of multiple proportions.
Sol.
Number of parts by mass of nitrogen that combines with one part of oxygen:
Oxide-1: = 63.64 1.75 36.36
Oxide-2: 46.67 0.875 53.33 =
The ratio of masses of nitrogen that combine with the same mass (i.e., one part by mass) of oxygen = 1.75 : 0.875 = 2 : 1
This is a simple whole number ratio. Hence, the given figures illustrate the law of multiple proportions.
Try yourself:
4. Does the formation of hydrocarbons explain the law of multiple proportions?
Example: ,10H4C ,12H5C 14H6C
Ans: No, it does not explain the law of multiple proportions as the parts by mass of H that combine with 12 parts by mass of C is not in a simple whole number ratio.
1.5.4 Law of Combining Volumes
This law was proposed by Gay-Lussac. The law states that “when gases under similar conditions of temperature and pressure chemically combine, they bear a simple ratio in their volumes. If the product is also a gas, the ratio of the volumes can be extended to its volume also.”
Example: Hydrogen chloride gas is produced from its gaseous elements in 1 : 1 volume ratio, as illustrated in Fig.1.1
Fig. 1.1 Illustration of Gay-Lussac’s law of formation of hydrogen chloride
The simple ratio between the volumes of H 2, Cl2, and HCl in the reaction is 1 : 1 : 2.
Hydrogen + Oxygen → Water
100 ml 50 ml 100 ml
100 ml of hydrogen combines with 50 ml of oxygen to give 100 ml of water vapour. The ratio of volumes of hydrogen and oxygen which combine together is 2 : 1.
Gay-Lussac’s discovery of integer ratio in volume relationship is actually the law of definite proportions by volume. The law of constant proportions stated earlier is definite proportions by mass. The Gay-Lussac’s law was explained by Amedeo Avogadro. Avogadro proposed that “equal volumes of gases at the same pressure and temperature contain equal number of molecules.” Later, Avogadro’s hypothesis is generalised as Avogadro’s law of gases.
Berzelius Hypothesis
Berzelius proposed that there is a relationship between, the volume of a gas and its atoms. He stated that “equal volumes of all gases contain equal number of atoms under similar conditions of temperature and pressure.”
Nitrogen + Hydrogen → Ammonia
1 vol 3 vol 2 vol
n atoms 3n atoms 2n compound atoms
1 2 atom 3 2 atoms 1 compound atom
This implies that one compound atom of ammonia gas is made up of 1 2 atom of nitrogen and 3 2 atoms of hydrogen. However, according to Dalton’s atomic theory, atoms are indivisible and are ultimate particles of elements. Therefore, the above hypothesis was rejected.
Avogadro’s Law
According to Avogadro’s hypothesis, “equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.”
The reaction between nitrogen and hydrogen can be explained on the basis of Avogadro’s law, as follows:
Nitrogen + Hydrogen → Ammonia
1 vol + 3 vol 2 vol
(By experiment)
n molecules + 3n molecules → 2n molecules
(By Avogadro’s law)
1 2 molecule + 3 2 molecules → 1 molecule (By dividing throughout by 2n)
One molecule of ammonia is made up of 1 2 molecule of nitrogen and 3 2 molecules of hydrogen. Since a molecule is made up of two
or more atoms, so, 1 2 molecule may contain one or more atoms. Thus, this result does not contradict Dalton’s atomic theory.
7. What is the volume ratio of the product gases in the decomposition of phosphorus pentachloride?
Sol. PCl5(g) → PCl3 (g) + Cl2( g)
Volume ratio of PCl3 and Cl2 is 1 : 1.
8. In what volume ratio does nitrogen gas and hydrogen gas combine to give ammonia under similar conditions of temperature and pressure?
Sol. N2 + 3H2 → 2NH3
Volume ratio of N2 and H2 = 1 : 3.
Try yourself:
5. Why does a balloon expand on blowing air?
Ans: When air is blown into a balloon, more air molecules enter into the balloon, thus expanding it.
1.5.5 Law of Reciprocal Proportions
This law was proposed by Ritcher. The law states that “the ratio of weights of two elements X and Y, which combine separately with a fixed weight of the third element Z, is simple multiple of the ratio of the weights in which X and Y combine directly with each other.”
Let us consider the example of two elements oxygen and hydrogen separately with the third element sulphur to form SO2 and H 2S. The elements combine directly with each other to form water, as shown in Fig.1.2. H H2O H2S SO2 S O
Fig. 1.2 Illustration of binary compounds between O, S, and H
The weight of hydrogen that combines with 32 parts of sulphur is 2.
The weight of oxygen that combines with 32 parts of sulphur is 32.
Therefore, the weight ratio of hydrogen and oxygen which separately combine with the same weight of sulphur in H 2S and SO 2 is 1 : 16.
When hydrogen and oxygen combine to form water, the ratio of the weights of combining elements is 1:8.
The importance of law of reciprocal proportions is that the law led to the understanding of the concept of chemical equivalents and equivalent weights.
9. Carbon dioxide contains 27.27% of carbon, carbon disulphide contains 15.79% of carbon and sulphur dioxide contains 50% of sulphur. Are these figures in agreement with the law of reciprocal proportions?
Sol. Yes,
1 g C will combine with S =
1 g C will combine with O =
=
∴ Ratio of masses of S and O, which combine with fixed mass of carbon (via 1 g) = 5.33 : 2.67 = 2 : 1. Ratio of masses of S and O which combine directly with each other = 50 : 50 = 1 : 1. Thus, the two ratios are simple multiples of each other.
Try yourself:
6. What is the difference between the law of multiple proportions and the law of reciprocal proportions?
Ans: In multiple proportions, two elements combine in more than one ratio of masses to form different compounds. In reciprocal proportions, two different elements combine separately with the same weight of the third element.
TEST YOURSELF
1. Choose the correct statement(s) among the following.
(1) Chemical equation is balanced according to the law of conservation of mass.
(2) Law of conservation is not applicable for nuclear reactions.
(3) Law of combining volumes was proposed by Gay-Lussac.
(4) All of the above
2. Two gaseous samples were analysed. One contained 1.2 g of carbon and 3.2 g of oxygen. The other contained 27.3% carbon and 72.7% oxygen. The experimental data is in accordance with the (1) law of conservation of mass
(2) law of definite proportions
(3) law of reciprocal proportions
(4) law of multiple proportions
3. Percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This proves the law of (1) constant proportions
(2) reciprocal proportions
(3) multiple proportions
(4) conservation of mass
4. “The total mass of reactants is always equal to the total mass of products in a chemical reaction.” This statement is known as (1) the law of conservation of mass
(2) the law of definite proportions
(3) the law of equivalent weights
(4) the law of combining masses
5. The reaction hydrogen (g) + oxygen (g) gives water vapour; the ratio of volumes is 2 : 1 : 2. This illustrates the law of
(1) conservation of mass
(2) combining weights
(3) combining volumes
(4) all of the above
6. Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates the (1) law of reciprocal proportions
(2) law of constant proportions
(3) law of multiple proportions
(4) law of equivalent proportions
7. One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combine together, the ratio of their masses will be governed by
(1) the law of definite proportions
(2) the law of multiple proportions
(3) the law of reciprocal proportions
(4) the law of conservation of mass
8. 14 g of element X combines with 16 g of oxygen. On the basis of this information, which of the followings is a correct statement?
(1) The element X could have an atomic weight of 7 and its oxide is XO.
(2) The element X could have an atomic weight of 14 and its oxide formula is X2O.
(3) The element X could have an atomic weight of 7 and its oxide is X 2O.
(4) The element X could have an atomic weight of 14 and its oxide is XO 2
9. Carbon and oxygen combine to form two oxides, carbon monoxide and carbon dioxide, in which the ratio of the weights
of carbon and oxygen are, respectively, 12 : 16 and 12 : 32. These figures illustrate the (1) law of multiple proportions
(2) law of reciprocal proportions
(3) law of conservation of mass
(4) law of constant proportions
10. Two elements X (at. mass 16) and Y (at. mass 14) combine to form compounds A, B, and C, in which the ratio of weights of Y combining with a fixed mass of X is 1 : 3 : 5, respectively. If 32 parts by mass of X combine with 84 parts by mass of Y in B, then in C, 16 parts by mass of X will combine with
(1) 14 parts by mass of Y
(2) 42 parts by mass of Y
(3) 70 parts by mass of Y
(4) 84 parts by mass of Y
11. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at STP. The data illustrates the (1) law of conservation of mass
(2) law of constant proportions
(3) law of multiple proportions
(4) law of reciprocal proportions
12. Which of the following pairs can be cited as an example to illustrate the law of multiple proportions?
(1) Na2O, K2O (2) CO, CO2
(3) SO2, SO3 (4) Both 2 and 3
13. In compound A, 1.00 g nitrogen unites with 0.57 g oxygen. In compound B, 2.0 g nitrogen combines with 2.28 g oxygen. In compound C, 3.00 g nitrogen combines with 5.13 g oxygen. These results obey which of the following laws?
(1) Law of constant proportion
(2) Law of multiple proportions
(3) Law of reciprocal proportions
(4) Dalton’s law of partial pressure
14. A sample of pure carbon dioxide, irrespective of its source, contains 27.27% carbon and 72.73% oxygen. The data supports the (1) law of constant composition
(2) law of conservation of mass
(3) law of reciprocal proportions
(4) law of multiple proportions
15. Hydrogen combines with oxygen to form H2O, in which 16 g of oxygen combines with 2 g of hydrogen. Hydrogen also combines with carbon to form CH4, in which 2 g of hydrogen combines with 6 g of carbon. If carbon and oxygen combine together, then they will show in the ratio of
(1) 6 : 16 (2) 6 : 18
(3) 1 : 2 (4) 12 : 24
16. Two elements X and Y have atomic weights of 14 and 16. They form a series of compounds
A, B, C, D, and E, in which the same amount of element X combines with Y in the ratio
1 : 2 : 3 : 4 : 5, respectively. If the compound A has 28 parts by weight of X and 16 parts by weight of Y, then the compoud C will have 28 parts by weight of X and
(1) 32 parts by weight of Y
(2) 48 parts by weight of Y
(3) 64 parts by weight of Y
(4) 80 parts by weight of Y
17. n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as, X + Y= R + S. The relation that can be established by the amounts of the reactants and the products will be
(1) n–m=p–q
(2) n+m=p+q
(3) n=m
(4) p=q
18. A sample of calcium carbonate (CaCO3) has the following percentage composition: Ca = 40%; C = 12%; O = 48%, If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate from another source will be
(1) 0.016 g (2) 0.16 g
(3) 1.6 g (4) 16 g
Answer Key
(1) 4 (2) 2 (3) 1 (4) 1 (5) 3 (6) 3 (7) 3 (8) 3
(9) 1 (10) 3 (11) 3 (12) 4 (13) 2 (14) 1 (15) 1 (16) 2 (17) 2 (18) 3
1.6 DALTON’S ATOMIC THEORY
The postulates of the Dalton’s theory are as follows:
1. Matter consists of indivisible atoms.
2. All the atoms of a given element have identical properties, including identical mass. Atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.
The smallest particle of an element which can take part in a reaction is called ‘atom’. It may or may not be capable of independent existence.
1.7 ATOMIC MASS AND MOLECULAR MASS
This topic deals with absolute mass and relative masses of atoms of elements and molecules of compounds.
1.7.1 Atomic Mass
The atomic mass of an atom is very small. Nowadays, mass spectrometer is used to determine atomic mass accurately. However, in old times, chemists determined the mass of one atom relative to another by experiments. Hence, the atomic masses of elements are expressed relative to another element, i.e., hydrogen, and the mass of hydrogen is arbitrarily taken as one.
Later on, reference element was changed to oxygen. Finally, since 1961, carbon-12 is used as a standard and its mass is taken as 12 atomic mass units (amu). One atomic mass unit is also known as dalton (Da).
One atomic mass unit is exactly equal to 1 th 12 of the mass of carbon-12 isotope.
1 amu = 1 12 × mass of one C–12 atom. It has no units, since it is a relative weight. Atomic mass of element
Massofoneatomofelement
Massof ×
= 12 1 12 C
10. If relative mass of He is taken as one unit, what is the relative mass of magnesium?
Sol. Relative masses of He and Mg are, respectively, 4 and 24. Compared to He as 1, that of Mg = 6.
When we use atomic masses of e lements in calculations, we use average atomic masses.
Average Atomic Mass
Fractional values of masses is due to the existence of isotopes and their relative abundance. The relative atomic mass of an element is equal to the average of atomic masses of all its isotopes.
Average atomic mass of xAyA12 E xy + = + , where A1, A2 are atomic masses of isotopes of an element x and y are the number of atoms.
11. The relative abundances of 12C and 13C are, respectively, 98.892 and 1.108. If the atomic masses of 12C and 13C are 12 u and 13.00335 u, respectively, calculate the average atomic mass of carbon.
Sol. Average atomic mass of carbon = (98.89212)(1.10813.00335) 12.011u (98.8921.108)
××× = +
1.7.2 Molecular Mass (Molecular Weight)
Molecular mass or molecular weight of a substance is defined as “the number of times a molecule of it is heavier than one amu.”
Molecular weight is also measured relatively.
It has no units.
Molecular weight = 12 1 12
Weight of one molecule of substance ×WeightofC
Molecular weight or molecular mass of a molecule of a substance is the sum of the atomic weights or atomic masses of all atoms present in the molecule.
Example: Molecular mass of Na 2CO3 is 2(23) + 1(12) + 3(15.995) = 105.985
Similarly, molecular weights or molecular masses of other molecules can be calculated.
If vapour density of a volatile substance is obtained, then molecular weight is taken as twice the vapour density.
Molecular mass = 2 × Vapour density
In Victor Meyer’s method, the volume of air displaced by a fixed mass of a volatile substance is determined.
The volume is translated to standard temperature and pressure conditions. The mass corresponding to 22400 cc of air gives the molecular mass.
Molecular mass
= 22,400Wgramsofsubstance
Volume(inml)ofairdisplacedatSTP ×
Minimum molecular weight (Cannizzaro’s principle):
It is calculated for one atom whose percentage is known.
Example: If a compound has 4% sulphur in it, its minimum molecular weight.
(symbol: mol) was introduced as a seventh base quantity in SI system.
2. Definition of mole : One mole is the amount of a substance that contains as many particles (or entities) as there are atoms in exactly 12 g (or 0.012 kg) of the C-12 isotope.
= At.wt.10032100 800
Givenpercent4 ×× ==
Minimum molecular weight
= NumberofatomsAt.wt.100 %mass ××
Formula Masses
Ionic compounds do not have exact independent molecules as their constitutent units. They are made up of their constituent ions and represented by a formula.
Therefore, for such compounds, formula masses but not molecular masses are used. For example, sodium chloride has a formula NaCl and its formula mass is 23 + 35.5 = 58.5 amu.
12. Molecular mass of a compound, Be3 Al x Si6O18 is 537 amu.
Find the value of x [At. Wt. : Be = 9, Al = 27, Si = 28, O = 16]
Sol. In the compound Be3 Al n Si6O18, sum of the atomic masses of all the atoms per molecule = 537 amu.
3(9) + n(27) + 6(28) + 18(16) = 537
27n = 537 – 483 = 54
Hence, n = 2
1.8 MOLE CONCEPT AND MOLAR MASS
1. Atoms and molecules are extremely small in size and even in a small amount of substance, their numbers and are very large. To handle such large numbers ‘mole’
3. One mole of all the pure substances always contain the same number of entities. To determine this number, accurate mass of a carbon-12 atom was determined by mass spectrometer. It is 1.992648 × 10–23 g. One mole of carbon weighs 12 g. Hence, the number of carbon atoms in one mole of carbon-12 23 12g/mol 1.99264810g/atom = × = 6.022 × 1023 atoms/mol.
This number is called Avogadro’s number (NA or NO).
4. Avogadro’s number is defined as the number of atoms present in one mole of any monoatomic element.
5. It is the number of molecules present in one mole of any molecular compound.
Gram Atom or Gram Atomic Weight
1. The weight in grams of an element which is numerically equal to its atomic mass is called gram atom or gram atomic weight.
Number of gram atoms = Weightingramsofelement GAW
Atomic weight of hydrogen is 1.007828 amu, one gram atom of hydrogen is 1.007825 g.
Atomic weight of carbon is 12 amu and the one gram atom of carbon is 12 g.
2. The mass of one mole of any monoatomic element or one gram atom of the element is one gram atomic weight.
Gram Molecular Weight or Molar Mass
The weight in grams of a substance which is numerically equal to its molecular mass is called gram molecule or gram molecular weight or molar mass. Examples of gram molecules are given in Table 1.5
Table 1.5 Gram molecules Compound
Nitrogen 28 amu 28 g
Carbon dioxide 44 amu 44 g
Glucose 180 amu 180 g
The mass of one mole of a molecular compound or one gram molecule of the compound is one gram molecular weight.
Molar Volume or Gram Molar Volume
The volume occupied by one mole or one gram molecular weight of any gaseous substance at STP (273 K and 1 atm) is called molar volume. Its value is 22.4 litres.
N2
Chemical Entities
The charge carried by one mole of electrons is called one faraday. The number of elementary chemical units in one mole of a substance is obtained by using Avogadro number. The number denoted by mole could be either molecules, atoms, ions, electrons, bonds, or any specified particles.
Number of moles =
Weightingramsof compound GMW (or) Weightingramsofelement GAW
Number of molecules = A Weightingramsofsubstance N Grammolecularweight
Number of atoms = A Weightingramsof anelement N Gramatomicweight ×
Number of chemical units denoted by molar mass of substances is given in Table 1.6 and some mole relationships are illustrated in Table 1.7.
Table 1.6 Mass and the number of particles
chloride
g 6.022 × 1023 atoms
6.022 × 1023 molecules
g 6.022 × 1023 molecules
The formation of ammonia from its elements in terms of mole concept is summarised in Table 1.8.
The Importance of Mole
Mole concept is important and useful as it accounts for the following:
1. One mole of an element represents one gram atomic weight of the element. It is called gram atom.
Table 1.7 Examples of mole relationships
Table 1.8 Formation of ammonia - mole concept
2. One mole of a substance represents one gram molecular weight of substance. It is called gram mole.
3. One mole of an element means the mass of 6.022 × 1023 atoms of the element.
4. One mole of a covalent substance means the mass of 6.022 × 10 23 molecules of the substance.
5. One mole of an ionic substance means the mass of 6.022 × 1023 formula units of the substance.
6. One mole of a gas or vapour occupies a volume called GMV. The GMV at STP conditions is 22.4 L.
7. The charge of one mole of electrons is called faraday. One faraday is given as 96,500 couloumbs.
8. The concept of mole represents Avogadro constant.
9. Density of a gas in g L –1 GMW = 22.4
10. Loschmidt number is the number of molecules present in one cc. of a gas at STP.
Loschmidt number A N 22400 =
11. Number of molecules in a molecular compound = Number of gram molecules × NA
12. Number of atoms in a pure substance = Number of molecules × atomicity
13. Number of moles in a compound = Mass GMW
14. Mole is extremely useful in calculations because it simplifies the work of a chemist.
13. Ho w many atoms are present in one cc of helium gas at STP?
Sol. At STP, 22400 cc = 6.022 × 10 23 atoms
At STP, one cc = ?
Number of atoms present in one cc at STP
14. Natural abundance of heavy water in water is 1 : 6000. How many heavy water molecules are present in one drop of water? (One mL water is 20 drops)
Sol. One mL water = One gram (Q d = 1 g per mL)
Weight of one drop of water = 1 20 g
Weight of D2O in one drop of water = 1 g 206000 ×
Number of moles of D2O = 2 2 Weight of DO GMW of DO = 1 20600020 ××
Number of D2O molecules = Number of moles × NA
Ans: Mass of Mg in 2 g chlorophyll = 2.5×2/100 = 0.05 g. Hence, the number of Mg atoms =
TEST YOURSELF
1. Number of molecules in one litre of oxygen at STP conditions is (1) 23 6.0210 32 × (2)
(3) 32 × 22.4 (4) 32 22.4
2. Which of the following has the least number of atoms?
(1) 0.5 g atom of Zn
(2) 0.645 g of Zn
(3) 0.25 mole of Zn
(4) 6.45×1020 amu of Zn
3. Air contains nitrogen and oxygen in the volume ratio of 4 : 1. The average molecular weight of air is (1) 30 (2) 28.8 (3) 28 (4) 14.4
4. The number of gram atoms of sulphur present in 3 moles of hydrogen sulphide is (1) 3 (2) 2 (3) 1.5 (4) 1.0
5. 1 mole of water vapour is condensed to liquid at 25 °C. Now, this water contains I) 3 moles of atoms
Number of heavy water molecules is 2.51 × 1017
Try yourself:
7. H ow many magnesium atoms would you expect to be in 2.0 g of chlorophyll, if it contains 2.5 percentage of megnesium by mass?
II) 1 mole of hydrogen molecules
III) 10 moles of electrons
IV) 6 g of oxygen
The correct combination is (1) (I) and (II) (2) (I) and (III) (3) (I) and (IV) (4) All are correct
6. One mole of CH4 contains (1) 6.02 × 1023 atoms of hydrogen (2) 4 gram atoms of hydrogen (3) 3 g of carbon
(4) 1.81 × 1023 molecules of CH4
7. The number of molecules in one litre of water is (density of water = 1 g/mL)
(1) 6 × 1023 / 22.4
(2) 3.33 × 1025
(3) 3.33 × 1023
(4) 3.33 × 1024
8. The ratio between the number of molecules in equal masses of CH4 and SO2 is (1) 1 : 1 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
9. The number of sulphur atoms present in 0.2 moles of sodium thiosulphate is (N = Avogadro number)
(1) 4N (2) 0.2N
(3) 0.4N (4) 0.1N
10. The number of nitrogen molecules present in 1 cc of gas at NTP is
(1) 2.67×1022 (2) 2.67×1021
(3) 2.67×1020 (4) 2.67×1019
11. A gaseous mixture contains oxygen and nitrogen in the ratio 1 : 4 by weight. The ratio of their number of molecules is
(1) 1 : 4 (2) 4 : 1
(3) 7 : 32 (4) 3 : 16
12. An a -particle changes into a helium atom. In the course of one year, the volume of helium collected from a sample of radium was found to be 1.12×10–2 mL at STP. The number of particles emitted by the sample of radium in the same time is
(1) 6 × 1017 (2) 3 × 1017
(3) 1.5 × 1017 (4) 1.2 × 1018
13. The mixture containing the same number of molecules as that of 14 g of CO is
(1) 14 g of nitrogen + 16 g of oxygen
(2) 7 g of nitrogen + 16 g of oxygen
(3) 14 g of nitrogen + 8 g of oxygen
(4) 7 g of nitrogen + 8 g of oxygen
14. The number of atoms of hydrogen present in 1.5 moles of H2O is
(1) 1 N (2) 2 N
(3) 3N (4) 0.5 N
15. Which of the following contains the maximum number of atoms?
(1) 10 g of CaCO3
(2) 4 g of hydrogen
(3) 9 g of NH4NO3
(4) 1.8 g of C6H12O6
16. Which contains more number of molecules?
(1) 1 mole of carbon dioxide
(2) 4 g of hydrogen
(3) 6 g of helium
(4) 33.6 litres of oxygen at STP
17. Which of the following is the heaviest?
(1) 50 g of iron
(2) 5 moles of nitrogen
(3) 0.1 gram atom of silver
(4) 1023 atoms of carbon
18. The density of a gas is 2, relative to nitrogen, under the same conditions. The molecular weight of the gas is
(1) 5.6 (2) 28
(3) 56 (4) 14
19. The density of a gas at STP is 1.5 g/L. Its molecular weight is
(1) 22.4 (2) 33.6 g
(3) 33.6 (4) 44.8
20. A copper plate of 20 cm × 10 cm is to be plated with silver of 1 mm thickness. The number of silver atoms required for plating is (density of silver = 10.8 g/cc)
(1) 1.2×1024 (2) 2.4×1024
(3) 1.2×1013 (4) 2.4×1023
21. Which of the following has the number of molecules present equal to those present in 16 grams of oxygen?
(1) 16 g O3 (2) 32 g SO2
(3) 16 g SO2 (4) All the above
22. What is the mole percentage of O 2 in a mixture of 7 g of N2 and 8 g of O2?
(1) 25% (2) 75%
(3) 50% (4) 40%
23. 7.5 g of a gas occupies 5.6 litres as STP. The gas is
(1) NO (2) N2O
(3) CO (4) CO2
24. Total number of sulphate ions present in 3.92 g of chromic sulphate is (Cr = 52, S = 32, O = 16)
(1) 1.8 × 1022
(2) 1.8 × 1023
(3) 1.2 × 1021
(4) 6 × 1023
25. Number of atoms in 558.5 grams of Fe (at. wt. of Fe = 55.85 g mol –1) is (1) twice that in 60 g of carbon
(2) 6.023 × 1022
(3) Half that in 8 g of He
(4) 558.6 × 6.023 × 1023
26. The mass of 1.5 × 10 26 molecules of a substance is 16 kg. Molecular mass of the substance is
(1) 64 g (2) 64 amu (3) 16 amu (4) 32 amu
Answer Key
(1) 2 (2) 4 (3) 2 (4) 1 (5) 4 (6) 2 (7) 2 (8) 2 (9) 3 (10) 4 (11) 3 (12) 2
(13) 4 (14) 3 (15) 2 (16) 2 (17) 2 (18) 3 (19) 3 (20) 1
(21) 2 (22) 3 (23) 1 (24) 1 (25) 1 (26) 2
1.9 PERCENTAGE COMPOSITION
A set of chemical symbols showing the elements present in the compound with relative proportions is called chemical formula.
1.9.1 Percent Composition of Elements
The composition of pure chemical compound is always fixed, according to the law of definite proportions.
The weight in grams of an element present in 100 grams of its compound is called weight percent of that element.
Weight percen t of an elem ent in a compound =
Weight of element in one mole of the compound
Gram molecular weight of compound
100 ×
Methods and equations for the estimation of different elements are listed in Table 1.9.
Table 1.9 Methods of estimating elements in a given substance (w)
Nitrogen Duma’s method
Nitrogen Kjeldahl’s method
Sulphur Carius tube method
Chlorine Carius tube method
Bromine Carius tube method
Iodine Carius tube method
Phosphorus Carius tube method
Oxygen Subtraction principle
of silver iodide (w 7)
15. What is the weight percent of oxygen in glucose?
Sol. Weight percent of oxygen in glucose = 616 10053.33% 180 × ×=
16. An organic compound has 8 percent of sulphur by weight. What is the molecular weight of the compound?
Sol. 8 parts of sulphur = 100 parts of compound
32 parts of sulphur = ?
The minimum molecular weight of the compound 32 100400 8 =×=
Molecular weight may be a multiple of 400, like 800, 1200, 1600, 2000, etc.
Try yourself:
8. What is the percentage by mass of oxygen in blue vitreol, CuSO45H2O? (Molecular mass = 249.6)
Hence, percentage of oxygen 916 10057.7% 1249.6 × =×= ×
Ans: One mole of O25H4CuSO contains 9 moles of oxygen atoms.
1.9.2 Chemical Formula
Formula represents the chemical composition of the substance. There are three kinds of formulae of compounds. They are: i) Empirical formula
ii) Molecular formula
iii) Structural formula
“Empirical formula of a compound is the simplest formula showing the relative number of atoms of different elements present in one molecule of the compound.”
“Molecular formula represents the actual number of atoms of different elements present in one molecule of the compound.”
For certain compounds, the molecular formula and the empirical formula may be one and the same.
The molecular formula of a compound may be same as empirical formula or whole number multiple of it. Thus,
The molecular formula = (Empirical formula) × n, where n is an integer 1,2,3,..... etc.
= Molecularweight n Empiricalformulaweight
If the vapour density of a volatile substance is known, its molecular weight can be calculated by using the following equation.
2 × Vapour density = Molecular weight
The empirical and molecular formulae are illustrated with some suitable examples in Table 1.10
Table 1.10 Empirical and molecular formulae of some substances
Empirical formula is same for compounds having same percentage composition of elements.
Empirical formula can be determined from the mass percentages of various elements present in a compound. The sequence of steps in the determination of empirical formula are as follows:
1) The weight percentage (or weight) of each constituent element is taken or is to be calculated.
2) The percent weight of each constituent is to be divided with its atomic weight to get relative number of atoms of each element.
3) The simplest whole number ratio of the values of step (2) is to be obtained. This may be done by dividing all values with the smallest among them. If it is not a whole number, then multiply them with a suitable integer to get whole number ratio.
4) The simplest atomic ratio obtained in (3) represents empirical formula.
17. An organic compound contains carbon, hydrogen, and oxygen, and nitrogen in the weight ratio 3 : 1 : 8 : 3.5. Calculate its empirical formula.
Sol. 3183.5 C:H:O:N::: 1211614 = = 0.25 : 1 : 0.5 : 0.25 = 1 : 4 : 2 : 1 The empirical formula is CH4 O2 N.
18. An organic compound, on analysis, was found to contain 16.27% carbon, 0.67% hydrogen, 72.2% chlorine. The VD of the compound is equal to 73.75. Calculate the empirical formula and molecular formula of the compound.
Sol.
Empirical formula of the compound = C2HCl3O
Empirical formula weight
= (2 × 12) + (1 × 1) + (3 × 35.5) + (1 × 16) = 147.5
Molecular weight = 2 × VD = 2 × 73.75 = 147.50
∴ Molecular formula = C2HCl3O
1.9.3 Eudiometric Method
The combustion of hydrocarbon is given by ()()()() x yg CH 1ml mlml 4 2 2g 2g
This is a method for finding the molecular formula of a gaseous hydrocarbon. The method involves the following steps:
1. A known volume of the gaseous hydrocarbon is mixed with an excess (known or unknown volume) of oxygen or air in the eudiometer over mercury.
2. The mixture is exploded by electric spark and then cooled so that water vapour condenses to liquid, the volume of which is negligible
amount of O2 may be found by absorbing it in alkaline pyrogallol. Thus, decrease in volume on adding KOH = volume of CO 2 produced, and decrease in volume on passing through alkaline pyrogallol = volume of O2 gas.
Values of x and y are then calculated from the following data:
i. Volume of O2 used per cc of hydrocarbon y xcc
ii. Volume of CO2 produced = x cc
iii. Contraction on explosion and cooling
3. KOH is inroduced, which absorbs CO2, and only unused O2 is left. Alternatively, the
19. 10 ml of gaseous hydrocarbon gives 40 ml of CO2 gas and 50 ml of H2O vapour. What is the formula of the hydrocarbon?
++→+
Sol. xy yy CHxOxCOHO mlxyyxmlml or yy mlxmlxmlml
Volume of CO2 = 10 x = 40 (given)
Volume of H2O vapour 1050 2 yml== (given) ∴ y = 10
Hence, the formula = Cx H y = C4H10
TEST YOURSELF
1. The weight percentage of oxygen in NaOH is (1) 40 (2) 6 (3) 8 (4) 20
2. Caffeine contains 28.9% by mass of nitrogen. If molecular mass of caffeine is 194, then the number of N atoms present in one molecule of caffeine is
(1) 3 (2) 4
(3) 5 (4) 6
3. 10 g of hydrofluoric acid gas occupies 5.6 litres of volume at STP. If the empirical formula of the gas is HF, then its molecular formula will be (Atomic mass of F = 19)
(1) HF (2) H 3 F 3
(3) H2F2 (4) H 4 F 4
4. An organic compound made of C, H, and N contains 20% nitrogen. What will be its molecular mass if it contains only one nitrogen atom in it?
(1) 70 (2) 140
(3) 100 (4) 65
5. A peroxidase enzyme contains 2% selenium (Se = 80). The minimum molecular weight of the enzyme is
(1) 1000 (2) 2000
(3) 4000 (4) 800
6. Haemoglobin contains 0.33% iron (Fe = 56). The molecular weight of haemoglobin is 68000. The number of iron atoms in one molecule of haemoglobin is (1) 2 (2) 3 (3) 4 (4) 5
7. A gaseous alkane requires five times its volume of oxygen under the same conditions for complete combustion. The molecular formula of the alkane is
(1) C2H6 (2) C4H10 (3) C3H8 (4) CH4
8. A dibasic acid containing C, H, and O was found to contain C = 26.7% and H = 2.2%. The vapour density of its diethyl ester was
found to be 73. The molecular formula of the acid is
(1) CH2O2 (2) C2H2O4 (3) C3H3O4 (4) C2H4O4
9. 0.36 g of an organic compound, on combustion, gave 1.1 g of CO2 and 0.54 g of H 2 O. The percentages of carbon and hydrogen in the compound are (1) 75, 25 (2) 60, 40 (3) 83.33, 16.67 (4) 77.8, 22.2
10. 40 ml of a hydrocarbon undergoes combustion in 260 ml of oxygen and gives 160 ml of CO2. If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is (1) C3H8 (2) C4H8 (3) C6H14 (4) C4H10
Answer Key
(9) 3 (10) 4
1.10 STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
Stoichiometry deals with the relative quantities of the substances taking part in a reaction.
1. Weight - Weight relationship
2. Weight - Volume relationship
3. Volume - Volume relationship
4. Mole - Mole relationship
Example: CH4(g) + 2O2 → CO2(g) + 2H2O
The above chemical reaction gives the following information.
1. The coefficients, one each for CH 4 and CO2 and the coefficients, 2 each for O2 and H2O, are called stoichiometric coefficients.
2. The qualities of the substances involved are as follows:
1. Relative number of moles involved 1 2 1 2
2. Relative number of molecules involved 1 2 1 2
3. Relative masses of the substances involved 16 g 2 ×
4. Relative volumes at STP
Thus, one mole of CH4 reacts with 2 moles of O2 to give one mole of CO2 and 2 moles of H2O.
1.10.1 Limiting Reagent
1. When two or more reactants are involved in a reaction with the quantities that are different from the quantities that are required by a balanced chemical equation, the quantity of product formed is governed by the reactant which is present in the least quantity.
2. In a reaction, the reactant which is completely consumed is called a limiting reactant or a limiting reagent. Other reactants are called excess reagents.
3. The quantity of limiting reagent decides the quantity of product formed in the reaction.
Example: Mg + 2 HCl → MgCl2 + H2
For this reaction, when 6 moles each of Mg and HCl are given, 6 moles of Mg demands 12 moles of HCl for th e reaction.
Hence, all the 6 moles of HCl are consumed by reacting with 3 moles of Mg. Therefore,
limiting reagent is HCl and excess reagent is Mg.
Finding Limiting Reagent in a Reaction
1. For each reactant, find the ratio between the given number of moles and its stoichiometric coefficient.
2. For whichever reactant the value obtained is the least, it is the limiting reactant.
3. The other reactant is excess reactant.
Example: Consider the reaction,
3BaCl2 + 2Na3PO4 → 6NaCl + Ba3(PO4)2
If th e number of moles of BaCl 2 and Na 3PO 4 taken is 10 each, ratio of moles of 2 10 BaCl3.33 3 ==→ lowest value.
Ratio of moles of 34 10 NaPO5 2 ==
Hence, the limiting reagent is BaCl 2
20. Certain mass of potassium chlorate is thermally decomposed and 3.36 L of O 2 is collected at STP. What will be the weight of the residue formed in the experiment?
Sol. Potassium chlorate, on decomposition, gives potassium chloride and oxygen.
2KClO3 → 2KCl + 3O2
3 moles of O2 ≡ 2 moles of KCl
(3 × 22.4) L of O2 at STP = (2 ×74.5) grams of KCl
The weight of residue (KCl) obtained when 3.36 L of O2 is collected at STP = × 3.36
322.4 × 2 × 74.5 = 7.45 grams
21. 0.4 mole of orthophosphoric acid and 1.0 mole of calcium hydroxide were allowed to react. Calculate the maximum number of moles of calcium phosphate formed.
Sol. The stoichiometric equation is
3Ca(OH)2+2H3PO4 → Ca3(PO4)2+3H2O
3 moles of Ca(OH)2 = 2 moles of H3PO4
1 mole of Ca(OH)2 = 0.67 mole of H3PO4
0.6 mole of Ca(OH)2 = 0.4 mole of H3PO4
H3PO4 is the limiting reagent. Hence, salt formed is dependent on the availability of acid only.
2 moles of H3PO4 = 1 mole of Ca3(PO4)2
0.4 moles of H3PO4 = ?
Number of moles of calcium phosphate formed = 0.4 × 1 2 = 0.2 mole
1.10.2 Concentration Methods
The concentration of a solution or the amount of a substance present in its given volume can be expressed in any of the following ways:
1. Mass per cent or weight per cent (w/w%)
2. Mole fraction
3. Molarity
4. Molality
5. Normality
1. Mass percent is obtained by using the following relation:
Mass percent = Mass of one component 100 Total mass of all components ×
2. Mole fraction (X) is the ratio of number of moles of a particular component to the total number of moles of all the components in the solution.
Mole fraction of a component =
Numberofmolesofthecomponent
Totalnumberofmolesofallcomponentsofsolution
If nA moles of a substance ‘A’ are dissolved in n B moles of a substance ‘B’ , then the mole fractions of A and B are given as:
3. Molarity (M) is the most widely used unit and it is denoted by M. It is defined as the number of moles of the solute in one litre of the solution. Thus,
Number
Molarity (M) = of moles of solute
Volume of solution in litre
Units of M = moles / L–1. 1 M HCl means 1 mole of HCl in 1 litre of solution, i.e., 36.5 g HCl in 1 litre of solution.
Molarity of a solution depends upon temperature because volume of a solution is temperature dependent.
Dilution formula: M 1V 1 = M 2V 2, where M1, M2 = initial and final molarities, V1, V2 = initial and final volumes of solution.
Suppose M1 = 0.2, V1 = 1000 ml
M2 = 1.0, V2 = ? 2 0.21000 V 200mL 1.0 × ∴==
Suppose we have 1M solution of NaOH and we want to prepare a 0.2M solution from it. 1M NaOH means 1 mole of NaOH present in 1 litre of the solution. For 0.2M solution, we require 0.2 moles of NaOH in one litre solution.
Suppose the solution prepared by dissolving 4 g NaOH in enough water to form 250 mL of the solution.
Numberofmolesofsolute(n)
Then molarity (M) w n MW
Volumeofsolutionin litres(v) ====
Molecularweight0.250L
4. Molality ( m ) is defined as the number of moles of solute present in one kg of solvent. It is denoted by m. ()
Numberofmolesof solute
Molality
Massofsolvent in kg = m
Units of m = moles/kg
Molality of a solution does not change with temperature since mass remains unaffected with temperature.
22. The density of 4% (w/v) NaOH solution is 1.02 g/mL. What is the molality of the solution?
Sol. 4% (w/v) NaOH solution contains 4 g of NaOH in 100 ml of the solution.
Density of the solution = 1.02 g/ml.
Weight of of the solute in 100 ml of the solution = 4 g
Weight of 100 ml of the solution = 100 × 1.02 = 102 g
Weight of solvent = w of solution – w of solute = 102 – 4 = 98 g.
Molality Solute W 1000 G.M.W ofsolventingrams Weight
=×= 41000 1.02m 4098
5. Normality (N) is defined as the number of gram equivalents of the solute present in one litre of a solution. Units of normality are eq L–1. It is dependent on temperature. As the temperature increases, the volume increases and normality decreases.
Normality (N) is given as
Number of gram equivalents of solute = Volume of solution in litres N (or)
Weight of solute
Gram equivalent weight × 1000 Volume of solution in ml
Equivalent weight of an acid is the number of grams of the acid which gives one mole of protons in water. It is the ratio of molecular weight of acid and basicity of acid.
(Number of H + ions furnished by one molecule of an acid is called basicity).
Equivalent weight of acid
= Molecularweightofacid
Basicityofacid
E qui valent weight of a base is the number of grams of the base which gives one mole
of hydroxyl ions in water. It is the ratio of molecular weight of base and acidity of base.
(The number of OH– ions furnished by one molecule of base is called acidity).
Equivalent weight of base
= Molecularweightofbase Acidityofbase
Equivalent weight of a salt it is the ratio of formula weight of salt and total number of units of positive or negative charges of the ionic substance of one formula unit.
Equivalent weight of salt
= Formulaweightofsalt
Totalnumberofpositive (ornegative)chargeunits
E quivalent weight of a salt in practice is taken as the sum of equivalent weights of ions represented in the empirical formula of salt.
Equivalent weight of salt = Equivalent weight of cation + Equivalent weight of anion
Dilution Law
Normality = Molarity × n f , nf = n–factor of the solute
Whenever two substances react, they always react in 1 : 1 ratio of their equivalents.
Number of equivalents of one reactant = Number of equivalents of other reactant
VANA = VBNB
VA(nf of A × MA) = VB (nf of B × MB)
23. Find the normality of oxalic acid solution containing 63 g of crystalline oxalic acid in 500 ml of solution.
Sol. Weight of solute = 6.3 g
GEW of solute = = 126 6.3g 2
Normality, N =×=×= ml W10006.31000 0.2N GEWV6.3500
24. Find the mass of Na2CO3 required to prepare 250 ml of 0.5 N solution.
Sol. Normality, N = × ml W1000 GEWV =× W1000 0.5 53250
Weight of Na2CO3 = 6.62 g
Try yourself:
9. Sulphuric acid used in lead storage cells is 5.0 M solution with density 1.3 g cm–3. What is the weight of the acid present in 1 kg of the solution.
Ans: 377 g
Volume Strength of H2O2 Solution(Adv)
The concentration of H 2 O 2 is usually represented in terms of volume. If a sample of H2O2 is labelled as ‘x volume’, it means that I volume of H2O2 solution gives ‘x volumes’ of O2 gas at STP on complete decomposition.
strength
Volume of O gas at STP Volume of HO solution
Consider the decomposition H 2O2 as
∴ 22400 mL of O2 gas is liberated by 68g of H2O2 solution.
∴ x mL of O2 gas will be liberated by 68x17x g 224005600 == of H2O2
It means that 17x g 5600 of H2O2 will present in 1 mL of solution.
∴1000 mL of solution constains H 2O2
= 17x17x 1000 56005.6 ×=
Number of grams of H2O2 per litre 17x 5.6
Strength (gL–1) = Normality × Equivalent weight.
17x34 N 5.62 =× (Q n-factor of H2O2=2)
x = 5.6 × N
∴ Volume strength of H 2 O 2 = 5.6 × Normality and volume strength of H 2O 2 = 11.2 × Molarity.
25. Calculate the volume strength of a 3% (W/V%) solutions of H2O2.
Sol. Step-1: To calculate the amount of H2O2 present in one litre of 3% solution.
100 mL of H2O2 solution contain H2O2 = 3 g
∴1000 mL of H2O2 solution will contain 22 3 HO100030g 100 =×=
Step-2: To calculate the volume strength Consider the chemical equation, 222 2 23468g 22.4litresatN.T.P.
2HO2HOO ×= →+
Now 68 g of H2O2 give O2 at N.T.P. = 22.4 litres
∴30 g of H2O2 give O2 at N.T.P.
22.4 309.88 68 =×= litres = 9880 mL.
But 30 g of H2O2 are present in 1000 mL of H2O2.
Hence, 1000 mL of H2O2 solution will gives O2 at N.T.P. =9880 mL.
∴ 1 mL of H2O2 solution will give O 2 at N.T.P. 9880 9.88 1000 == mL.
Hence, the volume strength of 3% H 2O2 solution = 9.88.
26. Find the mass of Na2CO3 required to prepare 250 mL of 0.5 N solution.
Sol. Normality, N = × ml W1000 GEWV
=× W1000 0.5 53250
Weight of Na2CO3 = 6.62 g
Percentage Labeling of Oleum(Advance)
Oleum or fuming sulphuric acid contains SO3 gas dissolved in sulphuric acid. When water is added to oleum, SO3 reacts with H2O to form H2SO4, thus mass of the solution increases.
SO3 + H2O → H2SO4
The total mass of H 2 SO 4 obtained by diluting 100 g of sample of oleum with desired amount of water, is equal to the percentage labeling of oleum.
% labeling of oleum = Total mass of H2SO4 present in oleum after dilution.
= mass of H2SO4 initially present + mass of H2SO4 produced on dilution.
27. Calculate the composition of 109% oleum.
Sol. Let the mass of SO3 in the sample be ‘w’ g, then the mass of H 2 SO 4 would be (100–w)g. On dilution,
SO3 + H2O → H2SO4
80 g 18g
Moles of SO3 in oleum w 80 == Moles of H2SO4
∴ Mass of H2SO4 formed on dilution
98w
80 =
Total mass of H 2SO 4 present in oleum after dilution
98w
80 = +(100–w) = 109
∴ w = 40.
Thus, oleum sample contains 40% SO3 and 60% H2SO4
Try yourself:
10. Sulphuric acid used in lead storage cells is 5.0 M solution with density 1.3 g cm–3. What is the weight of the acid present in 1 Kg of the solution.
Ans: 377 g
Degree of Hardness of Water(Advance)
Hard water is having soluble salts of calcium and magnetism which forms precipitates with the carboxylate ion present in soap.
There are two types of hardness
1. Temporary hardness:
It is due to the presence of bicarbonates of calcium and magnesium in water.
2. Permanent hardness:
It is due to the presence of chlorides and sulphates of Ca2+ and Mg2+ .
The concentration of the salts responsible for hardness is expressed in terms of degree of hardness.
Degree of hardness is defined as number of parts by mass of CaCO 3 (equivalent to the quantities of salts in the hard water) present in one million parts of mass of water.
Hardness of water
6 3 Mass of CaCO 10ppm Mass of water
28. One kilogram of a sample of hard water contains 1 mg of CaCl2 and 1 mg of MgCl2 Find out the total hardness in terms of parts of CaCO3 per 106 parts of water by mass.
Sol. (i) Mol. mass of CaCl2 = 111 g mol–1
Now, 111 g of CaCl2 ≡100 g CaCO3
∴1 mg of CaCl2 100 1 111 × mg of CaCO3 = 0.9 mg of CaCO3
(ii) Mol. mass of MgCl2 = 95 g mol–1
Now, 95 g of MgCl2 ≡ 100 g of CaCO3
∴1 mg of MgCl2
100 1 95 =× mg of CaCO3 ≡1.05 mg of CaCO3
Thus, 1 kg of hard water contains = 0.90 + 1.05 = 1.95 mg of CaCO3
Since, 1 kg of water = 103 g = 106 mg
Therefore, degree of hardness of water 6 6 1.95 101.95ppm 10 =×=
TEST YOURSELF
1 What is the weight of oxygen that is required for the complete combustion of 2.8 kg of ethylene?
(1) 6.8 kg (2) 6.4 kg
(3) 96.0 kg (4) 9.6 kg
2. 30 g of marble stone, on heating, produced 11 g of CO 2. The percentage of CaCO3 in marble is
(1) 75% (2) 80%
(3) 83.3% (4) 86.6%
3. The number of moles of CO2 produced when 3 moles of HCl react with excess of CaCO 3 is
(1) 1 (2) 1.5
(3) 2 (4) 2.5
4. The volume of CO2 obtained by the complete decomposition of one mole of NaHCO 3 at STP is
(1) 22.4 L (2) 11.2 L
(3) 44.8 L (4) 4.48 L
5. 5 g of a sample of magnesium carbonate, on treatment with excess of dilute hydrochloric acid, gave 1.12 L of CO 2 at STP. The percentage of magnesium carbonate in the mixture is
(1) 42 (2) 40 (3) 84 (4) 80
6. The volume of CO2 formed when 1 litre of O2 is reacted with 2 L of CO under the same condition is
(1) 1 L (2) 2 L (3) 3 L (4) 1.5 L
7. Air contains 20% by volume of oxygen. The volume of air required for the complete combustion of 1 L of methane under the same conditions is (1) 2 L (2) 4 L (3) 10 L (4) 0.4 L
8. When 20 ml of methane and 20 ml of oxygen are exploded together and the reaction mixture is cooled to laboratory temperature, the resulting volume of the mixture is (1) 40 ml (2) 20 ml (3) 30 ml (4) 10 ml
9. For the reaction A + 2B → C, 5 moles of A and 8 moles of B will produce (1) 5 moles of C (2) 4 moles of C (3) 8 moles of C (4) 13 moles of C
10. How many litres of CO 2 at STP will be formed when 100 ml of 0.1M H 2SO4 reacts with excess of Na2CO3?
(1) 22.4 (2) 2.24 (3) 0.224 (4) 5.6
11. When a sample of baking soda is strongly ignited in a crucible, it suffers a loss in weight of 3.1 g. The mass of baking soda is (1) 16.8 g (2) 8.4 g (3) 11.6 g (4) 4.2 g
12. 7 g of a sample of NaCl, on treatment with excess of silver nitrate, gave 14.35 g of AgCl. NaCl in the sample is (1) 80% (2) 50% (3) 65.8% (4) 83.5%
13. 18.4 g of a mixture of CaCO3 and MgCO3, on heating, gives 4 g of magnesium oxide. The volume of CO2 produced at STP in this process is (1) 1.12 L (2) 4.48 L (3) 2.24 L (4) 3.36 L
14. 8 g of sulphur is burnt to form SO 2, which is oxidised by chlorine water. The solution is treated with BaCl2 solution. The number of moles of BaSO4 precipitated is (1) 1 (2) 0.5 (3) 0.25 (4) 0.125
15. One mole of mixture of CO and CO2 requires exactly 20 g of NaOH to convert all the CO2 into Na 2 CO 3 . How many more grams of NaOH would it require for conversion into Na2CO3, if the mixture is completely oxidised into CO2?
(1) 80 g (2) 60 g (3) 40 g (4) 20 g
16. The weight of H 2O 2 to be decomposed to liberate 5.6 litres of O2 at STP is g.
(1) 34 (2) 3.4 (3) 17 (4) 1.7
17. A gas mixture contains acetylene and carbon dioxide. 20 litres of this mixture requires 20 litres of oxygen under the same conditions for complete combustion. The percentage by volume of acetylene in the mixture is (1) 50% (2) 40% (3) 60% (4) 75%
18. Acetylene can be prepared from calcium carbonate by a series of reactions. The mass of 80% calcium carbonate required to prepare 2 moles of acetylene is (1) 200 g (2) 160 g (3) 250 g (4) 320 g
19. 4.9 g of H2SO4 decomposes x g of NaCl to give 6 g of sodium hydrogen sulphate and 1.825 g of hydrochloric acid. The value of x is
(1) 6.92 (2) 4.65
(3) 2.925 (4) 1.41
20. One litre of a mixture of CO and CO 2 is passed over red hot coke when the volume is increased to 1.6 L under the same conditions
of temperature and pressure. The volume of CO in the original mixture is (1) 400 ml (2) 600 ml (3) 500 ml (4) 800 ml
Answer Key
(1) 4 (2) 3 (3) 2 (4) 2 (5) 3 (6) 2 (7) 3 (8) 2 (9) 2 (10) 3 (11) 2 (12) 4 (13) 2 (14) 3 (15) 2 (16) 3 (17) 2 (18) 3 (19) 2 (20) 1
TEST YOURSELF
1. The number of millimoles of H2SO4 present in 5 litres of 0.2 N H2SO4 solution is (1) 500 (2) 1000 (3) 250 (4) 0.5×10–3
2. The number of glucose molecules present in 10 ml of decimolar solution is (1) 6.0 × 1020 (2) 6.0 × 1019 (3) 6.0 × 1021 (4) 6.0 × 1022
3. 0.1 gram mole of urea is dissolved in 100 g of water. The molality of the solution is (1) 1 m (2) 0.01 M (3) 0.01 m (4) 1.0 M
4. Aqueous NaOH solution is labelled as 10% by weight. Mole fraction of the solute in it is (1) 0.05 (2) 0.0476 (3) 0.052 (4) 0.52
5. The density of 3 molal solution of NaOH is 1.110 g/ml. Molarity of the solution is (1) 1.97 M (2) 2.97 M (3) 3.45 M (4) 2.45 M
6. The volume of 0.025 M Ca(OH) 2 solution, which can neutralise 100 ml of 10–4 M H3PO4, is (1) 10 ml (2) 60 ml (3) 0.6 ml (4) 2.8 ml
7. When 100 ml of 0.06 M Fe(NO3)3, 50 ml of 0.2M FeCl3, and 100 ml of 0.26 M Mg(NO3)2 are mixed, the concentration of NO 3 – ions in the final solution is (1) 0.028 M (2) 0.32 M (3) 0.12 M (4) 0.28 M
8. The molality of 2% (W/W) NaCl solution is nearly (1) 0.02 m (2) 0.35 m (3) 0.25 m (4) 0.45 m
9. 100 ml of 2 M HCl solution completely neutralises 10 g of a metal carbonate. Then, the equivalent weight of the metal is
CHAPTER REVIEW
• The SI system has seven base units
• Precision refers to the closeness of various measurements for the same quantity.
• However, accuracy is the agreement of a particular value of the result.
■ According to the law of conservation of mass, in a chemical reaction, total mass of the products is equal to the total mass of the reactants.
■ According to the law of definite proportions, a given chemical substance (compound) always contains the same elements combined in a fixed proportion by weight.
■ According to the law of multiple proportions, if two elements chemically combine to give two or more compounds, then the weights of one element which combine with the fixed weight of the other element in those compounds bear a simple multiple ratio to one another.
■ Law of combining volumes was proposed by Gay-Lussac. This is applicable only for gases.
(1) 50 (2) 20 (3) 12 (4) 100
10. Molarity of 200 ml of HCl solution, which can neutralise 10.6 g of anhydrous Na2CO3, is (1) 0.1M (2) 1M (3) 0.6M (4) 0.75M
Answer Key
(1) 1 (2) 1 (3) 1 (4) 2 (5) 2 (6) 3 (7) 4 (8) 2 (9) 2 (10) 2
■ Avogadro’s hypothesis states that equal volumes of all gases contain equal number of molecules under the same conditions of temperature and pressure.
■ The smallest particle of an element that takes part in a chemical reaction is an atom.
■ The smallest particle of a substance that can exist in the free state is a molecule.
■ Atomic weight or atomic mass of an element is a relative mass and is expressed in atomic mass units or atomic weight units.
■ The latest standard for determining atomic masses is 6C12, which is assigned a mass of 12 amu.
■ One amu is 1/12 part of the mass of 6C12 atom.
■ One amu is also known as one dalton or one aston.
■ Atomic mass of an element is the average of the isotopic masses (in amu) of the isotopes present in it.
■ The law of combining volumes states that when gases under similar conditions of temperature and pressure chemically combine, they bear a simple ratio in their volumes.
■ Molecular weight or molecular mass is also a relative mass expressed in amu.
■ The numerical value of the molecular mass expressed in grams is called a grammolecular weight or a gram molecule or a gram mole or a molar mass or a mole of that substance.
■ One mole of any substance (or one mole of a mixture of substances) contains the same number of molecules, namely 6.023 × 1023 molecules.
■ The number of molecules in one mole of a substance is known as Avogadro number.
■ One mole of any gas or vapour (or a mixture of gases or vapour) at STP occupies a volume of 22.4 litres.
■ The volume occupied by one mole of a gas at STP is known as gram molar volume (GMV).
■ One mole is the numerical value of the molecular weight of the substance expressed in grams, or it is mass of 6.022 × 1023 molecules of the substance.
■ The numerical value of the atomic weight of an element expressed in grams is known as a gram atomic weight or a gram atom of that element.
■ One gram atom of carbon is 12 grams of carbon.
■ One gram atom of any element contains 6.022 × 1023 atoms of the element.
■ Number of gram-atoms of an element (n) Mass of element
Gram atomic weight =
■ One gram atom is the numerical value of the atomic weight of the element expressed in grams, or it is the mass of 6.022 × 10 23 atoms of the element.
■ A mole of molecules means 6.022 ×10 23 molecules; a mole of atoms means 6.022
×10 23 atoms; and a mole of ions means 6.022 ×1023 ions.
■ The value of one amu is 1.667 × 10 –24 g. This is known as an avogram.
■ The density of a gas varies with pressure and temperature. The densities of gases are generally expressed at STP in gram/litre.
■ Vapour density of a gas or vapour is given as 2
Density of the given compound
Vapour density = Density of H gas
Mol.Wtof compound 2 =
■ Molecular weight of a gas is the product of density of the gas at STP in g/L and 22.4 L.
■ Vapour density of a gas is the product of density of the gas at STP and 11.2.
■ Empirical formula of a compound is the simplest formula of the compound which indicates the ratio between the number of atoms of the different elements present in one molecule of the compound.
■ Molecular formula of a compound indicates the actual number of atoms of each element present in one molecule of the compound.
■ Compounds with same empirical formula contain same percent composition of various elements.
The masses of different elements present in 100 g of a compound is called the percentage composition of the compound.
■ The exact quantities of the reactants and products that appear in the balanced chemical equation are called stoichiometric quantities.
■ The reactant that limits the product formation is called limiting reagent and the concept is termed limiting factor.
Exercises
JEE MAIN
Level-I
Importance of Chemistry
Single Option Correct MCQs
1. In which medical treatment are the drugs Cisplatin and Taxol effective?
(1) Cancer therapy
(2) Cardiac therapy
(3) Physiotherapy
(4) Muscular therapy
2. In which medical condition is Azidothymidine (AZT) primarily used?
(1) AIDS
(2) Leprosy
(3) Skin pigment disorders
(4) Muscular disorders
Nature of Matter
Single Option Correct MCQs
3. Which of the following can be classified as an element?
(1) Water (2) Sodium
(3) Ammonia (4) Sugar
4. Which of the following is a pure substance?
(1) Air (2) Copper
(3) Salt solution (4) Brass
5. Which of the following is an example of a homogeneous mixture?
(1) Sand and water (2) Sugar solution
(3) Milk (4) oil and water
6. Which of the following is a characteristic of gases?
(1) Definite shape and definite volume
(2) Definite shape but variable volume
(3) Indefinite shape and definite volume
(4) Indefinite shape and indefinite volume
7. Which of the following processes involves a change from a liquid state to a gaseous state?
(1) Condensation (2) Freezing
(3) Evaporation (4) Sublimation
Numerical Value Questions
8. There are _____ fundamental physical states.
Properties of Matter and Their Measurement
Single Option Correct MCQs
9. Which of the following properties of a gas does not vary with pressure and temperature?
(1) Density (2) Volume of a mole
(3) Volume (4) Vapour density
10. The prefix zepto stands for (1) 10-15 (2) 109 (3) 10-21 (4) 10-12
11. The SI unit for luminous intensity is (1) mole (2) candela
(3) ampere (4) kelvin
12. The SI system has _______ base units. (1) six (2) seven (3) five (4) three
13. Which of the following is incorrect about SI units?
(1) Density in kg m–3
(2) Force in newtons
(3) Pressure in pascals
(4) Amount of the substance in mol L –1
Numerical Value Questions
14. The number of significant figures in 50000.020 × 10−3 is _______.
Uncertainty in Measurement
Single Option Correct MCQs
15. The proper value of significant figures in 38.0 + 0.0035 + 0.00003 is (1) 38 (2) 38.0035 (3) 38.00353 (4) 38.0
16. The actual product of 4.327 and 2.8 is 12.1156. The correctly reported answer will be
(1) 12 (2) 12.1
(3) 12.12 (4) 12.116
17. After rounding off 1.235 and 1.225, their answers will, respectively, be
(1) 1.23, 1.22 (2) 1.24, 1.123 (3) 1.23, 1.23 (4) 1.24, 1.22
18. The number of significant figures in electronic charge 1.602 × 10 –19 C is (1) 1 (2) 2 (3) 3 (4) 4
19. Which of the following statements is correct?
(1) There is no difference between precision and accuracy.
(2) A good precision always means good accuracy
(3) Accuracy means that all measured values of an experiment are close to the actual value.
(4) A measurement may have good accuracy but poor precision.
20. On dividing 0.25 by 22.1176, the actual answer is 0.011303. The correctly reported answer will be
(1) 0.011 (2) 0.01
(3) 0.0113 (4) 0.013
Numerical Value Questions
21. A student performs a titration with different burettes and finds the values 25.20 ml, 25.25 ml, and 25 ml. The average titer value is _______.
22. After rounding off 4.855 to three significant figures, the answer will be ______
23. The number of significant figures should be present in the answer for 4.0 × 4.364 will be
24. When 4000 is converted to scientific notation, the number of significant figures is ________.
25. The value in the multiplication of 1.1 with 1.111 is ______.
26. The number of significant figures in the sum of 8.01, 10.11, and 9.123 is _______.
Laws of Chemical Combinations
Single Option Correct MCQs
27. The law of conservation of mass holds good for all the following, except
(1) nuclear reactions
(2) endothermic reactions
(3) all chemical reactions
(4) exothermic reactions
28. The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This proves the law of
(1) constant proportions
(2) reciprocal proportions
(3) multiple proportions
(4) conservation of mass
29. In compound A, 1.00 g of nitrogen combines with 0.57 g of oxygen. In compound B, 2.00 g of nitrogen combines with 2.24 g of oxygen. In compound C, 3.00 g of nitrogen combines with 5.11 g of oxygen. These results obey which of the following laws?
(1) Law of constant proportions
(2) Law of multiple proportions
(3) Law of reciprocal proportions
(4) Dalton’s law of partial pressure
30. One part of an element A combines with two parts of B (another element). Six parts
of element C combine with four parts of element B. If A and C combine together, the ratio of their masses will be governed by the
(1) law of definite proportions
(2) law of multiple proportions
(3) law of reciprocal proportions
(4) law of conservation of mass
31. Law of combining volumes was proposed by
(1) Lavoisier (2) Gay Lussac
(3) Avogadro (4) Dalton
Daltons Atomic Theory
Single Option Correct MCQs
32. Dalton’s atomic theory could not explain the law of
(1) conservation of mass
(2) multiple proportions
(3) constant proportions
(4) gaseous volumes
33. Which of the following statements are incorrect postulates of Dalton’s atomic theory?
A) Atoms of different elements differ in mass.
B) Matter consists of divisible atoms.
C) Compounds are formed when atoms of different elements combine in a fixed ratio.
D) All the atoms of a given element have different properties, including mass.
E) Chemical reactions involve the reorganisation of atoms.
Choose the correct answer from the options given below.
(1) B, D, and E only
(2) C, D, and E only
(3) A, B, and D only
(4) B and D only
Atomic and Molecular Masses
Single Option Correct MCQs
34. Air contains nitrogen and oxygen in a volume ratio of 4 : 1. The average molecular weight of air is
(1) 30 (2) 28.8
(3) 28 (4) 14.4
35. It is given that the abundance of isotopes 54Fe, 56Fe, and 57Fe are 5%, 90%, and 5%, respectively. The atomic mass of Fe is (1) 55.85 (2) 55.95
(3) 55.75 (4) 56.05
36. The mass of 1.5 × 10 26 molecules of a substance is 16 kg. The molecular mass of the substance is (1) 64 g (2) 64 amu
(3) 16 amu (4) 32 amu
Mole Concept and Molar Masses
Single Option Correct MCQs
37. The number of molecules in one litre of water is (density of water = 1 g/mL)
(1) 6 × 1023 / 22.4 (2) 3.33 × 1025
(3) 3.33 × 1023 (4) 3.33 × 1024
38. The mixture containing the same number of molecules as that of 14 g of CO is
(1) 14 g of nitrogen + 16 g of oxygen
(2) 7 g of nitrogen + 16 g of oxygen
(3) 14 g of nitrogen + 8 g of oxygen
(4) 7 g of nitrogen + 8 g of oxygen
39. The charge present on one mole of electrons is
(1) 96500 coulombs (2) 1 coulomb
(3) 1.60 × 10-19 C (4) 0.1 faraday
40. Which of the following has the highest mass?
(1) 1 gram atom of iron
(2) 5 moles of N2
(3) 1024 carbon atoms
(4) 44.8 L of He at STP
41. The ratio between the number of molecules in equal masses of nitrogen and oxygen is (1) 7 : 8 (2) 1 : 9 (3) 9 : 1 (4) 8 : 7
42. 1 mole of water vapour is condensed to liquid at 25 °C. Now, this water contains i) 3 moles of atoms ii) 1 mole of hydrogen molecules iii) 10 moles of electrons iv) 16 g of oxygen
Choose the correct combination. (1) (i) and (ii) are correct. (2) (i) and (iii) are correct. (3) (i) and (iv) are correct. (4) All are correct.
43. If 1 gram of hydrogen contains 6×1023 atoms, then 20 grams of Ne contains (1) 6 × 1023 atoms (2) 12 × 1023 atoms (3) 24 × 1023 atoms (4) 1.5 × 1023 atoms
44. The charge present on 10 moles of electrons is (1) 10 faraday (2) 1 faraday (3) 100 faraday (4) 0.1 faraday
45. The weight of 0.1 moles of Na 2CO3 is (1) 106 g (2) 10.6 g (3) 5.3 g (4) 6.02 x 1022 g
46. Avogadro’s number of helium atoms has a mass of (1) 2 g (2) 4 g (3) 8 g (4) 4 × 6.02 × 1023 g
47. The ratio between the number of atoms in equal masses of methane and oxygen is (1) 1 : 5 (2) 5 : 9 (3) 9 : 10 (4) 5 : 1
48. The gas that is twice as dense as oxygen under the same conditions is (1) ozone (2) sulphur trioxide (3) sulphur dioxide (4) carbon dioxide
49. Forty grams of Calcium contains N atoms. Then 24 grams of Mg contains
(1) 6 × 1023 atoms (2) 12 × 1023 atoms (3) 24 × 1023 atoms (4) 1.5 × 1023 atoms
50. Number of electrons in 1.8 grams of H2O is (1) 6.02 × 1023 (2) 3.01 × 1023
(3) 0.602 × 1023 (4) 60.22 × 1023
51. Which of the following gases has the largest number of molecules?
(1) 0.5 g of H2 (2) 16 g of CO2 (3) 16 g of CH4 (4) 0.5 g of He
52. The number of molecules in 16 g of methane is
(1) 3.0 × 1023 (2) 23 16 10 6.02 ×
(3) 6.023 × 1023 (4) 23 16 10 3.0 ×
53. Number of molecules in one litre of oxygen at STP conditions is (1) 23 6.02 x10 32 (2) 23 6.02 x10 22.4 (3) 32 × 22.4 (4) 32 22.4
54. The atomic masses of two elements A and B are 20 and 40, respectively. If x g of A contains Y atoms, how many atoms are present in 2x g of B?
(1) 2y (2) y/2
(3) y (4) 4y
Numerical Value Questions
55. The number of atoms in 8 g of sodium is x × 10 23. The value of x is ____. (Nearest integer) [Given: NA = 6.02 × 1023 mol; Atomic mass of Na = 23 u]
56. The weight of methane, which occupies the same volume at STP as 7.5 g of ethane, is ___ g.
57. 7 g of nitrogen occupies a volume of 5 litres under certain conditions. Under the same conditions, one mole of a gas, having molecular weight 56, occupies a volume of _________ L.
Percentage Composition
Single Option Correct MCQs
58. A peroxidase enzyme contains 2% selenium (Se = 80). The minimum molecular weight of the enzyme is (1) 1000 (2) 2000 (3) 4000 (4) 800
59. Haemoglobin contains 0.33% iron (Fe = 56). The molecular weight of haemoglobin is 68000. The number of iron atoms in one molecule of haemoglobin is (1) 2 (2) 3 (3) 4 (4) 5
60. An element X forms two oxides. The formula of the first oxide is XO 2 . The first oxide contains 50% of oxygen. If the second oxide contains 60% oxygen, the formula of the second oxide is
(1) XO3 (2) X2O3 (3) X3O2 (4) X2O
61. 0.262 g of a substance gave, on combustion, 0.361g of CO2 and 0.147g of H2O. What is the empirical formula of the substance?
(1) CH2O (2) C3H6O (3) C3H6O2 (4) C2H6O2
62. A compound contains carbon and hydrogen in the mass ratio 3 : 1. The formula of the compound is (1) CH2 (2) CH3 (3) CH4 (4) C2H6
63. The empirical formula of acetic acid is the same as that of (1) sucrose (2) glucose (3) oxalic acid (4) formic acid
64. A compound contains 90% C and 10% H. The empirical formula of the compound is (1) C8H10 (2) C15H30 (3) C3H4 (4) C15H40
65. 5.6 g of an organic compound, on burning with excess oxygen, gave 17.6 g of CO 2 and 7.2 g of H2O. The organic compound is
(1) C6H6 (2) C4H8 (3) C3H8 (4) CH3COOH
66. An organic compound contains 40% of C, 13.33% of H, and 46.67% of N. Its empirical formula is (1) C2H2N (2) C3H7N (3) CH4N (4) CHN
67. The percentage of silica in sodium silicate is approximately (Atomic weight of Si = 28) (1) 25 (2) 40 (3) 50 (4) 60
68. A hydro carbon contains 80% by mass of carbon. The molecular formula of the hydro carbon is (1) C2H4 (2) C3H6 (3) C2H6 (4) C4H10
69. Analysis of chlorophyll shows that it contains 2.68% Mg. The number of magnesium atoms present in 2.4 g of chlorophyll is (1) 2.68 × 6 × 1021 (2) 2.68 × 6 × 1023 (3) 2.68 × 6 × 1020 (4) 2.68 × 6 × 1020/24
70. The weight percentage of nitrogen in urea (NH2CONH2) is (1) 38.4 (2) 46.6 (3) 59.1 (4) 61.3
Numerical Value Questions
71. An alkaloid contains 17.28% of nitrogen and its molecular mass is 162. The number of nitrogen atoms present in one molecule of the alkaloid is _____.
72. In diammonium hydrogen phosphate (NH4)2HPO4, the weight percentage of P2O5 is ________.(Nearest integer)
73. An enzyme contains 2% of sulphur. The molecular weight of the enzyme is 6400. How many sulphur atoms are present in that enzyme molecule?
74. A certain metal sulphide MS 2 is used extensively as a high-temperature lubricant. If MS 2 is 40.00% by mass of sulphur, the atomic mass of M is _____.
Stoichiometry and Stoichiometric Calculations
Single Option Correct MCQs
75. The number of molecules of CO 2 liberated by the complete combustion of 0.1 gram atoms of graphite in air is
(1) 1.01 × 1022 (2) 6.02 × 1023 (3) 6.02 × 1022 (4) 3.01 × 1023
76. When one mole of ozone completely reacts with SO2, the number of moles of SO3 formed is
(1) 1 (2) 2 (3) 3 (4) zero
77. Six grams each of magnesium and oxygen were allowed to react. Assuming that the reaction is complete, the mass of MgO formed in the reaction is
(1) 42 grams (2) 28 grams (3) 21 grams (4) 10 grams
78. The mass of CO2 obtained when 2 g of pure limestone is calcined is
(1) 44 g (2) 0.22 g
(3) 0.88 g (4) 8.8 g
79. X g of Ag was dissolved in HNO 3, and the solution was treated with excess NaCl when 2.87 g of AgCl was precipitated. The value of X is
(1) 1.08 g (2) 2.16 g (3) 2.70 g (4) 1.62 g
80. 1.2 g of Mg (at. mass 24) will produce MgO equal to (1) 0.05 g (2) 2 g (3) 40 mg (4) 4 g
81. The number of molecules of KI needed to produce 0.4 moles of K 2Hgl4 will be 4Kl + HgCl2 → K2HgCl4 + 2KCl (1) 1 (2) 3 (3) 16 (4) 1.6
82. 8Al + 30HNO3 → 8Al(NO3)3 + 3NH4NO3 + 9H2O
As per the given equation, the number of moles of aluminium metal that can be oxidised by one mole of HNO 3 is
(1) 8/3 (2) 8/30
(3) 3/8 (4) 30/8
83. The weight of oxygen required to completely react with 27 g of Al is
(1) 8 g (2) 16 g
(3) 32 g (4) 24 g
84. The weight of a pure sample of KClO3 to be decomposed in order to get 0.96 g of O 2 is
(1) 2.45 g (2) 1.225 g
(3) 9.90 g (4) none
85. 6 g of Mg reacts with excess of an acid. The amount of hydrogen produced will be
(1) 0.5 g (2) 1 g
(3) 2 g (4) 4 g
86. The number of moles of Fe2O3 formed when 5.6 L of O2 reacts with 5.6 g of Fe is
(1) 0.125 (2) 0.01
(3) 0.05 (4) 0.10
87. What volume of H 2 at NTP is required to convert 2.8 g of N2 into NH3?
(1) 2240 mL (2) 22400 mL
(3) 6.72 L (4) 224 L
88. The volume of CO2 obtained by the complete decomposition of one mole of NaHCO 3 at STP is
(1) 22.4 L (2) 11.2 L
(3) 44.8 L (4) 4.48 L
Numerical Value Questions
89. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na 3PO 4, calculate the maximum number of moles of Ba3(PO4)2 that can be formed. (Nearest integer)
90. 12 grams of a mixture of sand and calcium carbonate, on strong heating, produced 7.6 grams of residue. How many grams of sand are present in the mixture?
91. The volume of CO2 obtained by the complete decomposition of one mole of NaHCO 3 at STP is ______.
92. 0.1 mole of a hydrocarbon, on complete combustion, produced 17.6 g of CO 2. How many carbon atoms are present in each molecule of the hydrocarbon?
93. ‘ x’ g of calcium carbonate was completely burnt in air. The weight of the solid residue formed is 28 g. The value of ‘ x’ is _______.
94. How many moles of H2SO4 can be reduced to SO2 by 2 moles of aluminium?
95. Ammonia is oxidised by oxygen to give nitric oxide and water. The weight of water produced per gram of nitric oxide is _____ grams.
96. The mass of ammonia (in grams) produced when 2.8 kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is ____.
97. The minimum number of moles of O 2 required for the complete combustion of a mixture containing 1 mole of propane and 2 moles of butane is ____.
Level-II
Importance of Chemistry and Nature of Matter
Single Option Correct MCQs
1. (X) and (Y) are two pure substances. (X) can’t be decomposed into simpler substances by chemical reactions but (Y) can be decomposed into simpler substances. Identify (X) and(Y).
(1)
(2)
(3)
(4)
Properties of Matter - Measurement and Uncertainty in Measurement
Single Option Correct MCQs
2. Which of the following have same number of significant figures?
A. 0.00253 B. 1.0003
C. 15.0 D. 163
Choose the correct answer from the options given below.
(1) A, B, and C only
(2) C and D only
(3) B and C only
(4) A, C, and D only
3. 81.4 g sample of ethyl alcohol contains 0.002 g of water. The amount of pure ethyl alcohol (to the proper number of significant figures) is
(1) 91.398 g (2) 81.44 g
(3) 81.4 g (4) 81 g
Laws of Chemical Combinations
Single Option Correct MCQs
4. An unbalanced chemical equation is against (1) the law of gaseous volumes
(2) the law of constant proportions
(3) the law of mass action
(4) the law of conservation of mass
5. Which of the following reactions is not correct according to the law of conservation of mass?
(1) 2Mg(g)+O2(g)→ 2MgO(s)
(2) C3H8(g)+O2(g)→ CO2(g) +H2O(g)
(3) P4(s)+5O2(g) → P4O10(s)
(4) CH4(g)+2 O2(g)→ CO2(g)+2H2O(g)
6. The percentage of hydrogen in water and hydrogen peroxide is 11.2% and 5.94%, respectively. This illustrates the law of (1) constant proportions
(2) conservation of mass
(3) multiple proportions
(4) gaseous volume
7. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at STP. The data illustrates the (1) law of conservation of mass (2) law of constant proportions (3) law of multiple proportions (4) law of reciprocal proportions
8. Which one of the following combinations illustrates the law of reciprocal proportions?
(1) N2O3, N2O4, N2O5 (2) NaCl, NaBr, NaI (3) CS2, CO2, SO2 (4) PH3, P2O3, P2O5
Dalton’s Atomic Theory
Single Option Correct MCQs
9. Dalton’s atomic theory successfully explained:
(1) Law of conservation of mass
(2) Law of constant composition
(3) Law of multiple proportions
(4) All of the above
10. According to Dalton’s atomic theory, an atom can ____
(1) Be created
(2) Be destroyed
(3) Neither be created nor destroyed (4) Centrifuged
Atomic and Molecular Mass
Single Option Correct MCQs
11. The unified atomic mass unit has a value of
(1) 1.661 × 10–27 g
(2) 1.661 × 10–27 kg
(3) 1.661 × 10–25 g
(4) 1.661 × 10–25 kg
12. The atomic masses of two elements A and B are 20 and 40, respectively. If x g of A contains Y atoms, how many atoms are present in 2x g of B?
(1) 2Y (2) Y/2 (3) Y (4) 4Y
13. 0.2 mol of an alkane, on complete combustion, gave 26.4 g of CO 2 . The molecular weight of alkane is
(1) 16 (2) 30 (3) 44 (4) 58
14. The molar mass of a substance is 20 g. The molecular mass of the substance is
(1) 20 g (2) 20 amu
(3) 10 g (4) 10 amu
15. If m1 is the mass of 2 neutrons + 2 protons + 2 electrons and m2 is the mass of an α-particle, then
(1) m1 > m2
(2) m1 < m2
(3) m1 = m2
(4) m1 may be > or < m2, depending on its physical state.
16. The vapour density of a volatile chloride of a metal is 59.5 and the equivalent mass of the metal is 24. The atomic mass of the element will be
(1) 96 (2) 48 (3) 24 (4) 12
17. Mass of CO2 molecule in kg [approximately] (1) 44 (2) 44 × 10–3
(3) 73.04 × 10−27 (4) 63.04 × 10−3
18. Gram molecular weight of water is 18 g. The gram molecular weight of glucose is 180 g. Mass of 1 molecule of glucose is ( X ) and mass of 1 molecule of water is (Y). Relation between (X) and (Y) is
(1) X = Y (2) 10 X = Y
(3) X = 5 Y (4) X = 10 Y
19. A compound contains 7 carbon atoms, 2 oxygen atoms, and 1.0×10 −23 g of other elements. The molecular mass of compound is (NA = 6×1023)
(1) 122 (2) 116
(3) 148 (4) 154
20. 30 g of element x contains 18.069 × 10 23 atoms of x. The gram molar mass of x.
(1) 20 amu (2) 10 amu
(3) 10 g (4) 20 g
21. A certain compound has the molecular formula X 4O 6. If 10.0 g of the compound contains 5.62 g of X, the atomic mass of X is
(1) 62.0 amu (2) 48.0 amu
(3) 32.0 amu (4) 30.8 amu
Numerical Value Questions
22. An organic compound, on analysis, was found to contain 0.032% of sulphur. If one molecule contains two sulphur atoms, the molecular mass of organic compound is x × 105 Then, x is ______.
Mole Concept and Molar Mass
Single Option Correct MCQs
23. The correct increasing order of volume occupied by the following samples at NTP will be
A. Four gram atoms of nitrogen
B. One mole of methane
C. 14 grams of carbon monoxide
D. 1.5 grams atoms of helium
(1) C, D, B, A (2) B, C, A, D
(3) B, D, C, A (4) C, B, D, A
24. Which of the following gases will occupy the same volume that is occupied by 3 moles of CO2 at STP?
(1) 96 g of O2 (2) 2.8 g of N2
(3) 10 g of H2 (4) 72 g of SO3
25. Suppose, the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mol of XY2 weighs 10 g and 0.05 mol of X 3Y2 weighs 9 g, the atomic weights of X and Y are
(1) 40, 30 (2) 60, 40
(3) 20, 30 (4) 30, 20
26. Which of the following is heaviest?
(1) 50 g of iron
(2) 5 mol of nitrogen
(3) 0.1 g atom of silver
(4) 1023 atoms of carbon
27. Which of the following has the least number of atoms?
(1) 0.5 g atom of Zn
(2) 0.645 g of Zn
(3) 0.25 mol of Zn
(4) 6.45 × 1020 amu of Zn
28. Consider the following compounds:
A. 1 g of CO2 (mol. wt. = 44)
B. 1 g of CH3CHO (mol. wt. = 44)
C. 1 g of C3H8(mol. wt. = 44)
The correct order for the total atoms present in these compounds is
(1) A = B = C (2) A > B > C
(3) C > B > A (4) B > A > C
29. Study the following table: Compound (Molecular weight)
Which two compounds have least weight of oxygen? (molecular weights of compounds are given in brackets)
(1) II and IV (2) I and III
(3) I and II (4) III and IV
30. Number of moles of gas (A) is less than number of moles of gas (B) when equal masses of the gases are taken seperately under same conditions. Gas (A) and (B) are respectively,
(1) CH4 and SO2 (2) H2S and SO2
(3) NH3 and H2S (4) SO3 and CH4
31. The sample(s) containing the same number of Na atoms as there are Na atoms in 5.3 g of Na2CO3 is/are
(1) 4 g of NaOH
(2) 6.85 g of NaCl
(3) 0.25 mol of Na2SO4
(4) 5.6 g of Na3PO4
32. The mass of 1.5 × 1020 atoms of an element is 15 mg. The atomic mass of the element is
(1) 60 g (2) 60 mg
(3) 60 kg (4) 60 amu
33. 1 mol of water vapour is condensed to liquid at 25 °C. Now, this water contains
(i) 3 mol of atoms
(ii) 1 mol of hydrogen molecule
(iii) 10 mol of electrons
(iv) 16 g of oxygen
The correct combination is
(1) (i) and (ii)
(2) (i) and (iii)
(3) (i) and (iv)
(4) (i), (ii), (iii) and (iv)
34. One mole of any gas
A. occupies 22.4 L at STP
B. contains 3.05 × 1023 molecules
C. contains 6.023 × 1023 molecules
D. contains same number of molecules as 22 g of CO2
Correct statements among the given are
(1) B, D (2) A, C (3) B, C (4) A, D
35. Avogadro’s number is
(a) the number of atoms in gram atomic weight of substance
(b) the number of molecules in gram molecular weight of substance
(c) the number of atoms in 0.012 kg of C-12
(1) (a) and (b) only (2) (b) and (c) only (3) (c) only (4) (a), (b), and (c)
36. I) 4.4 g of CO2
II) 2.4 g of CH4
III) 10 g of SO3
Correct increasing order of number of atoms present in the above three samples is
(1) III, II, I (2) I, III, II (3) II, III, I (4) II, I, III
37. Correct the statement among the following.
A) Mass of one gram molecule oxygen is 32 g.
B) Mass of one gram atom oxygen is 16 g
C) Mass of one molecule oxygen is 32 amu.
D) Mass of one oxygen atom is 16 amu.
(1) A, B, C, and D (2) A, B, and C only (3) B and C only (4) A and B only
Numerical Value Questions
38. Henry thinks that a mole contains 6.023 × 1024 molecules. Hence, the mass of Henry’s mole of nitrogen (in grams) is ___
Percentage Composition
Single Option Correct MCQ
39. An organic compound having C, H, and S elements contains 4% sulphur. The minimum molecular weight of the compound is (1) 800 (2) 400 (3) 200 (4) 600
40. 60 g of a compound, on analysis, gave 24 g C, 4 g H, and 32 g O. The empirical formula of the compound is
(1) C2H4O2 (2) C2H2O2
(3) CH2O2 (4) CH2O
41. The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of the compound will be
(1) C2H2 (2) C3H3
(3) C4H4 (4) C6H6
42. A compound contains 5 g sulphur and 5 g oxygen atom. The empirical formula of the compound is
(1) SO (2) SO2 (3) S2O (4) SO3
43. Each 9.4 g of a compound contains 7.2 g carbon, 0.6 g hydrogen, and the rest is oxygen. The empirical formula of the compound is
(1) C3H3O
(2) C6H3O
(3) C6H6O
(4) C3H6O2
44. A gaseous hydrocarbon gives, upon combustion, 0.72 g of water and 3.08 g of CO 2 . The empirical formula of the hydrocarbon is
(1) C3H4 (2) C6H5
(3) C6H8 (4) C2H4
45. A carbon compound contains 12.8% of carbon, 2.1% of hydrogen, and 85.1% of bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula of the compound.
(1) CH3Br (2) CH2Br2
(3) C2H2Br2 (4) C2H3Br3
46. The relative number of atoms of different elements in a compound are as follows:
A = 1.33, B = 1 and C = 1.5. The empirical formula of the compound is
(1) A2B2C3 (2) ABC (3) A 8B6C9 (4) A 3 B3C4
47. 0.14 g of an element, on combustion, gives 0.28 g of its oxide. What is the element?
(1) Nitrogen (2) Carbon
(3) Fluorine (4) Sulphur
48. A compound contains carbon and hydrogen in the mass ratio 3 : 1. The formula of the compound is
(1) CH2 (2) CH3
(3) CH4 (4) C2H6
49. A sample of oleum is labelled as 118%. The number of moles of NaOH needed for complete neutralisation of 100 g oleum is
(1) 2.0 (2) 20 49
(3) 118 49 (4) 59 49
Numerical Value Questions
50. Calculate the percentage composition of oxygen in calcium nitrate.
Stoichiometry and Stoichiometric Calculations
Single Option Correct MCQs
51. The molality of 2% (W/W) NaCl solution is nearly (1) 0.02 m (2) 0.35 m (3) 0.25 m (4) 0.45 m
52. Assuming that sea water is a 3.5 wt% aqueous solution of NaCl, what is the molality of sea water?
(1) 2 molal (2) 2.5 molal (3) 0.62 molal (4) 1.5 molal
53. A 5.2 molal aqueous solution of methyl alcohol, CH 3OH, is supplied. What is the mole fraction of methyl alcohol in the solution?
(1) 0.190 (2) 0.086 (3) 0.050 (4) 0.100
54. 6 g of urea is dissolved in 90 g of water. The mole fraction of solute is (1) 1/5 (2) 1/50 (3) 1/51 (4) 1/501
55. Mole fraction of solute in an aqueous solution is 0.2, the molality of the solution is (1) 13.88 (2) 1.388 (3) 0.138 (4) 0.0138
56. 1 kg of 2 m urea solution is mixed with 2 kg of 4 m urea solution. The molality of the resulting solution is
(1) 3.33 m (2) 10 m (3) 1.67 m (4) 5m
57. Molarity and molality of a solution of a liquid (mol. mass = 50) in aqueous solution is 9 and 10, respectively. What is the density of the solution?
(1) 1 g/cc (2) 0.95 g/cc (3) 1.05 g/cc (4) 1.35 g/cc
58. A solution of acetic acid has molarity equal to 1.35 M and molality equal to 1.45 mol kg−1. The density of solution will be (1) 1.251 g L–1 (2) 1.125 g L–1 (3) 1.012 g L–1 (4) 0.994 g L–1
59. The number of ions present in 1 mL of 0.1 M CaCl2 solution is (1) 1.8 × 1020 (2) 6.0 × 1020 (3) 1.8 × 1019 (4) 1.8 × 1021
60. At 25 °C, for a given solution, M = m. Then, the correct relation between them if the temperature is increased to 50 °C, is
(1) M = m (2) M > m
(3) M = 2m (4) M < m
61. Normality of 250 mL of acidified solution containing 3.16 g of KMnO 4 (M = 158 g/mol) is
(1) 0.4 N (2) 1 N (3) 0.5 N (4) 0.2 N
62. Normality of 2% (W/V) of H2SO4 solution is nearly (1) 2 (2) 4 (3) 0.2 (4) 0.4
63. 500 mL of 1M H2SO4 completely neutralises 42 g of metal carbonate. Equivalent weight of metal is
(1) 24 (2) 12 (3) 30 (4) 23
64. 1 g Ca was burnt in excess of O 2 and the oxide was dissolved in water to make up one litre of solution. The normality of the solution is
(1) 0.04 N (2) 0.4 N
(3) 0.05 N (4) 0.5 N
65. Density of a solution is 0.5 g/cc. ( w/v)% of 40% (w/w) of the solution will be (1) 20% (2) 80%
(3) 15% (4) 60%
66. Sulphuric acid and orthophosphoric acid have the same molar mass. The ratio of the masses of these acids needed to neutralise the same amount of NaOH, if the sulphate and dihydrogen orthophosphate were formed in their separate reactions, is
(1) 1 : 2 (2) 2 : 1 (3) 1 : 3 (4) 1 : 1
67. The minimum amount of O 2 (g) consumed per gram of reactant is for the reaction (Given atomic mass: Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
(1) 2Mg(s)+O2→ 2MgO (s)
(2) 4 Fe (s) +3 O2(g)→ 2Fe2 O3 (s)
(3) P4(s)+5 O2(g)→ P4O10 (s)
(4) C3H8(g)+5 O2(g)→ 3CO2(g)+4H2O(l)
68. The number of moles of KI required to produce 0.4 mol K2HgI4 is 4KI + HgCl2→ K2HgI4 + 2KCl
(1) 1 (2) 3 (3) 16 (4) 1.6
69. Acetylene, C 2 H 2 , reacts with oxygen according to the unbalanced equation:
C2H2(g)+O2(g) → CO2(g)+H2O(g)
What is the O2/C2H2 ratio when this equation is correctly balanced?
(1) 2 1 (2) 3 1
(3) 4 1 (4) 5 2
70. 20 mL of nitric oxide combines with 10 mL of oxygen at STP to give NO2. The final volume will be
(1) 30 mL (2) 20 mL
(3) 10 mL (4) 40 mL
71. 2 9810 3 × g of tribasic acid is completely neutralised by 100 mL of 0.05M Ca(OH) 2 solution. The molecular weight of the acid is
(1) 49 (2) 74 (3) 37 (4) 98
Numerical Value Questions
72. Calculate the milliequivalents of HCl in its 3.65 g (at. wt. of H = 1, Cl = 35.5) is __.
73. Haemoglobin contains 0.34% of iron by mass. The number of Fe atoms in 3.3 g of haemoglobin is _______ ×10 19
Multiple Concept Questions
Single Option Correct MCQs
74. A compound has 40% of carbon by weight. If the molecular weight of the compound is 90, the number of carbon atoms present in 1 molecule of the compound is
(1) 3 (2) 2
(3) 1 (4) 5
75. Mass percentage of nitrogen in uracil is
(1) 70 (2) 50 (3) 80 (4) 25
76. What is the percentage of free SO 3 in a sample of oleum labelled a s 104.5%?
(1) 20% (2) 40%
(3) 60% (4) 80%
77. If 20 g water is added to 50 g of 109% oleum, the final composition of the sample is
(1) 50 g H2SO4
(2) 54.5 g H2SO4
(3) 54.5 g H2SO4 + 15.5 g H2O
(4) 54.5 g H2SO4 + 4.5 g SO3
78. The specific gravity of 98% H 2 SO 4 is 1.8 g/cc. 50 mL of this solution is mixed with 1750 mL of pure water. Molarity of the resulting solution is
(1) 0.2 M (2) 0.5 M
(3) 0.1 M (4) 1 M
79. The molarity of a sulphuric acid solution is 2.32 mol dm−3. If the density of solution is 1.14 g cm−3, the molality of the solution will be
(1) 2.54 mol kg–1 (2) 2.25 mol kg–1 (3) 2.62 mol kg–1 (4) 1.98 mol kg–1
80. What volume of 75% alcohol by weight (d = 0.8 g/cm3) must be used to prepare 159 cm3 of 30% alcohol by mass ( d = 0.90 g/cm3) (1) 67.5 mL (2) 56.25 mL (3) 44.44 mL (4) 4.44 mL
81. The amount of wet NaOH containing 15% water required to prepare 70 L of 0.5 N solution is (1) 1.65 kg (2) 1.4 kg (3) 16.5 kg (4) 140 kg
Level-III
Single Option Correct MCQs
1. 20.8 g of barium chloride is dissolved in 100 g of water. 14.2 g of sodium sulphate is dissolved in 100 g of water. Both solutions are mixed and 23.3 g of barium sulphate and 11.7 g of sodium chloride are formed. (At. wt. Ba = 137, Cl = 35.5, S = 32, Na = 23, O = 16). This data explains (1) law of conservation of mass (2) law of definite proportions
(3) law of constant proportions (4) law of multiple proportions
2. n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as X + Y → R + S. The relation that can be established in the amounts of the reactants and the products will be
(1) n – m = p – q
(2) n + m = p + q
(3) n = m
(4) p = q
3. A sample of calcium carbonate (CaCO 3 ) has the following percentage composition: Ca = 40%; C = 12%; O = 48%. If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate from another sourc e will be
(1) 0.016 g (2) 0.16 g
(3) 1.6 g (4) 16 g
4. Percentage composition of a natural sample of CuCO3 is as follows:
% Cu % O % C x y z
If the law of definite proportions is obeyed, then the correct composition of same elements in synthetic sample of CuCO 3 is
% Cu % O % C
(1) x y z
(2) >x >y >z
(3) <x <y <z
(4) >x <y >z
5. Which of the following statements indicates that the law of multiple proportions is being followed?
(1) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1 : 2.
(2) Carbon forms two oxides, namely CO 2 and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2 : 1.
(3) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(4) At constant temperature and pressure, 200 mL of hydrogen will combine with 100 mL of oxygen to produce 200 mL of water vapour.
6. Hydrogen combines with oxygen to form H2O, in which 16 g of oxygen combines with 2 g of hydrogen. Hydrogen also combines with carbon to form CH4, in which 2 g of hydrogen combines with 6 g of carbon. If carbon and oxygen combine together, then they will show in the ratio of
(1) 6 : 16 (2) 6 : 18
(3) 1 : 2 (4) 12 : 24
7. An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67%, and N = 46.67%, while the rest is oxygen. On heating, it gives NH 3 along with a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is
(1) CH3NCO
(2) CH3CONH2
(3) (NH2)2CO
(4) CH3CH2CONH2
8. 40 mL of a hydrocarbon undergoes combustion in 260 mL of oxygen and gives 160 mL of CO2. If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is
(1) C3H8 (2) C4H8
(3) C6H14 (4) C4H10
9. 10 mL of a gaseous hydrocarbon, on combustion, gives 40 mL of CO2 and 50 mL of H2O vapour under the same conditions. The hydrocarbon is
(1) C4H6 (2) C6H10
(3) C4H8 (4) C4H10
10. For the reaction, 2x + 3y + 4z → 5 w, initially, if 1 mole of x, 3 mol of y, and 4 mol of z is taken. 1.25 mol of w is obtained.
Then, the percentage yield of this reaction is
(1) 50% (2) 60%
(3) 12% (4) 25%
11. In which of the following solutions mole fraction of solute is highest? (density of water = 1 g/ml)
(1) 220 g of aqueous solution containing 1 mol of NaOH (mol. wt. = 40)
(2) 282 g of aqueous solution of urea containing 162 g of water
(3) 540 g of glucose added to 8 moles of water
(4) 4 mol of KOH dissolved in 126 g of water
12. Wood’s metal contains 50.0% bismuth, 25.0% lead, 12.5 % tin, and 12.5% cadmium by mass. What is the mole fraction of tin? (atomic mass: Bi = 209, Pb = 207, Sn = 119, Cd = 112)
(1) 0.202
(2) 0.158
(3) 0.182
(4) 0.221
13. The expression converting molality (m) into molarity (M) is
(where M2 is the molar mass of solute and r is the density of the solution)
(1) M =(1+ mM2) m r
(2) M = m r (1+mM2)
(3) M = mM2/ r
(4) M = m r/ M2
14. Relation between mol arity (M) and mole fraction is
(1) A AABA .1000d M .mm χ× = χ+χ
(2) B AABB 1000d M .mm χ×× = χ+χ
(3) A m M V100 ×χ = ×
(4) A M d 1 V χ =
15. Three solutions X, Y, and Z of HCl are mixed to produce 100 mL of 0.1 M solution. The molarities of X, Y, and Z are 0.07 M, 0.12 M, and 0.15 M, respectively. What respective volumes of X, Y, and Z should be mixed?
(1) 50 mL, 25 mL, 25 mL
(2) 20 mL, 60 mL, 20 mL
(3) 40 mL, 30 mL, 30 mL
(4) 55 mL, 20 mL 25 mL
16. One mole of KClO3 is heated in the presence of MnO2 (catalyst). The produced oxygen is used in burning of Al. Then, oxide of Al that will be formed is
(1) 2 mol (2) 1 mol
(3) 4 mol (4) 3 mol
17. One litre of a mixture of CO and CO 2 is passed over red hot coke, when the volume increased to 1.6 L under the same conditions of temperature and pressure. The volume of CO in the original mixture is
(1) 600 (2) 200
(3) 800 (4) 400
18. Iron has a density of 7.86 g cm −3 and an atomic mass of 55.85 amu. The volume occupied by 1 mol of Fe is
(1) 0.141cm3 mol–1
(2) 7.11 cm3 mol–1
(3) 4.28 × 104 cm3 mol–1
(4) 22.8 cm3 mol–1
19. The percentage of water of crystallisation in CaCl2·6H2O is
(1) 39.4 (2) 43.9
(3) 49.3 (4) 46. 5
20. 0.607 g of a silver salt of tribasic organic acid was quantitatively reduced to 0.37 g of pure Ag. What is the molecular mass of the acid?
(1) 207 (2) 210
(3) 531 (4) 324
21. An organic base is tetra-acidic. If, from every 10 g of the chloroplatinate salt of the base, 3.9 g of residue of platinum is obtained, then what will be the molecular mass of the base?
[Pt = 195]
(1) 180 (2) 360
(3) 90 (4) 270
22. The shape of tobacco mosaic virus (TMV) is cylindrical, having length and diameter 3000 Ao and 170 Ao, respectively. The density of the virus is 0.08 g/mL. The molecular weight of TMC is
(1) 3.28
(2) 5.44 × 10–24
(3) 5.44 × 10–18
(4) 3.28 × 106
23. Manganese forms non-stoichiometric oxides having the general formula MnOx . The value of x for the compound, which on analysis gave 64% by mass of Mn, is
(1) 1.16 (2) 1.83
(3) 2 (4) 1.93
24. An organic compound contains C, H, and N. When 9 volumes of a gaseous mixture containing the organic compound and just sufficient quantity of O 2 was exploded, 2 volumes of N2, 6 volumes of water vapour, and 4 volumes of CO 2 were produced. If all measurements are done under similar conditions of temperature and pressure, what will be the formula of the organic compound?
(1) C2H6N (2) C2H6N2
(3) C3H6N2 (4) C2H6N3
25. A hydrocarbon CnH2n yields CnH2n+2 by reduction. The molar mass of the product is 2.38% more than the reactant. The value of n is
(1) 8 (2) 4
(3) 6 (4) 5
26. A granulated sample of aircraft alloy (Al, Mg, Cu) weighing 8.72 g, was first treated with alkali and then with very dilute HCl, leaving a residue. The residue, after alkali boiling, weighed 2.10 g and the acid insoluble residue weighed 0.69 g. What is the composition of the alloy?
(1) Al = 75.9%, Mg = 16.2%, Cu = 7.9%
(2) Al = 16.2%, Mg = 75.9%, Cu = 7.9%
(3) Al = 7.9%, Mg = 16.2%, Cu = 75.9%
(4) Al = 75.9%, Mg = 19.2%, Cu = 4.9%
27. A mixture of ethane and ethene occupies 16.42 L at 1 atm and at 400 K. The mixture reacts completely with 51.2 g of O 2 to produce CO2 and H2O. Assuming ideal gas behaviour, calculate the mole fractions of C2H6 and C2H4 in the mixture.
(1) 0.3, 0.2
(2) 0.66, 0.33
(3) 0.75, 0.25
(4) 0.4, 0.6
28. x g of KHC2O4 requires 100 mL of 0.02 M KMnO 4 in acidic medium. In another experiment, y g of KHC2O4 requires 100 mL of 0.05 M Ca(OH)2. The ratio of x and y is
(1) 1 : 1 (2) 1 : 2
(3) 2 : 1 (4) 5 : 4
29. What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?
(1) 330 (2) 316
(3) 320 (4) 325
30. For the following reaction, A+2 B → C, if equal mass of A and B are taken, which of the following is correct?
[MA and MB are the molar masses of A and B, respectively]
(1) If MA = 2MB, then none of the reactant is left.
(2) If M A = M B , then A will be limiting reagent.
(3) If M B = M A , then B will be limiting reagent
(4) If A B 2 > M M then A wi ll be limiting reagent
Numerical Value Questions
31. A 300 mL bottle of soft drink has 0.2M CO2 dissolved in it. Assuming that CO2 behaves as an ideal gas, the volume of the dissolved CO2 at STP is ______ mL (nearest integer).
(Given: At STP molar volume of an ideal gas is 22.7 L mol–1
32. 2 mol of carbon reacts with oxygen in air or oxygen forming two oxides, CO and
THEORY-BASED QUESTIONS
Statement Type Questions
(Q.No. 1-10)
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
1. S–I : Sugar syrup is heterogeneous mixture.
S–II : Air is homogeneous mixture.
CO2, depending on the amount of oxygen available. If 19.2 g of CO is formed, then the percentage of carbon converted into CO is (Atomic weights of C = 12, O =16) ______.
33. 2 mol of a mixture of O 2 and O3 is reacted with excess of acidified solution of KI. The iodine liberated required 1L of 2 M hypo solution for complete reaction. The weight percentage of O3 in the initial sample is x.
Find x.
34. A gaseous mixture contains SO 3 (g) and CH4(g) in 5 : 1 ratio by mass. Calculate Q, where Q = 200 × ratio of total number of atoms present in SO3(g) to total number of atoms present in CH4(g).
35. Lithium nitride hydrolyses in water to produce a basic solution. What is the mass of HCl required to neutralise the 1 litre of solution, in which 2 mol of lithium nitride is dissolved?
36. One litre of 2M Na3PO4 and 1 litre of 1M BaCl 2 solutions are mixed. What is the normality of phosphate ions in the filtrate?
2. S-I : During a chemical reaction (at STP), the total volume remains constant in gaseous reactions.
S-II : During a chemical reaction, the total mass remains constant.
3. S–I : Formation of oxides of nitrogen explains the law of definite proportions.
S–II : Equivalent weight of nitrogen in oxides of nitrogen is always constant.
4. S-I : The number of atoms in a given mass of dioxygen (oxygen) and trioxygen (ozone) gases is same.
S-II : The number of atoms depends on atomic mass, not on molecular mass.
5. S-I : 12 C has atomic mass of exactly 12 atomic mass units (amu).
S-II : One amu is defined as a mass exactly equal to 1/12 of the mass of one carbon-12 atom.
6. S–I : Fructose and ethanoic acid have the same percentage composition of elements by weight.
S–II : Fructose and ethanoic acid have the same empirical formula
7. S–I : 16 g each of ozone and oxygen contain the same number of atoms.
S–II : Nitrogen cannot exhibit +5 oxidation state in its compounds.
8. S–I : 1 cc of nitrogen at STP contains 2.69 × 1019 molecules.
S–II : Molar volume of an ideal gas at STP contains Avogadro’s number of molecules.
9. S–I : 0.5 mol of P4O10 contains five gram atoms of oxygen.
S–II : 6.023 ×10 22 atoms of hydrogen can make 0.1 g atom of hydrogen.
10. Consider the given reaction and statements.
2A+3B → C
S–I : 4 3 mol of ‘C’ is always produced when 3 mol of ‘A’ and 4 mol of ‘B’ are added.
S-II : ‘B’ is the liming reactant for the given data.
Assertion and Reason Questions
(Q.No. 11-29)
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
11. (A) : One mole of CO 2 is formed when 1 mol of O2 is reacted with 1 mol of CO, according to the reaction
2CO+O2→ 2CO2
(R) : CO is the limiting reagent in the above reaction.
12. (A) : 3.4 g of NH 3 (g), on complete decomposition into N 2 and H 2 (g), produces 0.6 g of H2(g).
(R) : Law of conservation of mass is followed by the chemical reaction.
13. (A) : During the conversion of Fe 2O3 : Fe, the stoichiometric coefficients of iron are in the ratio of 1 : 2.
(R) : During a chemical reaction, atoms can neither be created nor destroyed.
14. (A) : Formation of C2H5OH and CH3OCH3 from respective elements obeys the law of definite proportions.
(R): Both compounds have the same molecular formula and the elements C : H : O are combined in the ratio of 12 : 3 : 8 by mass.
15. (A) : A certain element X forms three binary compounds with chlorine, containing 59.68 %, 68.95%, and 74.75% chlorine, respectively. This data illustrates the law of multiple proportions.
(R) : According to the law of multiple proportions, the relative amounts of an element combining with some fixed amount of a second element in a series of compounds is in the ratio of small whole numbers.
16. (A) : In Haber’s process, N2 and H2 combine in 1 : 3 volume ratio.
(R) : Gases combine in a simple volume ratio.
17. (A) : For reaction
2A(g)+ 3B (g) → 4C (g) + D (g), the vapour density remains constant throughout the course of the reaction.
(R) : In all gaseous chemical reactions, vapour density remains constant.
18. (A) : Atomic weight of an atom can never be in fraction.
(R) : Average atomic weight of chlorine is 35.5.
19. (A) : Atomic mass of Mg is 24.
(R) : An atom of magnesium is 24 times heavier than 1 th 12 of the mass of a carbon atom (C12)
20. (A) : Boron has relative atomic mass of 10.81.
(R) : Boron has two isotopes 10 5 B and 11 5 B, and their relative abundances are 19% and 81%, respectively.
21. (A) : 18 g of water vapour and 18 g of ice will not contain the same number of molecules.
(R) : Number of molecules is independent of temperature and pressure.
22. (A) : The empirical mass of ethene is half of its molecular mass.
(R) : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
23. (A) : Molality of a solution increases with temperature.
(R) : Molality expression does not involve any volume term.
24. (A) : 3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.
(R) : Molar mass of hydrated oxalic acid is 126 g.
25. (A) : Mass of a solution of 1 litre of 2 M H2SO4 [density of solution =1.5 g/mL] is greater than the mass of solution containing 400 g MgO which is labeled as 40% (w/w) MgO.
(R) : Mass of H2SO4 in 1 litre 2 M H2SO4 [density of solution =1.5 g/mL] is greater than the mass of MgO in 1 litre 40% (w/w)[density of solution =2 g/mL] solution.
26. (A) : The number of glucose molecules present in 10 mL of decimolar solution is 6.023×1020
(R) : Avagadro’s number of molecules present in 1 g molar volume of solution.
27. (A) : 22.4 L of N2 at NTP and 5.6 L O2 at NTP contain equal number of molecules.
(R) : Under similar conditions of temperature and pressure all gases contain equal number of molecules.
28. (A) : Combustion of 16 g of methane gives 36 g of water and 44 g of CO 2.
(R) : In the combustion of methane, water and CO2, are products.
29. (A) : A reactant that is entirely consumed when a reaction goes to completion is known as limiting reactant.
(R) : The amount of limiting reactant limits the amount of product formed.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Choose the correct statement(s).
(1) The mole is the amount of a substance containing the same number of chemical units as there are atoms in exactly 12 g of 12C.
(2) Avogadro’s number is the number of units in a mole.
(3) The weight of one gram atom of an element means its atomic weight in g.
(4) One gram atom of each element contains the same number of atoms.
2. A+2B + 3C AB2C3. Reaction of 6.0 g of atoms of B, and 0.036 mol of C yields 4.8 g of compound AB2C3. If the atomic masses of A and C are 60 and 80 amu, respectively, the atomic mass B of is (Avogadro’s constant = 6 × 1023)
(1) 70 amu (2) 60 amu
(3) 50 amu (4) 40 amu
3.
‘ x ’ grams of 50% pure potassium chlorate liberates 0.3 mol of oxygen on thermal decomposition. x grams of pure H 2 SO 4 requires g of NaOH for complete neutralisation
(mol wt. of potassium chlorate = 122.5)
(1) 20 g (2) 40 g
(3) 60 g (4) 50 g
4. 40 g of a sample of carbon, on combustion, left 10% of it unreacted. The volume of oxygen required at STP for this combustion reaction is
(1) 11.2 L (2) 22.4 L
(3) 44.8 L (4) 67.2 L
5. The molecular formula of a commercial resin used for exchanging ions in water softening is C 8 H 7SO 3 –Na + (molar mass: 206). What would be the maximum uptake of Ca+2 ions
by the resin, when expressed in mole per gram resin?
(1) 2 309 (2) 1 412
(3) 1 103 (4) 1 206
6. If the vapour density of a mixture of NO2 and N2O4 is 34.5, then the percentage abundance (by mol) of NO2 in the mixture is
(1) 50% (2) 25%
(3) 40% (4) 60%
7. Phosphine, on decomposition, produces phosphorus and hydrogen. When 100 mL of phosphine is decomposed, the change in volume under laboratory conditions is
(1) 50 mL increase
(2) 50 mL decrease
(3) 900 mL decrease
(4) 75 mL increase
8. 1.878 g of MBr x, when heated in a stream of HCl gas, was completely converted to chloride MCl x, which weighed 1.0 g. The specific heat of metal is 0.14 cal/g. Then, the molecular weight of metal bromide is (atomic weight of Br = 80)
(1) 285.5
(2) 185.7
(3) 85.7
(4) 216
9. Which of the following can explain the law of reciprocal proportions?
(1) H2O, CH4, CO2
(2) CO, CO2, C3O2
(3) HF, CF4, CH4
(4) N2O, NO2, N2O4
10. “Compounds are formed when atoms of different elements combine in a fixed ratio” is one of the statements of Dalton’s atomic theory. Which of the following laws is not related to this statement?
(1) Law of conservation of mass
(2) Law of definite proportions
(3) Law of multiple proportions
(4) Avogadro’s law
11. Acetic acid and glucose have the same
(1) empirical formula
(2) weight composition of elements
(3) ratio of masses of individual elements
(4) number of gram atoms of each element per mole
12. The pair of species having different percentage (mass) of carbon is
(1) CH3COOH and C6H12O6
(2) CH3COOH and C2H5OH
(3) HCOOCH3 and HCOOH
(4) C2H5OH and CH3OCH3
13. 80% carbon is present in an alkane, by weight. The possible conclusions are
(1) the empirical formula of the compound is CH3
(2) the minimum number of carbons in the molecule is 2
(3) the compound has gram atoms of C and H in 4 : 1 ratio
(4) this composition suits all alkanes
14. 10% w/v NaOH is same as
(1) 2.5 M (2) 2.5 N (3) 1.5 m (4) 10% w/w NaOH
15. 4.1 g of phosphorous acid H3PO3 is present in 250 mL of a solution. The strength of the solution is
(1) 0.2 M (2) 0.4 N
(3) 16.4% w/v (4) 8.2% w/v
16. Solution (s) containing 23 g HCOOH is/are
(1) W 46gof70%HCOOH V
(dsolution = 1.40 g/mL)
(2) 50 g of 10 M HCOOH (dsolution=1g/mL)
(3) W 50g of 25%HCOOH V
(4) 46 g of 5MHCOOH
(dsolution =1g} /mL)
17. If 400 mL of 0.1 M HCl, 500 mL of 0.3 M H 2SO 4 and 100 mL of 0.1 M KNO 3 are mixed, then what is correct for the resultant solution?
(1) [H+] = 0.19 M
(2) [Cl ] = 0.04 M
(3) [NO3–] = 0.01 M
(4) [K+] = 0.01 M
18. 100 mL of a solution contains 12 mg MgSO4. The concentration of the solution is
(1) 10–3 M
(2) 2 × 10–3 N
(3) 120 ppm
(4) 10–3 m
19. If 3.65% w/v HCl solution has density 1.0365 g/cc, then its concentration is (1) 1 M (2) 1 m
(3) 1 N (4) 0.5 M
20. If 1 mL of 1 M solution is mixed with 999 mL of pure water, then (1) 10–3 M solution is formed
(2) the mass of solute per mL decreases by 1000 times
(3) the quantity of solute decreases in the solution
(4) 10 mL of resultant solution contains 10–5 moles of solute
21. An aqueous solution of ammonia has molarity to 2 M. If the density of the solution is 1.534 g/mL, then identify the options in which the correct concentration terms are mentioned.
(1) Molality 4 3 = m
(2) () 34 %w/w 15.34 = (3) %(w/v) = 6.8
(4) mole fraction of NH3 = 3/128
22. 1 g atom of nitrogen represents (1) 6.02 × 1023 of N2 molecules
(2) 22.4 L of N2 at 1 atm and 273 K
(3) 11.2 L of N2 at 1atm and 273 K
(4) 14 g of nitrogen
23. Which of the following are correct statements?
(1) Unbalanced equations are against the law of conservation of mass.
(2) 1 mL of any gas at STP contains 2.69 × 10 19 molecules.
(3) The number of atoms present in 8 g of sulphur is same as that in 10 g of calcium.
(4) The oxidation number of ‘Cr’ in CrO5 is +10.
24. 8 g of O2 has the same number of molecules as in
(1) 11 g of CO2 (2) 22 g of CO2 (3) 7 g of CO (4) 14 g of CO
25. 6.023×1022 atoms of hydrogen can make (1) 0.05 mol of H2 molecules
(2) 0.1 g of hydrogen atoms
(3) 0.1 g molecules of hydrogen
(4) 0.1 g atoms of hydrogen
26. 1 mol of 143 7 N ions contains (1) 7 × 6.023 × 1023 electrons
(2) 7 × 6.023 × 1023 protons
(3) 7 × 6.023 × 1023 neutrons
(4) 14 × 6.023 × 1023 protons
27. We have 1.6 g CH4, 1.7g NH3, and 1.8 g H2O. Select the correct statement(s).
(1) There are equal number of moles of each reactant.
(2) Total number of atoms in CH4 > NH3 > H2O.
(3) Total number of H-atoms are in the ratio of 4 : 3 : 2.
(4) Total number of C-atoms in CH 4 < that of N-atoms in NH3 < that of O-atoms in H2O.
28. 18 grams of glucose contains
(1) 0.6 gram atom of carbon
(2) 0.6 gram molecule of hydrogen
(3) 0.6 gram molecule of CO2
(4) 1.2 gram atom of hydrogen
29. 3200 g sulphur is present in
(1) 9800 g H2SO4
(2) 20 mol H2SO4
(3) 100 mol H2SO4
(4) 6400 g of SO2
30. The molar mass of haemoglobin is about 65000 g mol−1. Every haemoglobin contains 4 iron atoms. Thus,
(1) iron content in haemoglobin is 0.35% by mass
(2) 1 mol of haemoglobin contains 56 g iron
(3) 1 mol of haemoglobin contains 224 g iron
(4) if iron content is increased to 0.56%, molar mass of haemoglobin would be higher than 65000 g mol −1
31. 1021 molecules are removed from 440 mg of CO2. It becomes
(1) 366 mg (2) 8.3 millimol
(3) 200 mg (4) 4.1 millimol
32. One takes 0.62 g of P 4 and 4.00 g of O 2 separately for each of the following reactions. (Atomic mass of P = 31; O = 16)
I. P4+5O2 → P4O10
II. P4+3 O2 → P 4 O6
Select the correct statements about I and II.
(1) P 4 is the limiting reagent in 1 and O2 is the limiting reactant in II.
(2) P 4 is the limiting reactant in II and O 2 is the limiting reactant in I.
(3) 1.42 g of P4O10 in I and 1.10 g of P4O6 in II are formed.
(4) P 4 is the limiting reactant in both I and II.
33. If 100 mL of O2 is subjected to silent electric discharge till volume of O 2 left and O 3 formed are equal, then
(1) decrease in volume is 20 mL
(2) volume of O3 formed is 40 mL
(3) volume of O2 reacted is 60 mL
(4) volume of mixture is 80 mL
34. Consider the following reactions:
I. (NH4)2SO4 +2 NaOH 40% → Na2SO3 + 2H2O +2 NH3
II. 80% 34 NHHClNHCl +→
If 4 g of NaOH is taken, then
(1) produced mole of NH4Cl (in reaction II) is 1.6 times of produced moles of Na2SO4 (in reaction I)
(2) reacting mole of HCl (in reaction II) is 20% lesser than original (NH4)2SO4 moles
(3) reacting mole of HCl (in reaction II) is lesser than reacting moles of NaOH (in reaction I)
(4) produced mass of NH4Cl is 2.71 g
35. When taken in an eudiometer tube operating at room temperature and pressure and subjected to complete reaction, in which of the following options, the contraction in volume is greater than or equal to 30% of the original volume? [Note: Produced H 2O is in liquid state]
(1) CO(g) and O2(g) taken in a molar ratio of 2 : 1
(2) 10 mL of CH4(g) and 30 mL of O2(g)
(3) N2(g) and H2(g) taken in a molar ratio of 3 : 1
(4) N2 and H2 taken in a molar ratio of 1 : 3
36. Which of the following requires six times their volume of oxygen for complete combustion?
(1) n-butane (2) 1-butene
(3) 2-butene (4) cyclobutane
37. Combustion of 2.24 L ethane at STP requires (1) 7.84 L of O2 (2) 0.35 mol of O2 (3) 11.2 g of O2 (4) 5.6 L of O2 at STP
38. 1 mol of iron reacts completely with 0.65 mol of O2 to give a mixture of only FeO and Fe2O3. Choose the correct statement(s).
(1) Ratio of number of moles of FeO to Fe2O3 is 3 : 2.
(2) Ratio of number of moles of FeO to Fe2O3 is 4 : 3.
(3) Higher mass of Fe2O3 is formed.
(4) The sum of the masses of Fe and O 2 equals the sum of the masses of FeO and Fe2O3.
39. What volume of 0.2 M Ba(MnO4)2 solution is required for the complete oxidation of 25 g of 89.6% pure FeC2O4 in acidic medium, according to the reaction
MnO4– + FeCr2O4 → Fe+3+Cr2O7-2+ Mn+2?
(1) 700 mL (2) 175 mL
(3) 350 mL (4) 200 mL
Numerical Value Questions
40. On heating two moles each of Li2CO3, K2CO3 and NaHCO3, how many grams of CO2 are evolved?
41. If the relative atomic mass of oxygen is 64 units, the molecular mass of CO becomes .
42. The empirical formula of an organic compound is CH 2O. Its vapour density is 45. The molecular formula of the compound is x. How many atoms are present in x?
43. A gas mixture contains C2H4 and CO2 and 40 L of this mixture require 30 L of O 2 for combustion. The percentage of CO 2 in the original mixture is
44. A certain metal sulphide MS 2 is used extensively as a high temperature lubricant. If MS2 is 40.00% by mass of sulphur, what is the atomic mass of M?
45. The molality of a 10% ( v / v ) solution of di-bromine solution in CCl 4 (carbon tetrachloride) is ‘x’.
x = × 10–2 M (Nearest integer)
[Given: Molar mass of Br 2 = 160 g mol−1
Atomic mass of Cl = 35.5 g mol –1
density of dibromine = 3.2 g cm –3
density of CCl4 = 1.6 g cm−3]
46. The mole fraction of glucose (C 6 H 12 O 6 ) in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is .
47. When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is × 10–2 M. (Nearest integer) (Molar mass of nitric acid is 63 g mol –1)
48. Density of methanol is 0.0008 kg L –1. What volume of methanol in mL is required to prepare 2.5 L of a 0.25 M solution of methanol?
49. The mole fraction of a solute in a 100 molal aqueous solution is × 10 –2 (Round off to the nearest integer).
50. An element X has the following isotopic composition.
The weighted average atomic mass of the naturally occurring element ‘X’ is closest to .
51. A gaseous mixture contains 40% O2, 40% N2, 10% CO2,10% CH4 by volume. Calculate the vapour density of the gaseous mixture.
52. A mixture of O2 and gas ‘Y’ (mol. wt. 80) in the mole ratio a : b has a mean molecular weight of 40. What would be the mean molecular weight, if the gases are mixed in the ratio b : a under identical conditions? (Gases are non-reacting)
53. Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL −1 at 257 °C and 100 mm Hg. The molar mass of geraniol is g/mol (Nearest integer) [Given: R = 0.082 Latm K–1mol–1]
54. What is the percentage of free SO 3 in an oleum that is labelled as 104.5% H 2SO4?
55. Calculate the molality of 1 litre solution of 93% H 2SO 4 (weight/volume). The density of the solution is 1.84 g/mL.
56. The molarity of HNO3 in a sample, which has density 1.4 g/mL and mass percentage of 63%, is . (Molecular weight of HNO3 = 63)
57. The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4.2H2O) in 250 mL of water in mol L−1 is x × 10−2. The value of ‘ x ’ is (Nearest integer) [Atomic mass: H:1.0, C:12.0, O: 16.0].
58. At NTP, how much volume in mL of NH 3 must be passed into 30 mL of 1 N H 2 SO 4 so that its normality becomes 0.2 N?
59. The number of nitrogen atoms in 681 g of C 7 H 5N 3O 6 is x × 10 21. The value of x is . N A = 6.02 × 10 23 mol –1 ) (Nearest integer)
60. What is the amount of Mg in grams to be dissolved in dilute H 2 SO 4 to liberate H2,which is just sufficient to reduce 160 g of ferric oxide?
61. 10 g impure NaOH is completely neutralised by 1000 mL of 1 N 10 HCl. Calculate the percentage purity of the impure NaOH.
Integer Value Question
62. Three isotopes of an element have mass number m, m+1, and m+2. If the ratio of abundance is 4 : 1 : 1 and mean mass is 1 m x + , then x is
63. Two elements A and B form 0.15 mol of A2B and AB3 type compounds. If both A2B and AB 3 weigh equally, then the atomic weight of A is times of the atomic weight of B.
64. Calculate the total moles of atoms of each element present in 122.5 g of KClO 3.
65. An organic compound containing carbon, hydrogen, and nitrogen have the percentage 40, 13.33, and 46.67, respectively. If its empirical formula is C x H y N z, then x + y + z is .
66. An element X forms two oxides. Formula of the first oxide is XO2. The first oxide contains 50% oxygen. If the second oxide contains 60% oxygen, the atomicity of the second oxide is
67. The ratio of the mass percentages of ‘C and H’ and ‘C and O’ of a saturated acyclic organic compound ‘X’ are 4 : 1 and 3 : 4, respectively. Then, the moles of oxygen gas required for complete combustion of two moles of organic compound ‘X’ is
68. 0.01 mol of a gaseous compound C 2H 2O x was treated with 224 mL of O2 at NTP. After complete combustion, the total volume of gases is 560 mL at NTP. On treatment with KOH solution the volume decreases to 112 mL. The value of x is
69. A sample of CaCO3 has 40% has Ca, 12% C and 40% O2 . The mass of Ca in 5 g of CaCO3 from another source will be g.
70. 1 M HCl and 2 M HCl are mixed in volume ratio of 4 : 1. What is the final molarity of HCl solution?
71. How many blood cells of 5 mL, each having [K +] = 0.1M, should burst into 25 mL of blood plasma having [K+]= 0.02M, so as to give final [K+]= 0.06 M?
72. One litre each of 1M Al 2 (SO 4 ) 3 and 1M BaCl 2 are mixed. What is the molarity of free sulphate ions in the resultant solution?
73. The mole fraction of a solute in a solution is 0.1 at 298 K. Molarity of this solution is the same as its molality. Density of this solution is 2 g/cc. The ratio of molecular weight of solute and solvent w w Msolute Msolvent is .
74. Number of moles of valence electrons present in 6.022 × 1023 NH4+ ion is .
75. Total 10 mol of KClO3 is decomposed by the following two parallel reactions. Calculate the moles of KClO4 produced if 3 mol of O2 is produced. 32
76. Due to partial corrosion of a piece of copper into cuprous sulphide, Cu2S, it gains weight. If the percentage of total copper that has undergone corrosion is 31.75 %, then the percentage gain in weight of piece of copper is ______.
77. Consider the reactions CO+O 2 → CO2; N2 + O2 → NO.
10 mL of the mixture containing CO and N2 required 7 mL oxygen to form CO 2 and NO, on combustion. The volume of N 2 in the mixture will be _____mL.
Passage-based Questions
Q(78-79)
Isotopes are the atoms of the same element; they have same atomic number but different mass numbers. Isotopes have different numbers of neutrons in their nucleus. If an element exists in two isotopes having atomic masses a and b in the ratio m : n, then the average atomic mass will be manb
mn + +
Different isotopes of same element have same position in the periodic table. The elements that have single isotope are called monoisotopic elements. Greater the percentage composition of an isotope, more will be its abundance in nature.
78. The isotopes of chlorine with mass number 35 and 37 exist in the ratio (if its average atomic mass is 35.5)
(1) 1 : 1 (2) 2 : 1
(3) 3 : 1 (4) 3 : 2
79. Atomic mass of boron is 10.81. It has two isotopes, namely 5 B 11 and 5 B x with their relative abundance of 80% and 20% respectively. The value of x is (1) 10.05 (2) 10 (3) 10.01 (4) 10.02
Q(80-81)
Vapour density
Mass of certain volume of gas (22.4L) at STP
Mass of same volume of H gas (22.4L) at STP
Vapour density of a compound is defined on the ratio of mass of a certain volume of gas to the mass of the same volume of hydrogen gas under identical conditions of temperature and pressure. 2
i.e., molecular mass of gas = vapour density × 2
Vapour density is a unitless quantity. It is unaffected by variation of temperature and pressure.
Vapour density may also be defined in terms of density.
Density of gas
Vapour density = Density of hydrogen
Mass of molecules of gas = Mass of molecules of hydrogen gas n n
80. At STP, 5.6 L of a gas weights 60 g. The vapour density of the gas is ______.
81. The average density of the universe on a whole is estimated at 3 × 10 –29 g/mL. If we assume that the entire mass is only H atoms, then what is the average volume (in litres) of space that contains one H-atom? (N = 6 × 1023)
Q(82-83)
A hydrocarbon gas (made up of only C-12 and H-1 isotopes) has 90% C.
82. The number of C-atoms present in the empirical formula of the gas is ____.
83. If empirical formula and molecular formula of the gas are same, then the sum of protons and neutrons per molecule of the gas is x then 10 x is ______.
Q(84-85)
A gaseous hydrocarbon consumed 5 times its volume of oxygen for combustion. The volume of CO 2 produced in the reaction is thrice the volume of hydrocarbon under the same conditions.
84. How much of water is produced by the combustion of 0.1 mol of the given hydrocarbon?
(1) 7.2 g (2) 3.6 g
(3) 14.4 g (4) 1.8 g
85. What is the ratio of molecular weight to empirical formula weight of the hydrocarbon?
(1) 1 (2) 2 (3) 3 (4) 4
Q(86-87)
Empirical formula is the simplest formula of the compound, which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula = (Empirical formula) × n.
86. Two metallic oxides contain 27.6% and 30% oxygen, respectively. If the formula of the first oxide is M3O4, then the formula of the second one will be
(1) MO (2) MO2 (3) M2O5 (4) M2O3
87. Which of the following formulae is the one odd out?
(1) C6H12O6 (2) C12H22O11
(3) HCOH (4) C5H10O5
Q(88-89)
Empirical formula is the simplest formula of the compound that gives the atomic ratio of various elements present in one molecule of the compound. However, the
molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula = (Empirical formula) × n. n = Molecular mass/Empirical formula mass. A compound may have the same empirical and molecular formula. Both these formulae are calculated using percentage composition of constituent elements.
88. Which of the following represents the formula of a substance that contains 50% oxygen?
(1) N2O (2) CO2
(3) NO2 (4) CH3OH
89. An oxide of iodine (at. wt. of iodine = 127) contains 25.4 g of iodine and 8 g of oxygen. Its formula could be
(1) I2O3 (2) I2O (3) I2O5 (4) I2O7
Q(90-91)
2 litre of 9.8% (w/w) H2SO4 (d = 1.5 g/mL) solution is mixed with 3 litres of 1 M KOH solution.
90. The number of moles of H 2SO4 added is (1) 1 (2) 2 (3) 3 (4) 0.5
91. The concentration of H+ if solution is acidic or concentration of OH– if solution is basic in the final solution is (1) 0 (2) 3/10 (3) 3/5 (4) 2/5
Q(92-93)
Avogadro’s law states that, under conditions of constant temperature and pressure, equal volume of gases contain equal number of particles. Experimental investigation shows that, at one atmosphere pressure and a temperature of 273 K, one mole of any gas occupies a volume which is very close to 22.4 L. Therefore, the number of moles in any gas sample can be found by comparing its volume at STP with 22.4 L.
92. At STP, 40 L of CO2 contains ___ mole.
93. Number of gram atoms of oxygen present in 0.3 gram moles of H2C2O4.2H2O is ___.
Q(94-96)
When we assume that one mole of a substance contains the same number of elementary entities as one mole of any other substance, we don’t actually need to know what that number is. Sometimes, however, we will need to work with the actual number of elementary entities in a mole of a substance. This number is called Avogadro’s number. N A = 6.022137 × 10 23 molecules. The unit mol–1, which we read as per mole, signifies that a collection of NA molecular level entities is equivalent to one mole at the macroscopic level. For example, a mole of carbon contains 6.02 × 1023 atoms of C. A mole of oxygen gas contains 6.02 × 1023 molecules of O2.
94. The number of atoms present in 8 g of ozone is
(1) NA (2) 3NA (3) NA/6 (4) NA/2
95. Which of the following is a reasonable value for the number of atoms in 1.00 g of helium?
(1) 0.25 (2) 4.0 (3) 4.1 × 10–23 (4) 1.5 × 1023
96. The dot at the end of a sentence has a mass of about one microgram. Assuming that the black stuff is carbon, the number of carbon atoms present in it is
(1) 5 × 1016 (2) 5 × 1017 (3) 8 × 1023 (4) 12 × 1023
Q(97-98)
Chlorine is a very strong oxidising agent and can oxidise several metals, non-metals, and compounds.
(at. wt. Fe = 56, Cl = 35.5, S = 32, H = 1)
97. A compound x is formed by adding 3.2 g of sulphur to sodium suphite in alkaline medium. What is the weight of chlorine required to oxidise the compound x?
98. What is the weight of the product formed when 3.55 g of chlorine reacts with iron?
Q(99-101)
In a reaction vessel, 100 g H2 and 100 g Cl2 are mixed and suitable conditions are provided for the reaction:
H2(g) +Cl2(g)→ 2HCl(g)
99. The amount of HCl formed in this reaction (at 100% yield) will be
(1) 102.8 g (2) 73 g
(3) 36.5 g (4) 142 g
100. The amount of excess reactant remaining is
(1) 50 g (2) 97.2 g
(3) 46 g (4) 64 g
101. The amount of HCl formed (at 90% yield) will be
(1) 36.8 g (2) 62.5 g
(3) 80 g (4) 92.53 g
Q(102-103)
32 2 (50% yield )
NH+ONO+HO → 22 (60% yield )
NO+HOHNO+NO →
NO+ONO → 223 (80%yield)
102. How many litres of NH3 at STP is required to produce 2.25 g of HNO3? (Round-off to nearest integer)
103. What volume of oxygen is required to oxidise 20 litres of ammonia (at STP) oxidised to HNO 3 , if yield is 100 %? (Round-off to nearest integer)
Q(104-106)
10 mol of SO2 and 4 mol of O2 are mixed in a closed vessel of volume 2 litres. The mixture is heated in the presence of Pt catalyst. Following reaction takes place:
2SO2(g) + O2(g) → 2SO3(g)
Assume that the reaction proceeds to completion.
104. Select the correct statement.
(1) SO2 is the limiting reagent. (2) O2 is the limiting reagent.
(3) Both SO2 and O2 are limiting. (4) This cannot be predicted.
105. Number of moles of SO 3 formed in the reaction will be (1) 10 (2) 4 (3) 8 (4) 14
106. Number of moles of excess reactant remaining is (1) 4 (2) 2 (3) 3 (4) 8
Matrix Matching Questions
107. Match Column I (molecule) with columnII(formula).
Column I Column II
A. NH3 I) EF = MF
B. N2H4 II) MF = (EF)2
C. N3H III) Maximum percentage of nitrogen by mass
D. C2N2 IV) Least percentage of nitrogen by mass
(A) (B) (C) (D)
(1) I II III IV
(2) II I III IV
(3) III I IV II
(4) IV III II I
108. Match the Column I (molecules) with column- II (composition).
Column I Column II
A. N2 I) 40% carbon by mass
B. CO II) empirical formula CH2O
C. C6H12O6 III) vapour density : 14
D. CH3COOH IV) 14 NA electrons in a mole
(A) (B) (C) (D)
(1) III, IV III, IV I, II I, II
(2) IV, III II, I I, III IV
(3) I II III IV
(4) II, I III IV,I IV
109. Match the Column I (concentration term) with Column II (Units)
Column I Column II
A. Molarity I) mol
B. Mole fraction II) Unitless
C. Mole III) mol L–1
D. Molality IV) mol Kg–1
V) Temperature dependent
(A) (B) (C) (D)
(1) III II IV I
(2) III I II IV
(3) III, V II I IV
(4) II III I IV
110. Match Column I (solution) with Column II (amount of solute present in the solution).
Column I Column II
A. 0.5 L of 1M H2SO4 I) 1 equivalent of solute
B. 6% w/v aqueous solution of urea II) 1 mol of solute
C. 100 mL of 0.2N aqueous Na2CO3 solution III) 1.06 g of solute
D. 518 g of 0.2 m aqueous solution of glucose IV) 0.1 mol of solute
V) 180 g of solute
(A) (B) (C) (D)
(1) I II III IV
(2) II I III IV
(3) I II III V
(4) I III II IV
111. Match Column I (solution) with Column II (concentration).
Column I
A. 100 mL of solution containing 0.98 g of H2SO4
B. 0.5 L solution containing 50 milli equivalents of H3PO4
C. 109 g of solution containing 9 g of glucose
Column II
I) Decinormal
II) Semimolal
III) Decimolar
IV) Semimolar
(A) (B) (C)
(1) III I IV
(2) I II III
(3) II I III
(4) III I II
112. When 200 mL 0.2 M NaCl and 200 mL
0.4 M BaCl 2 and 100 mL 0.2 M KCl are mixed, match Column I (the ion) with Column II (molarity)
Column I
Column II
(A) molarity of Na+ ion I) 0.04 M
(B) molarity of Ba+2 ion II) 0.08 M
(C) molarity of Cl– ion III) 0.16 M
(D) molarity of K+ ion IV) 0.44 M
(A) (B) (C) (D)
(1) II III IV I
(2) I II III IV
(3) II I III IV
(4) V II III I
113. Match Column I (equation)with Column II (equivalent weight).
Column I
Column II
A. BrO3– → Br- I) 3 M
B. P → PH 3 II) 6 M
C. S–2 → SO4-2 III) 8 M
D. Cr2O7–2 → Cr+3 IV) 5 M
(A) (B) (C) (D)
(1) II I III II
(2) IV I III II
(3) II I III IV
(4) IV I III IV
114. Match Column I (mole) with Column II (quantity)
Column I Column II
A. 0.5 mol of SO2(g) I) Occupy 11.2 L at 1 atm and 273 K
B. 1 g of H2(g) II) Weighs 24 g
C. 0.5 mol of O3(g) III) Total number of atoms =1.5×NA
D. 1 g molecule of O2(g) IV) Weigh 32 g
(A) (B) (C) (D)
(1) I, II, III, IV I, II I, II, III, IV IV
(2) III, IV I I, III IV
(3) I, III I, II I, II, III I, IV
(4) I, III, IV I I, II, III IV
115. Match Column I (compound) with Column II (equivalent quantity)
Column I Column II
A. One gram molecule of oxygen gas I) One mole of H2
B. Gram molar volume of H2 II) One mole of O2
C. 44 g of CO2 III) 22.4 L at STP
D. 18 g water IV) 3 NA atoms
(A) (B) (C) (D)
(1) II, III I, III II, III, IV I, IV
(2) II I III IV
(3) I, II II, III III, IV IV, II
(4) I, II, IV III, IV, II II, I, IV II, IV
116. Match Column I (compound) with Column II (percent carbon).
Column I Column II
A. CH4 I) 90% of carbon
B. C2H6 II) 75% of carbon
C. C2H4 III) 80% of carbon
D. C3H4 IV) 85.7% of carbon V) 60% of carbon
(A) (B) (C) (D)
(1) V I II IV
(2) II III IV I
(3) III II IV I
(4) II I V III
117. Match Column I (reaction) with Column II (limiting reagent)
Column I
A.
Column II
223 0.1 mol0.1 mol
N+3H2NH → (I) 4.16 × 10 –2 mol
B. 2 22 1 g 1 g H+2CCH → (II) 8.33 × 10 –2 mol
C. () 22 1 g 22.4L at NTP C+OCO → (III) 6.25 × 10 –2 mol
D. 222 1 g1 g 2H+O2HO → (IV) 6.67 × 10 –2 mol
(A) (B) (C) (D)
(1) II III IV I
(2) I II III IV
(3) III I IV II
(4) IV I II III
118. Match Column I (reaction) with Column II (limiting reagent).
Column I Column II
A. 223 0.2 mole0.7 mole N+3H2NH → (I) H2
B. 2 22 24 g 1 g H+2CCH → (II) C
C. 2 0.5 g 2.24 lit C+OCO → (III) O2
D. 222 2 g2 g 2H+O2HO → (IV) N2
(A) (B) (C) (D)
(1) IV I II III
(2) II III IV III
(3) III I II IV
(4) I II IV III
119. Match Column I with Column II. Column I Column II
A. 222 3 g22.68 g 2H + O2HO → (I) 25.5 g product is formed
B. 223 24.5g5.5g N + 3H2NH → (II) 0.25 g of a reactant is left
C. 22 1.4g43g H + Cl2HCl → (III) H2 is the limiting reagent
D. 24 20g 6.375g C + 2HCH → (IV) 41.12 g product
(A) (B) (C) (D)
(1) I II IV III
(2) II I IV III
(3) I IV II III
(4) I III IV II
BRAIN TEASERS
Single Correct MCQs
1. One gram atom of element ‘X’ has 0.444 times the mass of one gram atom of element ‘Y’. Atomic mass of ‘X’ is 2.96 times the atomic mass of C12. The atomic weight of ‘Y’ is
(1) 80 (2) 31.4
(3) 46.67 (4) 40
2. The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. One molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of the compound C XHYOZ is
(1) C2H4O3
(2) C3H6O3
(3) C2H4O
(4) C3H4O2
3. To 1 L of 1.0 M impure H2SO4 sample, 1.0 M NaOH solution was added and a plot was obtained, as follows:
FLASHBACK
( Previous JEE Questions )
JEE Main
1. Mass of methane required to produce 22 g of CO2 after complete combustion is ______g.
(C = 12.0, H = 1.0, O = 16.0) (2024)
2. 1 mole of PbS is oxidised by “X” moles of O3 to get “Y” moles of O2. Then, the value X + Y = __________. (2024)
3. Number of moles of methane required to produce 22 g of CO2(g), after combustion is x × 10–2 moles. The value of x is (2024)
The percentage purity of H2SO4 and the slope of the curve, respectively, are
(1) 75%, –1/2
(2) 75%, –1
(3) 50%, –1/3
(4) 50%, –1/4
4. 76 g of a silver salt of dibasic acid, on heating, left a residue of 54 g silver. Silver salt contains Ag, C and O only and C and O in mole ratio of 1 : 2. Then find the mass of CO2 gas liberated during ignition of 76 g silver salt. (Ag = 108)
(1) 2.2 g (2) 0.22 g
(3) 44 g (4) 22 g
4. A sample of CaCO3 and MgCO3. Weighing 2.21 g is ignited to a constant weight of 1.152 g. The composition of the mixture is: (Given molar mass in g mol–1 CaCO3.100, MgCO3:84) (2024)
(1) 1.187 g CaCO3 + 1.023 g MgCO3
(2) 1.023 g CaCO3 + 1.023 g MgCO3
(3) 1.187 g CaCO3 + 1.187 g MgCO3
(4) 1.023 g CaCO3 + 1.187 g MgCO3
5. A protein ‘A’ contains 0.30% of glycine (molecular weight 75). The minimum molar
mass of the protein ‘A’ is ______________ ×103 g mol–1[nearest integer]. (2022)
6. CNG is an important transportation fuel. When 100 g CNG is mixed with 208 g oxygen in vehicles, it leads to the formation of CO2 and H2O and produces large quantity of heat during this combustion. Then, the amount of carbon dioxide produced (in grams) is _______ (nearest integer). Assume CNG to be methane. (2022)
7. The moles of methane required to produce 81 g of water after complete combustion is ____ × 10–2 mol. (Nearest integer)
(2022)
8. 4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is x × 10 –1 The value of x is ____. (Rounded off to the nearest integer)
(2021)
CHAPTER TEST – JEE MAIN
Section A
1. Which of the following reactions does not obey the law of conservation of mass?
(A) C2H4(g)+O2(g) → CO2(g) + H2O(g)
(B) KHCO3 + 2HCl → KCl + H2O + CO2(g)
(C) BaCl2(aq)+H2SO4(aq) → B a SO4(ppt)+2HCl(aq)
(D) PbO(s)+C(s) → Pb(s) + CO 2(g)
(1) A, B, and C
(2) B, C, and D
(3) A, C, and D
(4) A, B, and D
2. The number of millimoles of solute present in 0.5 L of 0.02 M aqueous solution is (1) 0.01 (2) 10
(3) 100 (4) 1
9. Complete combustion of 750 g of an organic compound provides 420 g of CO2 and 210 g of H2O. Then, percentage composition of carbon and hydrogen in organic compound is 15.3 and ______ respectively. (Round off to nearest integer) (2021)
10. A 6.50 molal solution of KOH (aq.) has a density of 1.89g cm–3. The molarity of the solution is_____ mol dm–3. (Round off to the nearest Integer). [Atomic masses: K: 39.0 u; O : 16.0 u; H : 1.0 u] (2021)
11. 10.30 mg of O 2 is dissolved in a litre of sea water of density 1.03 g/mL. The concentration of O2 in ppm is ___________.
(2020)
3. The density of an aqueous 49% H 2 SO 4 solution is 1.02 g/cm3. Identify the correct relation between molarity (M) and molality (m) for the same solution at 25 °C.
(1) M = m
(2) M > m
(3) M < m
(4) M = 2 m
4. While determining the empirical formula for a given substance, the percentage by mass of each element is divided by its atomic mass. This ratio gives
(1) relative number of moles
(2) simplest molar ratio
(3) simplest whole number ratio
(4) simplest whole number molar ratio
5. 1 g of CaCO 3 is dissolved in 99 g of H 2O. The mass percentage of CaCO 3 in solution is
(1) 10 (2) 0.1
(3) 100 (4) 1
6. The number of hydrogen atoms present in 22 u of C3 H 8 is (1) 8 (2) 4
(3) 2 (4) 1
7. The weight of a mixture of 2 ×1023 molecules of A and 3 ×1023 molecules of B is 130 g. If the molecular mass of B is 180, then the molecular mass of A is (Avogadro number = 6×1023)
(1) 120 u (2) 60 u
(3) 90 u (4) 180 u
8. Match Column I (solution) with Column II (concentration).
Column I
A. 200 mL of aqueous solution containing 0.02 mol of solute.
Column II
I) Mole fraction of solute = 0.1
B. 222 g of aqueous solution containing 60 g of urea. II) Seminormal solution
C. 20 mL of aqueous solution containing 0.5 g of CaCO3 III) Mole fraction of solute = 0.9
D. 1.18 kg aqueous solution containing 180 g of glucose IV) Decimolar V) 1 molal solution
(A) (B) (C) (D)
(1) IV I II V
(2) III I II IV
(3) IV II I III (4) IV III II I
9. The cost of 9 g of glucose is Rs 0.9. What is the cost of 9 mol of glucose?
(1) Rs 1620 (2) Rs 162
(3) Rs 16.2 (4) Rs 1.62
10. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
(A) : Both 100 g of calcium carbonate and 44 g of carbon dioxide contain the same number of carbon atoms.
(R) : Both contain 1 g-atom of carbon, which contains 6.023 ×10 23 carbon atoms.
In light of the given statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A),
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) (A) is true but (R) is false,
(4) (A) is false but (R) is true.
11. Given below are two statements
S–I : A pair of compounds, C2H5OH and CH3OCH3, does not obey the law of constant proportions.
S–II : The combining ratio of C : H : O in these compounds by mass is 12 : 3 : 8.
In light of the given statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
12. A compound is made up of two elements, ‘X’ and ‘Y’. The percent by mass composition of these two elements in the compound is 60% ‘X’ and 36% ‘Y’. The relative number of moles of these elements in the compound is 1.25 and 1.8 respectively. Atomic masses of ‘X’ and ‘Y’, respectively, are (1) 48 and 20 (2) 20 and 48 (3) 2 and 48 (4) 48 and 2
13. A mixture contains 1 mol each of calcium nitride and sodium phosphide. The mixture, on complete hydrolysis, liberates gases with the molar ratio of
(1) 1 : 1 (2) 2 : 1
(3) 3 : 2 (4) 1 : 3
14. 100 cm3 of acetylene and 100 cm3 of oxygen are mixed and subjected to combustion in a closed vessel under identical conditions. Identify the correct statement from the following.
(1) 40 mL of C2H2 is a limiting reagent.
(2) 120 mL of CO2 is formed.
(3) 60 mL of C2H2 is burnt in oxygen.
(4) 80 mL of CO2 is formed.
15. Given below are two statements.
S–I : Mass of 6.02 ×1022 molecules of CO2 is heavier than the mass of 6.02 ×1022 molecules of H2O.
S–II : 1 mole of any covalent compound contains 6.02 ×1023 molecules.
In light of the given statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
16. 8 mol of N2 and 5 moles of O2 are allowed to react in a vessel according to the reaction 2N2 + 5O2 → 2N2O5. The reaction proceeds until 6 moles of N2 are left unreacted. Then, the number of moles of product formed is
(1) 1 (2) 2
(3) 6 (4) 5
17. Consider the following statements about a compound:
(A) A molecule of a compound has atoms of different elements.
(B) A compound can’t be separated into its constituent elements by physical methods of separation.
(C) A compound retains the physical properties of its constituent elements.
(D) The ratio of atoms of different elements in a compound is fixed. The correct statements are
(1) A and B only
(2) C only
(3) A, B, and C only
(4) A, B, and D only
18. Given below are two statements.
S–I : 2 mol of water and 3 mol of CO 2 are produced when 30 g of C 2 H 6 completely burns in oxygen.
S–II : 30 g of ethane requires 112 g of pure oxygen for complete combustion.
In light of the given statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
19. Which of the following is the semi-normal solution?
(1) 0.5 M H2SO4 (2) 2.5 M H2SO4 (3) 0.25 M H2SO4 (4) 0.05 M HCl
20. The mass of one molecule of a compound (containing C, H, and O) is 15 times heavier than 1 12 th mass of the C-12 isotope atom. E mpirical formula of the compound is CH2O. Number of carbon atoms present in one molecule of the compound is (1) 2 (2) 6 (3) 4 (4) 8
Section B
21. Caffeine contains 28.9% of nitrogen. The molecular weight of caffeine is 194. The number of nitrogen atoms present in one molecule of caffeine is _____.
CHAPTER TEST – JEE ADVANCED
2020 P1 Model
Section-A
[Single Option Correct MCQs]
1. In compound A, 1.00 g nitrogen combines with 0.57 g oxygen. In compound B, 2.00 g nitrogen combines with 2.24 g oxygen. In compound C, 3.00 g nitrogen combines with 5.11 g oxygen. These results obey which of the following laws?
(1) Law of constant proportions
(2) Law of multiple proportions
(3) Law of reciprocal proportions
(4) Dalton’s law of partial pressure
2. One mole of an element contains 4.8×10 24 electrons. If the elements exists as diatomic, then choose the correct option about the element.
(1) Atomic number = 16
(2) Gram atomic weight = 8 g
(3) Gram molecular weight = 16 g
(4) Gram molecular weight = 32 g
22. 200 mL of pure oxygen is subjected to electric discharge. 15% of oxygen is converted into ozone. The volume of ozonised oxygen is 10(10+x). Find the value of x.
23. A mixture of Na2CO3 and NaHCO3, having a total weight of 100 g on heating, produced 11.2 L of CO2 under STP conditions. The percentage of Na2CO3 in the mixture is (x).
Find the value of 8 x
24. Aspirin has the formula C 9 H 8 O 4 . How many atoms of oxygen are there in a tablet weighing 360 mg is x × 10 21. Then 4.816 x value is_____ (NA = 6.02 × 1023).
25. The measured density at NTP of He is 0.1784 g/L what is the weight (in g) of one mole of He?
3. 8 mol of SO2 and 14 mol of O2 were passed over the catalyst to produce 6 mol of SO3. The ratio of SO2 and O2 moles in the mixture is
(1) 11 : 2 (2) 2 : 11
(3) 2 : 6 (4) 11 : 6
4. On heating 2.44 g of hydrated barium chloride to dryness, 2.08 g of anhydrous salt remained. Formula of hydrated salt is (1) BaCl2.3H2O (2) BaCl2.2H2O (3) BaCl2. H2O (4) BaCl2.6H2O
Section-B
[Multiple Correct Option MCQs]
5. To 50 L of 0.2 N NaOH, 5 L of 1 N HCl and 15 L of 0.1 N FeCl3 solutions are added and ignited. Identify the correct statement from the following.
(1) Mass of Fe2O3 precipitated = 40 g
(2) Normality of NaOH left unreacted = 0.05
(3) Equivalents of NaOH reacting with HCl = 5
(4) Equivalents of NaOH used for FeCl3 = 1.5
6. 124 g of P4 contains
(1) 4NA atoms of phosphorus
(2) 4NA molecules of phosphorus
(3) NA molecules of phosphorus
(4) 4 atoms of phosphorus
7. For the reaction aA + bB→ cC+ dD, initially ‘X’ moles of reactant ‘A’ and ‘Y’ moles of reactant ‘B’ are taken. Identify the correct statement(s) from the following.
(1) Yb X > – reactant ‘A’ is a limiting reagent.
(2) Yb X < – reactant ‘B’ is a limiting reagent.
(3) X a Y < – reactant ‘B’ is a limiting reagent.
(4) X a Y < – reactant ‘A’ is a limiting reagent.
Section-C
[Numerical Value Quations]
8. A regular cube of metal measures 10 cm on the edge and has a density of 0.84 g/cc. If the cube contains 1.68×1025 atoms of the metal, the atomic mass of the metal is _______u.
(NA = 6 ×1023)
9. An organic compound contains 49% C, 2.7% H, and 48.2%Cl by mass, and its molecular weight is 147 u. Atomicity of the compound is _____.
10. 105 mL of pure water at 4 °C is saturated with ammonia, producing 40% mass of ammonia. The total weight of the solution after saturation is _________ g.
11. The mass of one molecule of urea is m × 10–23 g. Value of m is ______ (NA= 6 × 1023).
12. A 100 mL sample of brackish water was made of ammonical and the sulphide ion in solution was titrated against 15 mL of 0.02 M AgNO 3 solution. The concentration of H2S in the water is ______ ppm.
13. 320 mL of 0.4 M MnSO 4 is completely oxidised by 16 mL of KMnO4 of unknown normality, each forming a Mn+4 oxidation state. The normality of KMnO 4 solution is ______.
Section-D
[Passage-based Questions]
Q(14-15)
Chlorine is a very strong oxidising agent and can oxidise several metals, non-metals, and compounds.
(At. wt. Fe = 56, Cl = 35.5, S = 32, H = 1)
14. A compound x is formed by adding 3.2 g of sulphur to sodium suphite in alkaline medium. What is the weight of chlorine required to oxidise the compound x?
15. What is the weight of the product formed when 3.55 g of chlorine reacts with iron?
Q(16-17)
Into hot concentrated solution of KOH containing 33.6 g of KOH, chlorine gas is passed till all the KOH is completely reacted. Evaporation of the solution gives a solid.
16. The solid is dissolved in water and treated with AgNO 3 . What is the weight of precipitate formed?
17. The solid is heated in the presence of MnO2 What is the weight of the gas liberated ?
ANSWER KEY
JEE Main
(51) 2 (52) 3 (53) 2 (54) 3 (55) 1 (56) 1 (57) 4 (58) 3 (59) 1 (60) 4 (61) 1 (62) 4 (63) 2 (64) 3 (65) 1 (66) 1 (67) 2 (68) 4 (69) 4 (70) 2 (71) 4 (72) 100 (73) 12 (74) 4 (75) 1
Level-III (1) 1 (2) 2 (3) 3 (4) 1 (5) 2 (6) 1 (7) 3 (8) 4 (9) 4 (10) 1 (11) 4 (12) 3 (13) 2 (14) 2 (15) 4 (16) 2 (17) 4 (18) 2 (19) 3 (20) 2 (21) 1 (22) 4 (23) 4 (24) 2 (25) 3 (26) 1 (27) 4 (28) 2 (29) 2 (30) 1 (31) 1362 (32) 34 (33) 60 (34) 160 (35) 292 (36) 2
Theory-based Questions (1) 4
JEE Advanced Level
Brain Teasers
Chapter Test JEE-Mains
Chapter Test JEE-Advanced
