Homework Packet Answers – Unit 5 1) C

15) C

2) A

16) A

3) A or B

17) A

4) B

18) B

5) A

19) B

6) B

20) A. At 537 torr, 25 °C and 0.205 L, there must be 5.92 x 10-3 moles of diethyl ether. With a density of 0.708 g/mL, 10 mL of liquid diethyl ether is 9.55 x 10-2 moles. Therefore most of the diethyl ether must be in liquid form.

7) D 8) A 9) B 10) A. F2 B. ClO2 C. HCl D. C3H8 E. PH3 F.PH3 11) A. Water: H-bonding B. Propanone: Dipole-dipole C. Pentane: Dispersion 12) Both enthalpy of vaporization and boiling point are in the order of water, propanone, pentane. This demonstrates that H-bonding in water is the strongest intermolecular force and that the dispersion forces in pentane are the weakest. 13) Pentane and water would not mix. The forces of attraction between pentane and water molecules are not strong enough to break the intermolecular forces between water molecules themselves. 14) Enthalpy of mixing is positive when more energy is required to pull water molecules apart than is gained as a result of the water’s attraction to propanone. There is no way to tell from the information given which force will predominate.

B. 9.55 x 10-2 moles of diethyl ether at 25 °C and 537 torr takes up a volume of 3.31 L if it is all vapor with no liquid. C. The pressure of 9.55 x 10-2 moles of diethyl ether vapor at 25 °C in a 5.00 L container will be 355 torr. 21) First solve the Clausius–Clapeyron equation for C. Use the known pressure and temperature of water at its boiling point (1 atm, 100 °C) with the known enthalpy of vaporization (40.67 kJ/mol) to find the value for C of 13.11. Use this value in the Clausius–Clapeyron equation at 81 °C to find the pressure of 0.492 atm. 22) A 23) A 24) D 25) C 26) D

27) As solvent temperature increases, dissolved gas molecules are squeezed out by the energetic solvent molecules and leave the solvent for the vapor. The increase kinetic energy of the gas molecules causes more of the gas molecules to escape the solvent for the vapor phase. 28) Cooler water contains more oxygen. 29) 46,300 ppm. Assume the 100 kg is mass of dead sea solution, not pure water. 30) 14.6 M, 57.8 m, X = 0.510 31) D 32) C 33) D 34) B 35) E 36) C 37) D 38) D 39) E 40) C 41) B 42) 200 g/mol 43) 0.16 M assuming i = 2.0, which it doesn’t. i is probably closer to 1.8 so the concentration is probably closer to 0.19 or 0.20 M. For a nonelectrolyte, the concentration would be 0.314 M. 44) Concentration of FeCl3 = 2.19 M. If FeCl3 is totally dissociated, i could be as high as 4. If so, use 2.19 M x 4 = 8.78 M. π = 214 atm. 45) 27.4 atm

46) Using Kf = 1.86 °C/m, m = 2.88 m. Total moles of ions is 0.144 moles. Assuming i = 2, there are 0.0719 moles of ionic compound. Molar mass is 74.5 g/mol, which is KCl. 47) Calculate the mole fraction of I2. This is the same as the decrease in the mole fraction of CHCl3 that produces a change of 150 mm Hg in the vapor pressure of CHCl3 (0.147). Subtract from 1 to find the mole fraction of CHCl3 (0.853). Multiply moles of CHCl3 by the ratio of the mole fractions of I2 to CHCl3 to find the moles of I2. Convert to grams of I2 (693 g). 48) Use ΔT and the freezing point depression constant for cyclohexane to calculate the molality of the solution (0.322 m). Use mass of solvent calculated from the volume and density and the molality to find moles of solute (0.0188 mol). Use moles of solute and grams of solute to find molar mass of solute (168 g/mol). Use percent composition to find empirical formula (C3H2NO2). Compare molecular mass to empirical mass to find multiplier of 2. Formula is C6H4N2O4. 49) A. C3H2Cl B. 147.3 g/mol C. C6H4Cl2 D. 0.937 E. 140 mm Hg

Homework Packet 5 Answers

AP Chem stuffs