g gg11,,mm g gg  mol mol  58 mL mLOIML   46 46.068 18 .V 015 a of ..37 ;; V V V  0.8041  46..068 068 18 18 ..015 015 a37 100 gFor For 95.02% ABW, the density was found to be g/mL according toggmL Table V58 the 22,,m m mol mol  18 mL mL 95.02% ABW, the density was found to be 0.8041 g/mL according to Table V of the OIML mol mol mol mol mol mol  58.37of ethanol ;  . 05 V 124 V mL 1,m  real volume is: and water  .363 mL . gThe masses 0..7893 58 58..37 37 ;; V  V V11,,mm 2,m V22,,mm  0 7893 00.1 mol mol g g mol mol 100 g g gg gg mL mL 0.7893 0.9982 100 0.8041 0 0 . . 7893 7893 0 0 . . 9982 9982 tables. Thus, the real volume is: . The masses of ethanol and water  124 363 mL mL Thus, the realmL tables. volume is: The masses of ethanol and water mL  124.363 mL mL mL. mL mL g g 0 . 8041 Equation Equation 77 isis valid valid only only for for mixtures mixtures of of an an ideal ideal ss 0.8041mL spectively. The number of moles of ethanol and water are: mL Equation 7 is valid only for mixturesEquation of an ideal such asmixtures that mix of benzene Equation 77 valid only for for mixtures of of an aideal such asand Equation 7isis issolution valid valid only only for mixtures of of an ansolution ideal ideal solution solution such such toluene. toluene. But But when when the the components components interact interact suc sucaa .and 98 g areare 95495 5 g respectively. The number of moles of ethanol and water are: and 5 g respectively. The number of moles of ethanol and water are: that of a mix of benzene and toluene. But when the components toluene. But when the components interact such as for ethanol and water, then the molar of ethanol  2.0626 mol ; n2   0.2764 mol ; mole fractiontoluene. toluene. But Butinvolumes when when the theneed components components interact interact such as as for for ethan ethan volumes need to be be replaced replaced by bysuch the the partial partial mola mola g .95 95 02.02 4partial .98 g gneed to be replaced by the interact, g such as for ethanol andto water, then the molar volumes 4 . 98 g 18 . 015 volumes molar volumes (Equations 5a, 5b). ; mole fraction of ethanol in 2 . 0626 ; n   mol n   0 . 2764 mol volumes volumes need need to to be be replaced replaced by by the the partial partial molar molar volumes volumes (Eq (Eq 1 n1  ; mole of ethanol5a,in l mol  2.0626 mol ; 2 n2  need gto be replaced  0.2764 by themol partial molarfraction volumes (Equations g g   g 46 . 068 18 . 015 18 or.068 x1=0.8818. both equations 3.015 and 4,mol weHere is calculated as x1=nof  is calculated 1/(n 1+n2),46 mol mol 5b). mole fraction ethanol in this composition as ,V22 (Eq. (Eq. 5b) 5b) to to represent represent partial partial m m Here we use use VV11,V   we mol  Using V ,V (Eq. 5b) to represent partial molar volumes which are composition Here we use 1 be 2 58.145 Here we use , (Eq. 5b) to represent partial molar volumes V V ,V ,V (Eq. (Eq. 5b) 5b) to to represent represent partial partial molar molar volumes volumes Here Here we we use use artial molar ethanol and water to and 14.817 mL/mol =n /(n +n ), or x =0.8818. Using both equations 3 and 4, we this composition is calculated as x x1=n1volumes /(n1+n2), orfor x1=0.8818. Using both equations 3 and 4, we 1 12+n2), or 1 x1=0.8818. 11 22  Using both equations 3 and 4, we this composition is calculated as1 x11=n1/(n  which are composition dependent while (Eq.7) 7) V dependent dependent while while using using 7) to to repres repres V11,,mm,,,VV22,,mm to using (Eq. (Eq. molar volumes for ethanol and water tomolar be ethanol e volumedetermine for 100 the g ofpartial 95.02% ABW calculated using partial volumes is: determine the partial molar volumes for and water to be 58.145 and 14.817 mL/mol determine the partial volumes ethanol andwhile water towhich be 58.145 and 14.817 mL/molmolar dependent while molar using (Eq. 7) V1for to represent molar volumes which are composition , Vdependent m m , 2 , to to represent represent molar vol vol dependent while using using (Eq. (Eq. 7) 7) V V , , V V represent molar volumes are composition independent. 58.145 and 14.817 mL/mol, respectively. The volume for 100 g of m m 11,,m 22,,m respectively. TheThe volume forfor 100100 g ofg of 95.02% ABW calculated using partial molar volumes is: is: independent. independent. respectively. volume 95.02% ABW calculated using partial molar volumes 95.02 % ABW calculated using partial molar volumes is: independent. Example 2: 100 mL of 95 % (w/w) ABW Ethanol independent. Equation 5a independent. 4.025 mL     Example 2: 2: ABW 100 100 mL mL0.8041 of of 95% 95% (w/w) (w/w) ABW ABWtoEthanol Ethanol TheEthanol density ofExample 95 % (w/w) is g/mL according Equation 5a n1 V .025 mL 100 mL Equation 5a n11 V1nExample 2 V n22 V2 124  2: 124 .025 mLof 95% (w/w) ABW Example Example 2: 2: 100 100 mL mL of of 95% 95% (w/w) (w/w) ABW ABW Ethanol Ethanol of the OIML tables. The total mass is 80.41 g; the masses en this numerical value and the real volume, 124.363 mL is 0.27 %.Table TheVaerrors The The density density of of 95% 95% (w/w) ABW ABW isis 0.8041 0.8041 g/mL g/mL ac a EQUATION 5a a (w/w) of ethanol andaccording water are 76.39 g and V 4.02 g, respectively. NumberThe of the OIML tables. The density this ofequations. 95% (w/w) ABW isand 0.8041 g/mL to Table rrors inherited in the use of the regression TheThe error between numerical value the real volume, 124.363 mL is 0.27 %. The errors The The density density of of 95% 95% (w/w) (w/w) ABW ABW is is 0.8041 0.8041 g/mL g/mL according according to to Ta Ta error between this numerical value and the real volume, 124.363 mL=1.66 is the 0.27 %.and Thenof errors total totaland mass mass isis 80.41 80.41 g; g; the masses masses of=0.223 ethanol ethanol and and of moles of ethanol water are: mole The error between this total numerical value and the real masses volume, of ethanol 1 2 mass is 80.41 g; the and water are 76.39 gnand 4.02 g respectively. come from the errors inherited in the use of the regression equations. total total mass mass is is 80.41 80.41 g; g; the the masses masses of of ethanol ethanol and and water water are are 76. errors inherited intwo the usemole. of the regression equations. =1 Number of ofethanol moles moles of ofxethanol ethanol and and water waterthe are: are: nn76. The moleNumber fraction of is = 0.8814 . Using 11=1 124.363to mLrepresent is 0.27come %. the Thefrom errors come fromafter the errors inherited be generalized totalthe volume mixing but 1 Number of moles of ethanol and miscible water are: n =1.66 mole and n =0.223 mole. The mole fraction 1 2 Number Number of of moles moles of of ethanol ethanol and and water water are: are: n n =1.66 =1.66 mole mole and and 1 1 regression equations (Equations 3 and 4), partial molar volumes in the use of the regression equations. of of ethanol ethanol is is xx11 == 0.8814. 0.8814. Using Usingbut the the regression regression eq e s. We can re-write Equation 5a as:5a Equation cancan beisbe generalized to to represent thethe total volume after mixing two miscible Equation represent total volume mixing but = 0.8814. regression equations (Equations 3two &regression 4),miscible partialequations molar of5a ethanol x1generalized forethanol ethanol and are: after Equation 5a can be generalized to represent the total Using volumethe of = 0.8814. 0.8814. Using Using the the regression equations (Equ (Equ of ethanol is is xxwater 11 =  interacting liquids. We cancan re-write Equation 5a 5a as:as: volumes m m interacting liquids. We volumesfor for ethanol ethanol and water are: are: V  V11  58 58..143 143  water   after mixing two miscible but interacting liquids. We re-write can rewriteEquation mL mLand volumes for ethanol and water are: respectively .   V 58 . 143 , V 14 . 82 mL mL respectively. 1 2 water  V2 volumes for for ethanol ethanol and water are: are:mol V V11  58 58..143 143 ,,V V 1 Equation 5b volumes Equation 5a as: mol and     mol mol 22 Vtotal Equation Using Using Equation Equation 5b 5b in in5b which which partial molar molar volume volume Using Equation 5b in which partial molar volumes are considered, V  n1 nV1 1V1n2 n V2 2 V2 Equation 5b partial Equation in whichofpartial molar volumes are considered, we have V =99.72 mL, an asses (95 g for ethanol andtotal 5 gUsing for water) and5b densities pure ethanol and total Using Using Equation Equation 5b 5b in in which which partial partial molar molar volumes volumes are are consider conside we have Vtotal=99.72 mLof , an error of If 0.28 %. If Equation 7 is used, error error of 0.28%. 0.28%. If Equation Equation 77 isis used used which which ignor ignor EQUATION 5b error ofmasses 0.28%. If(95 Equation 7 is used ignores the composition dependence, we have 3 g/mL for ethanol andIf ρ=0.9982 g/mL for water) mixture, the we only use (95 g in for ethanol andand 5which g5calculated for water) densities of pure ethanol and which ignores the and composition dependence, we have ignores Vtotal =100.82 error error of 0.28%. 0.28%. If If Equation Equation 7 7 is is used used which which ignores the the compo compo If we only use masses g this for ethanol gof for water) and densities of pure ethanol and VVtotal =100.82 mL, mL, an an error error of of 0.82 0.82 %. %. total=100.82 If we only usewater masses(ρV(95.02 g forg/mL ethanol and 4.98 g0.82 for ρ=0.9982 mL, anfor error ofand , an error of for 0.82 %. xture becomes =(ρ0.7893 for ethanol g/mL water) in this mixture, calculated VVmL =100.82 =100.82 mL, mL, an an error error of ofthis 0.82 0.82 %. %. thethe water total ==100.82 0.7893 g/mL ethanol and%.ρ=0.9982 g/mL for water) in mixture, calculated total total water) and densities of pure ethanol and water (ρ= 0.7893 g/mL Example Example 3: 3: Mixing Mixing 100 mL of of 95% 95% (v/v) (v/v) ethanol ethano volume of of the mixture becomes Example Mixing 100 mL 100 of 95mL % (v/v) volume mixture Example Mixing 100 mL ofthe 95% Example (v/v) ethanol with3:144 mL ofof water for ethanol and ρ=0.9982 g/mLthe for 3: water) in becomes this mixture, Example 3: 3: Mixing Mixing 100 100 mL mL of 95% 95% (v/v) (v/v) ethanol ethanol with with 144 144 mL mL ethanol with 144 mL of water calculated volume of the mixture becomes: When When 100 100 mL mL of of 95% 95% (v/v) (v/v) ABV ABV and and 144 144 mL mL of of w w When4100 mL of 95% (v/v) ABV andWhen 144 mL of mL water are mixed, the resulting ABV is 40% (v/v) When 100 mL of 95 % (v/v) ABV andand 144 mL ofmL water are mixed, w1 w2 95 . 02 . 98 g g When 100 100 mL of of 95% 95% (v/v) (v/v) ABV ABV and 144 144 mL of of water water are are mixed mixe w 95.02 g  4.98 g  125.37 mL ABV ABV40 to toaccording the the paper paper written written by by Spedding, Spedding Equation 6according 1 w2 1according the Spedding, resultingEquation ABVetisal (v/v) ABV, to the Spedding, paper written1et according g125 .37 mL 1.. In .%6In that specific article, al, ABV the paper writtenABV by 1  2 w10.7893 used the following formula to find that the final volume w2 g g 95 .02 4.98 g g g to In tha tha ABV according according to to the paper paper written written by by Spedding, Spedding, et et al al used the following formula to find that the final vo 0 . 9982 6  mL  mL  125.37 mLby Spedding,used 1 2  0.7893 et al.1 Equation In that specific article, Spedding, et al., used 0.9982 the following formula to find that the final volu mL mL mL. mL. to find that the final volume of mixing was 1  2 0.7893 g 0.9982 g the following formula mL mL mL. EQUATION 6 TheThe error becomes 0.81% which is about 3 times larger than thethe error using partial molar 237.5 mLerror instead of the 244 mL.partial error becomes 0.81% which is about 3 times larger than using the molar 95 % 95% xV   100 mL  237 volume information. The error becomes 0.81 %, which is which about three times 3larger xV   100 mL.5mL 237.5 mL The error becomes 0.81% is about times larger than error using the partial molar dilutethe dilute 95% volume information. 40 % 40 % xV   100 mL  237 .5 mL than the error using the partial molar volume information. dilute information. 4.98 gthatvolume 40% Equation 6 can be transformed to become: Equation 6  Note  125 . 37 mL Note that Equation 6 can be transformed to become: Note that Equation 6 can be transformed to become: This article, uses the partial molar volume approach to fi This article, uses the partial molar volume approa This article uses the partial molar volume approach to find out 0.9982 g Note that Equation 6 can be transformed to become:   This article, uses the partial molar volume approach L mL w1 w2 w1 M 1 w2 M 2 volume of 7mixing. w w w M w M n  V1,m  n2  V2,m  the resulting Equation Vtotal   95% (v/v), density of this of ABV is ABV 0.8114 g/mL, ag With 95%the the density this is 0.8114 Equation 7(v/v), Vtotal  1  w2 M  w1 w1 M  M2  w2 1 M n1  V1,m  n2 V2,m WithWith  95 % (v/v), the density of this ABV is 0.8114 g/mL , and 1 2 1 1 2 2 a,b 1% which is about 3 times than 1 larger 2 1 the With 95% (v/v), the density of moles this of ABV is 0.8114(ng/m of the OIML table. Number of of ethanol V92.41 1) 7table. Vtotal  M 12 error   2 2 using  2 the  n1partial  V1,m molar nABW Va,b of Equation the OIML Number moles 1  2  M 1  1 2  Vis 2 ,m % according to Tables Va,b of the OIML table. Numberof ethan a,b M M 1 EQUATION   V accordingly: of mass the OIMLmolar table. Number of moles of ethanol 2 1 2 71 accordingly: of moles ofratio ethanol (n1) andtowater (n2)mass are calculated accordingly: where M1M andand M2M areare the molar masses of2 ethanol andand water. The of where the molar masses of ethanol water. The ratio of mass to molar mass 1 2 accordingly: where M1 and Mof are the molar masses of ethanol and water.The The ratio of the molar 2become: the number moles of the specific component. mass to the density is mass an be is transformed to where M and M are the molar masses of ethanol and water. The ratio of mass to molar 1 2 n  1 . 628 mol , n  8 . 321 mol is the number of moles specific component. The ratio of1the molar to the n1  1mass .6282mol , n2density  8.321ismol ratio of mass to molar mass is of thethe number of moles of the specific thethe molar TheThe molar volumes forfor purecomponent. solutions of each component ofmol ethanol isvolume. the number of moles of Theeach ratio molar to8.the is nof 1.628 , n2 and 321density mol 1 the  of the  molar volume. molar volumes of component of mass ethanol and The ratio molar mass tothe thespecific densitypure is the solutions molar M 1 wcomponent. M The mole fraction of ethanol in this mixture is x1mixture = 0.164. Thus, 2 2 The mole fraction of ethanol in this is x1= 0.164. are (as also noted above): The mole fraction of ethanol in this mixture is x1= Equation 7 water   n  V  n  V the molar Thesolutions molar volumes for pure solutions of each component of ethanol and volume. The molar pure of each component 1 also 1volumes ,m volume. 2 for2 ,above): m water are (as noted the partial molar volumes for ethanol and water in this case are is x1= 0. 1 M 2  2 The mole fraction of ethanol in this mixture ethanolethanol and water this case then: gare are and in water in thisare case are then: of ethanol and water alsonoted noted above): above): also   46water .068 18.015 g g then: g (as(as   46 . 068 18 . 015 ethanol and water in this case are then: mol mol mL mL   58 37   mass 18 05 V 1,mmasses V2,m of mol g .58 g.18 he molar water. The ;ratio .37 mL ; V2,m mass   togmolar .05 mL V 1,m   of ethanol mol mol 46.mol 068and 18 . 015 g molmL mol V1 54 mL/mol ; V 2 ; -17V.777 mL/mol 0.7893 0;.9982 mol  58.37 mol g g V.592  54 .592 mL/mol .777 mL/mol mL 2  17 - 18 1.05 the V02.,to m  mL 01.,7893 9982 mL of the specificVcomponent. The ratio of the molar mass density is m mol mol mL mL g g V  54 . 592 mL/mol ; V 2  17.777 mL/mol 1 0.7893 0.9982 The The final final volume using Equation 5, now shows Vtotal =shows e molar volumes for pure solutions of each component of ethanol and mL mL volume using Equation 5, now Vtotal = 236 The final volume using Equation 5, 236.774 now shows Vto Equation 7 is valid only for mixtures of an ideal solution such ascloser that ofwhat aofmix of benzene andcalculated. mL, to Spedding, et al. , had The real lesson Equation 7 is valid only for mixtures of an ideal solution such as that a mix of benzene and The final volume using Equation 5, now shows V d above): total al., had calculated. The real lesson learnedlearned from this exath al., had calculated. The realand lesson from toluene. ButBut when components interact such for ethanol water, then molar Equation 7the is valid only for mixtures of such anasideal solution such as that ofthen aisthe mix ofmolar benzene learnedand from this example that after mixing 100 mL of 95 % ABV toluene. when the components interact as for ethanol and water, the al., had calculated. The real lesson learned from this g ABVmLand 144 mL of water, the final volume is final not 244 mL ABV and mL of the water, the final volume is not  to be 18 .replaced 015 volumes need by by the partial molar volumes 5a, 5b). andfor 144 of water, the144 final volume is molar not 244 mL. The toluene. But when the components interact such (Equations as ethanol and water, then mol volumes need to be replaced the partial molar volumes (Equations 5a, 5b). mL mL ABV and 144 mL of water, the final volume is not 24 58.37 ; V2,m   18.05 40.00-40.12% depending upon ‘true’ correctbe alcohol willABV be 40.00-40.12 % ABVwhich depending becontent 40.00-40.12% ABV depending upon whichcalcu ‘true mol volumes molpartial molar volumes togbe replaced by the (Equations 5a, 5b).   need 0 . 9982 ABV depending upon which ‘true’ upon which be true40.00-40.12% calculated volume is used.  mL V1 ,V2 (Eq. 5b)5b) to represent partial molar volumes which are composition Here wewe useuse Example 4: composition Mixing 100 mL100 of 95% (v/v) with 10w V1 ,V2 (Eq. to represent partial molar volumes which are Here Example 4: Mixing mL of 95%ethanol (v/v) ethanol     Example 4: Mixing 100 mL of 95% (v/v) ethanol wi V ,V (Eq. 5b) to represent partial molar volumes which are composition Here we use   1 2 such as that of a mix of benzene and for mixtures of an idealusing solution dependent while 7) V to represent molar volumes which are composition 1,mV, V2,,m This example is quite useful when we need to know the f WWW.ARTISANSPIRITMAG.COM   (Eq.(Eq. 83 This example is quite useful when we need to kno dependent while using 7) to represent molar volumes which are composition V   m components interact such as for ethanol1,mand2,water, then the molar This example isare quite useful when we need know dependent while using (Eq. 7) V1,m , V2,m to represent molar volumes which composition solutions with different ABV values. Most professionals independent. solutions with different ABV values. Mosttoprofess placed byindependent. the partial molar volumes (Equations 5a, 5b).