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Calculate the depression in the freezing point of water when 10 g of CH_(3)CH_(2)CHClCOOH is added to 250 g of water. K_(a)=1.5xx10^(-3),K_(f)=1.86" K kg mol"^(-1) |
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Answer» Solution :`"MOLAR MASS of "CH_(3)CH_(2)CHClCOOH=15+14+13+35.5+45="122.5 g mol"^(-1)` `"10 g of "CH_(3)CH_(2)CHClCOOH=(10)/(122.5)" mole"=8.16xx10^(-2)" mole"` `therefore"Molality of the solution (m)"=(8.16xx10^(-2)" mole")/("250 g")xx"1000 g kg"^(-1)=0.3264` If `alpha` is the degree of dissociation of `CH_(3)CH_(2)CHClCOOH`, then `{:(,CH_(3)CH_(2)CHClCOOH,hArr,CH_(3)CH_(2)CHClCOO^(-),+,H^(+)),("Initial CONC.","C mol L"^(-1),,"0",,"0"),("At eqm.",""C(1-alpha),,""Calpha,,Calpha):}``therefore""K_(a)=(Calpha.Calpha)/(C(1-alpha))~=Calpha^(2)or alpha=SQRT((K_(a))/(C))=sqrt((1.4xx10^(-3))/(0.3624))=0.065` The CALCULATE van't Hoff factor : `{:(,CH_(3)CH_(2)CHClCOOH,hArr,CH_(3)CH_(2)CHClCOO^(-),+,H^(+)),("Initial moles","1",,,,),("Moles at eqm.",""1-alpha,,""alpha,,alpha):}` `i=(1+alpha)/(1)=1+alpha=1+0.065=1.065=1.065" ,"DeltaT_(f)=iK_(f)m=(1.065)(1.86)(0.3264)=0.65^(@)`. |
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