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Calculate the depression in the freezing point of water when 10 g of CH_(3)CH_(2)CHCl COOH is added to 250 g of water. K_(a)=1.4xx10^(-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+12+16+16+1` `= 122.5 g mol^(-1)` Therefore, No. of moles present in 10 g `CH_(3)CH_(2)CHClCOOH` `=(10 g)/(122.5 g mol^(-1))` = 0.0816 mol It is given that 10 g of `CH_(3)CH_(2)CHClCOOH` is added to 250 g of water. Therefore, Molality of the solution, `=(0.0186)/(250)xx1000` `=32.64 " mol kg"^(-1)` Let a be the degree of dissociation of `CH_(3)CH_(2)CHClCOOH`. `CH_(3)CH_(2)CHClCOOH` UNDERGOES dissociation according to the following EQUATION : `{:("Initial",CH_(3)CH_(2)CHClCOOH,hArr,CH_(3)CHClCOO^(-),+,H^(+)),("conc. At",C,,O,,O),("equilibrium",C(1-alpha),,C alpha,,C alpha):}` `K_(a)=(C^(2)x^(2))/(C(1-x))` Since x is very small with respect to 1 so x can be ignored, `1-alpha =1` Now, `K_(a)=(cx^(2))/(1)` `x=((K_(4))/(C ))^(0.5)` `= sqrt((1.4xx10^(-3))/(0.3264)) (because K_(a)=1.4xx10^(_3))` = 0.0655 Again, `{:("Again,",CH_(3)CH_(2)CHClCOOH,hArr,CH_(3)CHClCOO^(-),+,H^(+)),("initial","1 mol",,0,,0),("equilibrium",1-x,,x,,x):}` i(Vant Hoff factor) `= 1-x+x+x=1+x` `= 1+0.0655` = 1.0655 Hence, the depression in the freezing point of water is given as : `Delta T_(f)=i.K_(f).m` `=1.0655xx1.86" K kg mol"^(-1)xx0.3264 " mol kg"^(-1)` = 0.65 K |
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