  Question

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. [$$Ka = 1.4 \times 10^{-3}, K_{f} = 1.86\ K\ kg\ mol^{-1}$$]

Solution

The molar mass of $$2$$-chloro butanoic acid is $$122.5$$ g/mol. Number of moles $$\displaystyle = \dfrac {10}{122.5} =0.0816$$ molMolality of solution $$\displaystyle = \dfrac {0.0816 \times 1000}{250} = 0.3265$$ mLet $$\displaystyle \alpha$$ be the degree of dissociation and c be the initial concentration. The concentration after dissociation is as shown.$$\displaystyle \underset c (1 - \alpha)}{CH_3CH_2CHClCOOH} \rightleftharpoons \underset c \alpha}{CH_3CH_2CHClCOO^-} + \underset c \alpha } H^$$The equilibrium constant expression is $$\displaystyle K = \dfrac {c \alpha \times c \alpha }{c(1-\alpha)} = c \alpha^2$$$$\displaystyle \alpha = \sqrt {\dfrac {K}{c}} = \sqrt {\dfrac {1.4 \times 10^{-3}}{0.3265} } = 0.065$$Calculation of vant Hoff factor:$$\displaystyle \underset {\displaystyle (1 - \alpha)}{CH_3CH_2CHClCOOH} \rightleftharpoons \underset {\displaystyle \alpha}{CH_3CH_2CHClCOO^-} + \underset }{ \alpha } H^$$$$\displaystyle i = \dfrac {1- \alpha + \alpha + \alpha }{1} =1 + \alpha = 1 + 0.065 = 1.065$$The depression in the freezing point $$\displaystyle \Delta T_f = iK_fm = 1.065 \times 1.86 \times 0.3265 = 0.647 ^o$$ChemistryNCERTStandard XII

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