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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 $$\displaystyle = \dfrac {10}{122.5} =0.0816$$ mol

Molality of solution $$\displaystyle = \dfrac {0.0816 \times 1000}{250} = 0.3265$$ m

Let $$\displaystyle \alpha $$ be the degree of dissociation and c be the initial concentration. The concentration after dissociation is as shown.
$$\displaystyle \underset {\displaystyle c (1 - \alpha)}{CH_3CH_2CHClCOOH} \rightleftharpoons \underset {\displaystyle c \alpha}{CH_3CH_2CHClCOO^-} + \underset {\displaystyle c \alpha } H^+$$

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:

$$\displaystyle \underset {\displaystyle  (1 - \alpha)}{CH_3CH_2CHClCOOH} \rightleftharpoons \underset {\displaystyle  \alpha}{CH_3CH_2CHClCOO^-} + \underset {\displaystyle }{ \alpha } H^+$$

$$\displaystyle i = \dfrac {1- \alpha + \alpha + \alpha }{1} =1 + \alpha = 1 + 0.065 = 1.065 $$

The depression in the freezing point $$\displaystyle \Delta T_f = iK_fm = 1.065 \times 1.86 \times 0.3265 = 0.647 ^o$$

Chemistry
NCERT
Standard XII

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