12.25 g of CH_3CH_2CHClCOOH is added to 250 | vedclass

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(A) Step $I$: Calculation of molality $(m)$Molar mass of $CH_3CH_2CHClCOOH = 122.5 \ g \ mol^{-1}$.Molality $(m)$ = $\frac{	ext{Mass of solute}}{	ex

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$0.789$

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(A) Step $I$: Calculation of molality $(m)$Molar mass of $CH_3CH_2CHClCOOH = 122.5 \ g \ mol^{-1}$.Molality $(m)$ = $\frac{	ext{Mass of solute}}{	ext{Molar mass of solute} 	imes 	ext{Mass of solvent in kg}} = \frac{12.25 \ g}{122.5 \ g \ mol^{-1} 	imes 0.250 \ kg} = 0.40 \ m$.Step $II$: Calculation of degree of dissociation $(\alpha)$For a weak acid,$K_a = C\alpha^2$ (assuming $\alpha \ll 1$).$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.44 	imes 10^{-3}}{0.4}} = \sqrt{36 	imes 10^{-4}} = 0.06$.Step $III$: Calculation of van't Hoff factor $(i)$For dissociation $CH_3CH_2CHClCOOH ightleftharpoons CH_3CH_2CHClCOO^{-} + H^{+}$,$i = 1 + \alpha = 1 + 0.06 = 1.06$.Step $IV$: Calculation of depression in freezing point $(\Delta T_f)$$\Delta T_f = i 	imes K_f 	imes m = 1.06 	imes 1.86 	imes 0.40 = 0.789 \ K$ (or $^{\circ}C$).

$12.25 \ g$ of $CH_3CH_2CHClCOOH$ is added to $250 \ g$ of water to make a solution. If the dissociation constant of the above acid is $1.44 \times 10^{-3}$,the depression in the freezing point of water in $^{\circ}C$ is ($K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$)

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