The freezing point (in {}^{circ}C) of a solut | vedclass

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(A) The dissociation of $K_3[Fe(CN)_6]$ is given by: $K_3[Fe(CN)_6] \rightarrow 3K^{+} + [Fe(CN)_6]^{3-}$.Since $1 \ mol$ of $K_3[Fe(CN)_6]$ produces

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$-2.3 	imes 10^{-2}$

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(A) The dissociation of $K_3[Fe(CN)_6]$ is given by: $K_3[Fe(CN)_6] ightarrow 3K^{+} + [Fe(CN)_6]^{3-}$.Since $1 \ mol$ of $K_3[Fe(CN)_6]$ produces $4 \ mol$ of ions,the van't Hoff factor $i = 4$.The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$.Here,molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1} = \frac{0.1 	imes 1000}{329 	imes 100} = \frac{1}{329} \ mol \ kg^{-1}$.$\Delta T_f = 4 	imes 1.86 	imes \frac{1}{329} \approx 0.0226 \ K \approx 0.023 \ K$.Since the freezing point of pure water is $0 \ {}^{\circ}C$,the freezing point of the solution is $T_f = 0 - \Delta T_f = -0.023 \ {}^{\circ}C = -2.3 	imes 10^{-2} \ {}^{\circ}C$.

The freezing point (in ${}^{\circ}C$) of a solution containing $0.1 \ g$ of $K_3[Fe(CN)_6]$ (Mol. Wt. $329$) in $100 \ g$ of water $(K_f = 1.86 \ K \ kg \ mol^{-1})$ is

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