Calculate the freezing point (in {}^{circ}C) | vedclass

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Question

(A) The dissociation of potassium ferricyanide is: $K_{3}[Fe(CN)_{6}] \rightarrow 3 K^{+} + [Fe(CN)_{6}]^{3-}$.Since $1 \ mol$ of salt gives $4 \ mol$

Vedclass · Accepted Answer

$- 2.3 	imes 10^{-2}$

Vedclass · Answer

(A) The dissociation of potassium ferricyanide is: $K_{3}[Fe(CN)_{6}] ightarrow 3 K^{+} + [Fe(CN)_{6}]^{3-}$.Since $1 \ mol$ of salt gives $4 \ mol$ of ions,the van't Hoff factor $i = 4$.The depression in freezing point is given by: $\Delta T_{f} = i 	imes K_{f} 	imes m$.Molality $m = \frac{	ext{mass of solute}}{	ext{molar mass}} 	imes \frac{1000}{	ext{mass of solvent in } g} = \frac{0.1}{329} 	imes \frac{1000}{100} = \frac{1}{329} \ mol \ kg^{-1}$.$\Delta T_{f} = 4 	imes 1.86 	imes \frac{1}{329} \approx 0.0226 \ K \approx 2.26 	imes 10^{-2} \ K$.Since $T_{f} = T_{f}^{\circ} - \Delta T_{f}$ and $T_{f}^{\circ} = 0 \ {}^{\circ}C$,the freezing point is $0 - 0.0226 = - 2.26 	imes 10^{-2} \ {}^{\circ}C$,which is approximately $- 2.3 	imes 10^{-2} \ {}^{\circ}C$.

Calculate the freezing point (in ${}^{\circ}C$) of a solution obtained by dissolving $0.1 \ g$ of potassium ferricyanide (molecular weight $= 329$) in $100 \ g$ of water. Given that $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$.

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