What is the freezing point (in ^circ C) of a | vedclass

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(A) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. First,calculate the molality $(m)$: $m = \frac{	ext{mass of

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$-2.3 	imes 10^{-2}$

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(A) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. First,calculate the molality $(m)$: $m = \frac{	ext{mass of solute}}{	ext{molar mass}} 	imes \frac{1000}{	ext{mass of solvent in g}} = \frac{0.1}{329} 	imes \frac{1000}{100} = \frac{1}{329} \approx 0.00304 \ mol \ kg^{-1}$. $K_3[Fe(CN)_6]$ dissociates as $K_3[Fe(CN)_6] ightarrow 3K^{+} + [Fe(CN)_6]^{3-}$,so the van't Hoff factor $(i)$ is $4$. Now,calculate $\Delta T_f$: $\Delta T_f = 4 	imes 1.86 	imes 0.00304 \approx 0.0226 \ K$. Since the freezing point of pure water is $0^\circ C$,the freezing point of the solution is $0 - 0.0226 = -0.0226^\circ C$,which is approximately $-2.3 	imes 10^{-2} \ ^\circ C$.

What is the freezing point (in $^\circ C$) of a solution containing $0.1 \ g$ of $K_3[Fe(CN)_6]$ (molar mass = $329 \ g \ mol^{-1}$) in $100 \ g$ of water? (Given $K_f = 1.86 \ K \ kg \ mol^{-1}$)

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