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(B) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f = 0 - (-0.063) = 0.063 \ K$. Using the formula $\Delta T_f = i 	imes K

Vedclass · Accepted Answer

$67$

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(B) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f = 0 - (-0.063) = 0.063 \ K$. Using the formula $\Delta T_f = i 	imes K_f 	imes m$,we have $0.063 = i 	imes 1.86 	imes 0.011$. Calculating $i$: $i = \frac{0.063}{1.86 	imes 0.011} = \frac{0.063}{0.02046} \approx 3.079$. For $K_3[Fe(CN)_6]$,the dissociation is $K_3[Fe(CN)_6] ightarrow 3K^+ + [Fe(CN)_6]^{3-}$,so $n = 4$. The degree of dissociation $\alpha$ is given by $\alpha = \frac{i - 1}{n - 1}$. Substituting the values: $\alpha = \frac{3.079 - 1}{4 - 1} = \frac{2.079}{3} = 0.693$. Percentage of dissociation $\approx 69.3 \%$,which is closest to $67 \%$.

What is the percentage of dissociation of a $0.011 \ m$ aqueous solution of $K_3[Fe(CN)_6]$ having a freezing point of $-0.063 \ ^oC$? (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$ for water)

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