The depression in freezing point for 0.01 m | vedclass

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Question

(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. Given: $\Delta T_f = 0.0744 \ K$,$K_f = 1.86 \ K \ kg \ mol^

Vedclass · Accepted Answer

$K_{3}[Fe(CN)_6]$

Vedclass · Answer

(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. Given: $\Delta T_f = 0.0744 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$. Substituting the values: $0.0744 = i 	imes 1.86 	imes 0.01$. $i = \frac{0.0744}{0.0186} = 4$. For complete dissociation of $K_x[Fe(CN)_6]$,the van't Hoff factor $i$ is equal to the number of ions produced,which is $x + 1$. $x + 1 = 4 \Rightarrow x = 3$. Therefore,the molecular formula is $K_3[Fe(CN)_6]$.

The depression in freezing point for $0.01 \ m$ aqueous solution of $K_{x}[Fe(CN)_6]$ is $0.0744 \ K$. The molal depression constant for solvent is $1.86 \ K \ kg \ mol^{-1}$. If the solute undergoes complete dissociation,what is the correct molecular formula for the solute?

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