The freezing point depression of 0.001 M of | vedclass

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(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes M$. Given $\Delta T_f = 5.58 	imes 10^{-3} \ K$,$K_f = 1.86 \ K

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$2$

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(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes M$. Given $\Delta T_f = 5.58 	imes 10^{-3} \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $M = 0.001 \ M = 10^{-3} \ M$. Calculating the van't Hoff factor $(i)$: $i = \frac{\Delta T_f}{K_f 	imes M} = \frac{5.58 	imes 10^{-3}}{1.86 	imes 10^{-3}} = 3$. The complex salt dissociates as $A_x B_y[Fe(CN)_6] ightarrow xA^{n+} + yB^{m+} + [Fe(CN)_6]^{4-}$. Since the total number of ions produced is $i = 3$,and the complex anion $[Fe(CN)_6]^{4-}$ counts as $1$ ion,the total number of cations $(x+y)$ must be $3 - 1 = 2$. Given the charge balance: $x(n) + y(m) = 4$. Possible combinations for $x+y=2$ with $n, m$ as positive integers: $1$. If $x=1, y=1$,then $n+m=4$. Possible $(n, m)$ pairs are $(1, 3)$ and $(2, 2)$. $2$. If $x=2, y=0$ (not possible as $B$ is mentioned) or $x=1, y=1$ with different valencies. Thus,there are $2$ possibilities for the cation types.

The freezing point depression of $0.001 \ M$ of $A_x B_y[Fe(CN)_6]$ is $5.58 \times 10^{-3} \ K$. If the oxidation state of $Fe$ is $+2$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$,then the total number of possibilities for different types of $A$ and $B$ cations are

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