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(B) The depression in freezing point $\Delta T_f$ is given by $\Delta T_f = K_f \cdot m$,where $m$ is the molality.Since the mass of solvent $(1 \ kg)

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$4: 1$

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(B) The depression in freezing point $\Delta T_f$ is given by $\Delta T_f = K_f \cdot m$,where $m$ is the molality.Since the mass of solvent $(1 \ kg)$ and the mass of solute $(1 \ g)$ are the same for both,the molality $m$ is inversely proportional to the molar mass $M$ $(m = \frac{n}{W_{solvent}} = \frac{mass}{M \cdot W_{solvent}})$.Therefore,$\frac{\Delta T_x}{\Delta T_y} = \frac{M_y}{M_x}$.Given $\frac{\Delta T_x}{\Delta T_y} = \frac{1}{4}$,we have $\frac{1}{4} = \frac{M_y}{M_x}$.This implies $\frac{M_x}{M_y} = \frac{4}{1}$,so $M_x : M_y = 4 : 1$.

Two solutions $A$ and $B$ are prepared by dissolving $1 \ g$ of non-volatile solutes $X$ and $Y$ respectively in $1 \ kg$ of water. The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$. The ratio of molar masses of $X$ and $Y$ is.

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