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(D) The formula for depression in freezing point is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality. Molality $m = \frac{W_{solute} 	imes 1

Vedclass · Accepted Answer

$25 \ u, 50 \ u$

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(D) The formula for depression in freezing point is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality. Molality $m = \frac{W_{solute} 	imes 1000}{M_{solute} 	imes W_{solvent} 	ext{ (in g)}}$. For $XY$: $5.333 = 2 	imes \frac{10 	imes 1000}{M_{XY} 	imes 50} \implies M_{XY} = \frac{20000}{5.333 	imes 50} \approx 75 \ g/mol$. So,$X + Y = 75$ (Equation $1$). For $XY_3$: $2.2857 = 2 	imes \frac{10 	imes 1000}{M_{XY_3} 	imes 50} \implies M_{XY_3} = \frac{20000}{2.2857 	imes 50} \approx 175 \ g/mol$. So,$X + 3Y = 175$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(X + 3Y) - (X + Y) = 175 - 75 \implies 2Y = 100 \implies Y = 50 \ u$. Substituting $Y = 50$ in Equation $1$: $X + 50 = 75 \implies X = 25 \ u$. The atomic weights are $X = 25 \ u$ and $Y = 50 \ u$.

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