JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Solutions

Equal freezing-point depressions mean equal molalities, since both are aqueous solutions with the same $k_f$. Writing molality as $\frac{\text{mass of solute}\times 1000}{\text{molar mass}\times \text{mass of water}}$ and taking $100\text{g}$ of each solution, the $4\%$ solution of $X$ holds $4\text{g}$ of $X$ in $96\text{g}$ water and the $12\%$ solution of $Y$ holds $12\text{g}$ of $Y$ in $88\text{g}$ water. Setting the molalities equal, $\frac{4}{A\times 96}=\frac{12}{M_Y\times 88}$, which gives $M_Y=\frac{12\times 96}{4\times 88}A\approx 3A$.