Two 5 molal solutions are prepared by dissolving a non-electrolyte,...

The relationship between molar masses of the two solvents is $\mathrm{M}_{\mathrm{X}}=\frac{3}{4} \mathrm{M}_{\mathrm{Y}}$The relative lowering of vapour pressure of the two solutions is $\left(\frac{\Delta \mathrm{P}}{\mathrm{P}}\right)_{\mathrm{X}}=\mathrm{m}$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}}\right)_{\mathrm{Y}}$But, the relative lowering of vapour pressure of solutions is directly proportional to the mole fraction of solute. Given 5 molal solution, means 5 moles of solute are dissolved in$1 \mathrm{~kg}$( or$1000 \mathrm{~g}$) of solvent. The number of moles of solvent$=\frac{1000 \mathrm{~g}}{\mathrm{M}}$The mole fraction of solute$=\frac{5}{1000 / \mathrm{M}}=\mathrm{M} \times \frac{5}{1000}$hence$\mathrm{M}_{\mathrm{X}} \times \frac{5}{1000}=\mathrm{m} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000}$. Substitute equation (i) in equation (ii)$$\begin{aligned}&\frac{3}{4} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000}=\mathrm{m} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000}\\ &\mathrm{~m}=\frac{3}{4} \end{aligned}$$