JEE Main 2018 — Solutions Question with Solution

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) $$\begin{array}{rl}& \frac{3}{4}\times {\mathrm{M}}_{\mathrm{Y}}\times \frac{5}{1000}=\mathrm{m}\times {\mathrm{M}}_{\mathrm{Y}}\times \frac{5}{1000}\\ & \mathrm{m}=\frac{3}{4}\end{array}$$