Boiling point of a $2 \%$ aqueous solution of a non-volatile solute A is equal to the boiling point of $8 \%$ aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is

For $\mathbf{A}: 100 \,\mathrm{gm}$ solution $\rightarrow 2 \,\mathrm{gm}$ solute $\mathrm{A}$<br/><br/> $\therefore$ Molality $=\frac{2 / \mathrm{M}_{\mathrm{A}}}{0.098}$ <br/><br/> For B : $100 \,\mathrm{gm}$ solution $\rightarrow 8 \,\mathrm{gm}$ solute $\mathrm{B}$<br/><br/> $$ \begin{aligned} &\therefore \text { Molality }=\frac{8 / \mathrm{M}_{\mathrm{B}}}{0.092} \\\\ &\because\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{A}}=\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{B}} \end{aligned} $$<br/><br/> $\therefore$ Molality of $\mathrm{A}=$ Molality of $\mathrm{B}$<br/><br/> $$ \therefore \frac{2}{0.098 \mathrm{M}_{\mathrm{A}}}=\frac{8}{0.092 \mathrm{M}_{\mathrm{B}}} $$<br/><br/> $\frac{2}{98} \times \frac{92}{8}=\frac{M_A}{M_B}$<br/><br/> $\frac{1}{4.261}=\frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}}$<br/><br/> $\therefore \mathrm{M}_{\mathrm{B}}=4.261 \times \mathrm{M}_{\mathrm{A}}$