Boiling point of a 2% aqueous solution of a nonvolatile 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 A : 100 gm solution $\to$ 2 gm solute A

$\therefore \text{ Molality }=\frac{2/{\mathrm{M}}_{\mathrm{A}}}{0.098}$

For B : 100 gm solution $\to$ 8 gm solute B

$\begin{array}{l}\therefore \text{ Molality }=\frac{8/{\mathrm{M}}_{\mathrm{B}}}{0.092}\\ \because {\left({\mathrm{\Delta T}}_{\mathrm{B}}\right)}_{\mathrm{A}}={\left({\mathrm{\Delta T}}_{\mathrm{B}}\right)}_{\mathrm{B}}\\ \therefore \text{ Molality of }\mathrm{A}=\text{ Molality of B }\\ \therefore \frac{2}{0.098{\mathrm{M}}_{\mathrm{A}}}=\frac{8}{0.092{\mathrm{M}}_{\mathrm{B}}}\\ \frac{2}{98}\times \frac{92}{8}=\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}}\\ \frac{1}{4.261}=\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}}\\ \therefore {\mathrm{M}}_{\mathrm{B}}=4.261\times {\mathrm{M}}_{\mathrm{A}}\end{array}$