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Q.
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 nonvolatile solute B. The relation between molecular weights of $A$ and $B$ is.
For $A : 100\, gm$ solution $\rightarrow 2 \,gm$ solute A
$\therefore \text { Molality }=\frac{2 / M _{ A }}{0.098}$
For B : $100\,gm$ solution $\rightarrow 8$ gm solute B
$\therefore \text { Molality }=\frac{8 / M _{ B }}{0.092}$
$ \because\left(\Delta T _{ B }\right)_{ A }=\left(\Delta T _{ B }\right)_{ B }$
$ \therefore \text { Molality of } A =\text { Molality of } B$
$ \therefore \frac{2}{0.098 M _{ A }}=\frac{8}{0.092 M _{ B }} $
$ \frac{2}{98} \times \frac{92}{8}=\frac{ M _{ A }}{ M _{ B }} $
$ \frac{1}{4.261}=\frac{ M _{ A }}{ M _{ B }} $
$\therefore M _{ B }=4.261 \times M _{ A }$