Boiling point of a 2 % aqueous solution of a nonvolatile solute A...

For $A$ : $100gm$ solution $\to 2gm$ solute $A$ $∴$ Molality $= \dfrac{2 / {M}_{A}}{0 . 098}$

For $B$ : $100gm$ solution $\to 8gm$ solute $B$ $∴$ Molality $= \dfrac{8 / {M}_{B}}{0 . 092}$

$∵ {\left({\Delta T}_{B}\right)}_{A} = {\left({\Delta T}_{B}\right)}_{B}$

$∴$ Molality of $A =$ Molality of $B$

$∴ \dfrac{2}{0 . 098{M}_{A}} = \dfrac{8}{0 . 092{M}_{B}}$

$\dfrac{2}{98} \times \dfrac{92}{8} = \dfrac{{M}_{A}}{{M}_{B}}$

$\dfrac{1}{4 . 261} = \dfrac{{M}_{A}}{{M}_{B}}$

$∴ {M}_{B} = 4 . 261 \times {M}_{A}$