JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Solutions

Elevation of boiling point follows $\Delta T_b=K_{b}m$. For AB, $2.7=0.5m$ gives molality $m=5.4{\text{mol kg}}^{-1}$; with 1 g dissolved in 0.015 kg of water, $\frac{1/M_{AB}}{0.015}=5.4$, so $M_{AB}=12.35{\text{g mol}}^{-1}$. For $\mathrm{AB}{}_2$, $1.5=0.5m$ gives $m=3{\text{mol kg}}^{-1}$ and $\frac{1/M_{A{B}_2}}{0.015}=3$, so $M_{A{B}_2}=22.22{\text{g mol}}^{-1}$. Writing the atomic masses of A and B as $a$ and $b$, $a+b=12.35$ and $a+2b=22.22$. Subtracting gives $b=9.87$ and $a=2.47\text{amu}$. Expressed in the required units, the atomic mass of A is $24.7\times 10^{-1}\approx 25\times 10^{-1}$ amu.