JEE Main 2025 — Solutions Question with Solution

For AB

$$\begin{array}{rl}& \Delta {\mathrm{T}}_{\mathrm{b}}=2.7\mathrm{K}\\ & 2.7=1\times 0.5\times \mathrm{m}\\ & \mathrm{m}=\frac{27}{5}\end{array}$$

Let molar mass of $AB=x$.

So $\frac{1/x}{15}\times 1000=\frac{27}{5}$

$$x=12.34$$

For ${\mathrm{AB}}_2$

$$\begin{array}{rl}& \Delta {\mathrm{T}}_{\mathrm{b}}=1.5\mathrm{K}\\ & 1.5=1\times 0.5\times \mathrm{m}\\ & \mathrm{m}=3\end{array}$$

Let molar mass of ${\mathrm{AB}}_2=\mathrm{y}$

So $\frac{1/\mathrm{y}}{15}\times 1000=3$

$$\begin{array}{rl}& y=\frac{1000}{45}\\ & y=22.22\end{array}$$

Now let a and b be atomic masses of A and B respectively, then

$$\begin{array}{rl}& \mathrm{A}+\mathrm{b}=12.34\text{...... (i)}\\ & \mathrm{A}+2\mathrm{b}=22.22\text{...... (ii)}\\ & \mathrm{B}=22.22-12.34=9.88\end{array}$$

Now $\mathrm{a}=12.34-9.88=2.46$

$$=24.6\times 10^{-1}=25\times 10^{-1}$$