JEE MAIN - Chemistry (2025 - 2nd April Evening Shift - No. 22)
When 1 g each of compounds AB and $\mathrm{AB}_2$ are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in $a m u$ ) is____________ $\times 10^{-1}$ (Nearest integer)
(Given : Molal boiling point elevation constant is $0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
Explanation
For AB
$$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \\ & 2.7=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=\frac{27}{5} \end{aligned}$$
Let molar mass of $A B=x$.
So $\frac{1 / x}{15} \times 1000=\frac{27}{5}$
$$x=12.34$$
For $\mathrm{AB}_2$
$$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \\ & 1.5=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=3 \end{aligned}$$
Let molar mass of $\mathrm{AB}_2=\mathrm{y}$
So $\frac{1 / \mathrm{y}}{15} \times 1000=3$
$$\begin{aligned} & y=\frac{1000}{45} \\ & y=22.22 \end{aligned}$$
Now let a and b be atomic masses of A and B respectively, then
$$\begin{aligned} & \mathrm{A}+\mathrm{b}=12.34 \quad\text{...... (i)}\\ & \mathrm{~A}+2 \mathrm{~b}=22.22 \quad\text{...... (ii)}\\ & \mathrm{~B}=22.22-12.34=9.88 \end{aligned}$$
Now $\mathrm{a}=12.34-9.88=2.46$
$$=24.6 \times 10^{-1}=25 \times 10^{-1}$$
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