Q20 / 75 · JEE Main 2025 (2 Apr Shift 2) · Chemistry
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 amu) is ____ $\times 10^{-1}$ (Nearest integer)
(Given : Molal boiling point elevation constant is $\left.0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
(Given : Molal boiling point elevation constant is $\left.0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
Correct answer: 25
Solution
For AB
$\begin{aligned}<br />& \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \\<br>& 2.7=1 \times 0.5 \times \mathrm{m} \\<br>& \mathrm{~m}=\frac{27}{5}<br />\end{aligned}$
Let molar mass of $\mathrm{AB}=\mathrm{x}$.
$\begin{aligned}<br />& \text { So } \frac{1 / x}{15} \times 1000=\frac{27}{5} \\<br>& x=12.34<br />\end{aligned}$
For $\mathrm{AB}_2$
$\begin{aligned}<br />& \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \\<br>& 1.5=1 \times 0.5 \times \mathrm{m} \\<br>& \mathrm{~m}=3<br />\end{aligned}$
Let molar mass of $\mathrm{AB}_2=\mathrm{y}$
So $\frac{1 / \mathrm{y}}{15} \times 1000=3$
$\begin{aligned}<br />& y=\frac{1000}{45} \\<br>& y=22.22<br />\end{aligned}$
Now let a and b be atomic masses of A and B respectively, then
$\begin{aligned}<br />& \mathrm{A}+\mathrm{b}=12.34 \quad...(i)\\<br>& \mathrm{~A}+2 \mathrm{~b}=22.22 \quad...(ii) \\<br>& \mathrm{~B}=22.22-12.34=9.88 \\<br>& \text { Now } \mathrm{a}=12.34-9.88=2.46 \\<br>& =24.6 \times 10^{-1}=25 \times 10^{-1}<br />\end{aligned}$
$\begin{aligned}<br />& \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \\<br>& 2.7=1 \times 0.5 \times \mathrm{m} \\<br>& \mathrm{~m}=\frac{27}{5}<br />\end{aligned}$
Let molar mass of $\mathrm{AB}=\mathrm{x}$.
$\begin{aligned}<br />& \text { So } \frac{1 / x}{15} \times 1000=\frac{27}{5} \\<br>& x=12.34<br />\end{aligned}$
For $\mathrm{AB}_2$
$\begin{aligned}<br />& \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \\<br>& 1.5=1 \times 0.5 \times \mathrm{m} \\<br>& \mathrm{~m}=3<br />\end{aligned}$
Let molar mass of $\mathrm{AB}_2=\mathrm{y}$
So $\frac{1 / \mathrm{y}}{15} \times 1000=3$
$\begin{aligned}<br />& y=\frac{1000}{45} \\<br>& y=22.22<br />\end{aligned}$
Now let a and b be atomic masses of A and B respectively, then
$\begin{aligned}<br />& \mathrm{A}+\mathrm{b}=12.34 \quad...(i)\\<br>& \mathrm{~A}+2 \mathrm{~b}=22.22 \quad...(ii) \\<br>& \mathrm{~B}=22.22-12.34=9.88 \\<br>& \text { Now } \mathrm{a}=12.34-9.88=2.46 \\<br>& =24.6 \times 10^{-1}=25 \times 10^{-1}<br />\end{aligned}$
Hence, the correct answer is 25.