Q18 / 75 · JEE Main 2026 (21 January Shift 1) · Chemistry
Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and $\mathrm{PQ}_{2}$. When 1 g of PQ is dissolved in 50 g of solvent ${ }^{\prime} \mathrm{A}^{\prime}, \Delta \mathrm{T}_{\mathrm{b}}$ was 1.176 K while when 1 g of $\mathrm{PQ}_{2}$ is dissolved in 50 g of solvent ${ }^{\prime} \mathrm{A}^{\prime}, \Delta \mathrm{T}_{\mathrm{b}}$ was 0.689 K. ($\mathrm{K}_{\mathrm{b}}$ of ' $\mathrm{A}^{\prime}=5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$). The molar masses of elements P and Q (in $\mathrm{g} \mathrm{mol}^{-1}$) respectively, are :
- A$70,110$
- B$65,145$
- Correct answerC$25,60$
- D$60,25$
Solution
Using $\Delta T_b = K_b \times \frac{w}{M} \times \frac{1000}{W}$
For PQ: $1.176 = 5 \times \frac{1}{M_{PQ}} \times \frac{1000}{50} = \frac{100}{M_{PQ}}$
$M_{PQ} = \frac{100}{1.176} = 85$ g/mol
For $PQ_2$: $0.689 = \frac{100}{M_{PQ_2}}$
$M_{PQ_2} = \frac{100}{0.689} = 145$ g/mol
$P + Q = 85$ ... (i)
$P + 2Q = 145$ ... (ii)
Subtracting: $Q = 60$ g/mol
From (i): $P = 85 - 60 = 25$ g/mol
For PQ: $1.176 = 5 \times \frac{1}{M_{PQ}} \times \frac{1000}{50} = \frac{100}{M_{PQ}}$
$M_{PQ} = \frac{100}{1.176} = 85$ g/mol
For $PQ_2$: $0.689 = \frac{100}{M_{PQ_2}}$
$M_{PQ_2} = \frac{100}{0.689} = 145$ g/mol
$P + Q = 85$ ... (i)
$P + 2Q = 145$ ... (ii)
Subtracting: $Q = 60$ g/mol
From (i): $P = 85 - 60 = 25$ g/mol
Hence, the correct answer is option: 'C'.