JEE Main 2026 — Jan 21, Shift 1 — Chemistry solution

JEE Main2026Jan 21, Shift 1Chemistry

Elements P and Q form two types of non-volatile, non-ionizable compounds $P Q$ and $P Q_{2}$ . When 1 g of $P Q$ is dissolved in 50 g of solvent 'A', $Δ T_{b}$ was 1.176 K, while when 1 g of $P Q_{2}$ is dissolved in 50 g of solvent 'A', $Δ T_{b}$ was 0.689 K. ( $K_{b}$ of 'A' = 5 K kg mol $^{- 1}$ ). The molar masses of elements P and Q (in g mol $^{- 1}$ ) respectively, are:

A
70, 110
B
65, 145
C
60, 25
D
25, 60

Solution

$Δ T_{b} = K_{b} \cdot m$ . For $P Q$ : $1.176 = 5 \cdot \frac{1}{M _{1}} \cdot \frac{1000}{50} \Rightarrow M_{1} = 85.03$ .

For $P Q_{2}$ : $0.689 = 5 \cdot \frac{1}{M _{2}} \cdot \frac{1000}{50} \Rightarrow M_{2} = 145.13$ .

$M_{P} + M_{Q} = 85.03$ and $M_{P} + 2 M_{Q} = 145.13$ , giving $M_{Q} = 60.1 \approx 60$ and $M_{P} = 24.93 \approx 25$ .

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