JEE Main 2025 Apr 2 Shift 2, Chemistry Q24: Colligative Properties

JEE Main2025Apr 2, Shift 2Chemistry

When 1 g each of compounds AB and $A B_{2}$ 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 _______ $\times 1 0^{- 1}$ (Nearest integer)

(Given : Molal boiling point elevation constant is 0.5 K kg mol $^{- 1}$ )

Your answer

Solution

Using $Δ T_{b} = K_{b} \cdot m$ with $K_{b} = 0.5 K kg mol^{- 1}$ and a fixed solvent mass of 15 g:

$m = \frac{moles of solute}{0.015 kg} = \frac{1000/ M}{15}$ for a 1 g sample of molar mass $M$ .

For AB: $Δ T_{b} = 2.7 K \Rightarrow m = 5.4 \Rightarrow M_{A B} = \frac{1000}{15 \times 5.4} \approx 12.35$ .

For $A B_{2}$ : $Δ T_{b} = 1.5 K \Rightarrow m = 3.0 \Rightarrow M_{A B_{2}} = \frac{1000}{15 \times 3.0} \approx 22.22$ .

Let the atomic masses of A and B be $a$ and $b$ . Then:

$a + b = 12.35$ ... (i)

$a + 2 b = 22.22$ ... (ii)

Subtracting (i) from (ii): $b = 9.88$ . Substituting back: $a = 12.35 - 9.88 = 2.47$ amu.

Expressing as $\times 1 0^{- 1}$ : $a \approx 24.7 \times 1 0^{- 1} \approx 25 \times 1 0^{- 1}$ .

$∴$ Nearest integer $= 25$ .

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JEE Main 2025 Apr 2 Shift 2, Chemistry Q24: Colligative Properties

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