JEE Main 2025 Apr 3 Shift 1, Chemistry Q16: Colligative Properties

JEE Main2025Apr 3, Shift 1Chemistry

2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is :

(Given : Ebullioscopic constant of water = $0.52$ K kg mol $^{- 1}$ )

Solution

Both solutes are non-electrolytes ( $i = 1$ ). Total moles of solute particles: $2 + 2 = 4$ mol in 0.5 kg of solvent.

Total molality: $m = \frac{4}{0.5} = 8$ mol kg $^{- 1}$ .

Elevation: $Δ T_{b} = K_{b} m = 0.52 \times 8 = 4.16$ K.

Boiling point of water = 373.15 K, so

$T_{b} (soln) = 373.15 + 4.16 \approx 377.3 K .$

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