JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Solutions

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For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take

K_{b} = 0.76 K kg mol^{- 1}

)

(A)724✓ Correct
(B)740
(C)736
(D)718

Answer: (A)

Solution

From boiling point elevation: ΔTb = Kb × m, so 2 = 0.76 × m, giving m = 2.63 mol/kg. Using Raoult's law for dilute solution: (P⁰ - Ps)/P⁰ = (m × 18)/1000. With P⁰ = 760 mm Hg: (760 - Ps)/760 = (2.63 × 18)/1000 = 0.047. Solving: Ps = 760 - 35.7 ≈ 724 mm Hg. The answer is (A).

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option

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