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 )
- (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