2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25 . The solution showed a boiling point elevation by 2 . Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is mm of Hg (nearest integer) [Given : Molal boiling point elevation constant of water (Kb) = 0.52 K kg mol-1, 1 atm pressure = 760 mm of Hg, molar mass of water = 18 g mol-1]
Answer
707
Solution
Tb = Kb m ⇒ m = 20.52 = 3.846 mol kg-1
Dilute solution: p - pp = nsolutensolvent = m 181000
∵ at its boiling point, p of water = 1 atm = 760 mm Hg
⇒ p - p = 760 × 3.846 181000 = 52.6 mm Hg
∴ p = 760 − 52.6 = 707.4 ≈ 707 mm Hg
Answer: 707