A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is . (Nearest integer) [Given: boiling point of the pure solvent = 319.5 K, Kb of the solvent = 0.3 K kg mol-1]
Answer
5
Solution
Elevation of boiling point: Tb = 320 - 319.5 = 0.5 K
Tb = Kb m ⇒ m = 0.50.3 = 53 mol kg-1
Moles of solute, n = m Wsolvent = 53 0.040 = 115 mol
Relative lowering of vapour pressure (Raoult's law): p - pp = nn + N
⇒ 760 - 750760 = 176 = nn + N
⇒ n + N = 76n ⇒ N = 75n = 75 115 = 5
∴ Moles of solvent = 5
Answer: 5