At 100∘C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb=0.52, the boiling point of this solution will be
101∘C
Given WB=6.5g, WA=100g,
ps=732 mm, Kb=0.52, T0b=100∘C, p∘=760mm
p∘−psp0=n2n1⇒760−732760=n210018
⇒ n2=n2×1000760×18=0.204 mol
ΔTb=Kb×m
Tb−100∘C=Kb×n2×1000wA(g)
Tb−100∘C=0.25×0.2046×1000100=1.06
Tb=100+1.06=101.06∘C