Question

An aqueous solution freezes at -0.2°C. What is the molality of the solution? Determine also (i) elevation in boiling point (ii) lowering in vapour pressure at 25°C, given that Kf = 1.86°C Kg mol-1, Kb = 0.512°C Kg mol-1 and vapour pressure of water at 25°C is 23.756 mm.

Answer 1

molality=0.1075 (i)0.055℃ (ii) 0.046mm

Explanation:

∆Tf=Kf×m

m=∆Tf/Kf. =0.2÷1.86=0.1075

(i) ∆Tb=Kb.m.

∆Tb=0.512×0.1075=0.055℃