A solution of a non-volatile solute in water freezes at −0.40oC. The vapour pressure of pure water at 298 K is 23.51 torr. For water, Kf=1.86 K mol−1 kg. Thus, vapour pressure of the solution (in torr)

For (w1m1) mole of non-volatile solute in w2 g solvent, depression in freezing point is ΔTf=1000Kfw1m1w20.40=1000×1.86w1m1w2∴w1m1w2=401000×186By Raoult's law, po−pspo=n1n1+n2=n1n2=w1m1m2w21−pspo=40×181000×186=72186001−ps23.51=3.87×10−3ps=19.64 torr