Question

calculate the vapor pressure of water above a solution prepared by adding 16.2g of lactose, C12H22O11, to 105.7g of water at 338K.

Answer 1

From Rauolt’s law, we know, relative lowering of vapour pressure is equal to the mole fraction of the solute.

that means

no of moles of water = 105.7 / 18 = 5.87 moles

no of moles of water = 16.1 / 342 = 0.047 moles

mole fraction (x2 ) of solute = 0.047 mol / ( 5.87 mol + 0.047 mol ) =0.047 / 5.91 = 0.0079

The vapor pressure of water at 338 K is 187.5 torr, i.e P1^(o) = 187.5 torr, we need to calculate P1.

P1 -187.5 / 187.5 = 0.0079

P1 -187.5 = 0.0079*187.5 = 1.5

P1 = 1.5+187.5 = 189

the vapour pressure of water above (Lactulose solution) is 189 torr.