The vapor pressure of water at 338 K is 187.5 torr. Calculate the vapor pressure of water above a solution prepared by adding 16.2 g of lactose (C12H22O11) to 105.7 g of water at 338 K.
The molar mass of lactulose is =(12*12+22*1+16*11) = 342
From Rauolt’s law, we know, relative lowering of vapour pressure is equal to the mole fraction of the solute.
Mathematically, `(P_1 - P_1^o)/P_1^o = x_2 `
Here, p1(o) = 187.5 torr,
P1 has to be determined.
x2 = mole fraction of the solute = `(16.2/342)/(16.2/342+105.7/18)`
=0.047368/(0.047368+5.87222) = 0.008
Putting these values, in the vapour pressure equation, (p1-187.5)/187.5 = 0.008
or, p1 - 187.5 = 1.5
or, p1 = 187.5 + 1.5
So, the vapour pressure over of water above lactulose solution is 189 torr.