Question

a) The vapor pressure of pure water at 30°C is 31.8 torr. An aqueous solution of urea had a vapor pressure of 29.3 torrat the same temperature. What is the mole fraction of urea in the solution?

b) How many grams of sucrose, C12H22O11(MW=342), must be dissolved in 552 g H2O (MW=18.016) to give a vapor pressure 2.0 torr lower than of pure H2O at 20°C? The VP of H2O at 20°C is 17.5 torr.

Answer 1

Solution:

According to Rault's law, relative lowering of vapour pressure is equal to mole fraction of solute.

P1° - P1 / P1° = X2

Where, P1° = vapour pressure of pure solvent

P1 = vapour pressure of solution

X2 = mole fraction of solute

Part A)

Given, P1° = 31.8 torr

P1 = 29.3 torr

Thus,

P1° - P1 / P1° = X2

31.8 torr - 29.3 torr / 31.8 torr = X2

X2 = 0.0786

Part B)

P1° - P1 / P1° = X2 = wM /mW

Where,

w = mass of sucrose =?

m = molar mass of sucrose = 342 g mol-1

W = mass of water = 552 g

M = molar mass of water = 18.016 g mol-1

P1° = 17.5 torr

P1° - P1 = 2.0 torr

Thus,

2 / 17.5 = w x 18.016 g mol-1 / 342 g mol-1 x 552 g

315.28 x w = 552 g x 342 x 2

w = 1197.56 g

Hence, mass of sucrose = 1197.56 g