Question

2) Vapour Pressure of Solutions of Non-Volatile or Volatile Solutes

A) 48.490 g of a non-volatile solute is dissolved in 455.0 g of water.
The solute does not react with water nor dissociate in solution.
Assume that the resulting solution displays ideal Raoult's law behaviour.
At 40°C the vapour pressure of the solution is 54.453 torr.
The vapour pressure of pure water at 40°C is 55.324 torr.
Calculate the molar mass of the solute (g/mol).

B) Now suppose, instead, that 48.490 g of a volatile solute is dissolved in 455.0 g of water.
This solute also does not react with water nor dissociate in solution.
The pure solute displays, at 40°C, a vapour pressure of 5.532 torr.
Again, assume an ideal solution.
If, at 40°C the vapour pressure of this solution is also 54.4

Answer 1

A) mass of non volatile m1 = 48.49 gm

Molar mass of non volatile solute = M1

Moles of solute n1 = m1/M1 = 48.49/M1

Water treated as solvent.

Mass of water m2 = 455 gm

Molar mass of water M2 =18 gm/mol

Moles of water, n2 = m2/M2 =455/18 = 25.277 mol

Mole fraction of solvent = Xsolvent = n2/(n1+n2) = 25.277/(48.49/M1 + 25.277)

At 40C, the vapor pressure of solution, Psolution= 54.453 torr

Vapor pressure of pure water(solvent),   Psolvent=55.324 torr

Roults law, solute doesn't react with water nor dissociate with solution.

Psolution = Xsolvent* Psolvent

54.453 = 25.277/(25.277+48.96/M1) * 55.324

25.277 + 48.96/M1 = 25.68

M1 = 121.48 gm/mol

Molar mass of solute M1 = 121.48 gm/mol