Question

16.636 g of a non-volatile solute is dissolved in 410.0 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 90°C the vapour pressure of the solution is 517.01 torr.
The vapour pressure of pure water at 90°C is 525.80 torr.  The molar mass of the solute is 43.0 g/mol.

Now suppose, instead, that 16.636 g of a volatile solute is dissolved in 410.0 g of water.
This solute also does not react with water nor dissociate in solution.
The pure solute displays, at 90°C, a vapour pressure of 52.58 torr.
Again, assume an ideal solution.
If, at 90°C the vapour pressure of this solution is also 517.01 torr.
Calculate the molar mass of this volatile solute.

Answer 1

vapor pressure of solution = vapor pressure of water * molefraction

         517.01                      = 525.8* mole fraction

mole fraction of water = 517.01/525.8 = 0.983
no of moles of water = W/G.M.Wt = 410/18 = 22.8moles
mole fraction of water = no of moles of water/no of moles of water + no of moles of solute
                0.983             = 22.8/22.8+x

                 0.983*(22.8+x) = 22.8

                        x = 0.394
no of moles of solute = 0.394 moles
molar mass of solute = 16.636/0.394   = 42.22g/mole