Question

16.177 g of a non-volatile solute is dissolved in 290.0 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 54.506 torr. The vapour pressure of pure water at 40°C is 55.324 torr. Calculate the molar mass of the solute (g/mol).

Now suppose, instead, that 16.177 g of a volatile solute is dissolved in 290.0 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 54.506 torr. Calculate the molar mass of this volatile solute.

Answer 1

raoults law

   P0-P/P0 = i*Xsolute

    P0-P/P0 = n2 / n1+n2

i = vanthoffs factor of solute = 1

P = vapor pressure of water above the solution = 54.506 torr

p0 = vapor pressure of pure water at this temperature = 55.324 torr

   n1 = no of mol of solvent = (290/18) = 16.11 mol

   n2 = no of mol of solute particles = (16.177/x) mol

(55.324-54.506)/55.324 = ((16.177/x)/((16.177/x)+16.11)

x = molarmass of solute = 66.91 g/mol

from daltons law

Ptotal = Psolute + pSolvent

         = Xsolute*P0Solute + Xsolvent*P0solvent

    54.506 = ((16.177/x)/((16.177/x)+16.11)*5.532 + ((16.11)/((16.177/x)+16.11)*55.324

x = molarmass of solute = 60.11 g/mol