General Chemistry II

Question

The addition of 44.6g of a nonvolatile solute to 2.30 mol of benzene lowers the vapor pressure of benzene from 105.8 torr to 97.0 torr. Calculate the molar mass of the solute.

Answer 1

This is an application of Raoult’s Law that states that the vapor pressure of a solution is given by the mole fraction of solvent in the solution multiplied by the vapor pressure of the pure solvent at a given temperature. In equation form, this is

P = X • P^o  where P is the vapor pressure of the solution, P^o is vapor pressure of pure solvent, and X is the mole fraction of solvent.

P = 97.0 torr

P⁰ = 105.8 torr

X is then 97.0 torr / 105.8 torr = 0.917

X is also given by moles solvent / (moles solute + moles solvent), and moles solute is solute mass divided by the molar mass of the solute which we are trying to find.

So, X = (Moles solvent) / (moles solute + moles solvent) that in equation form is

X = M_solvent / (M_solute + M_sovent)

Solve for M_solute to get

M_solute = M_solvent (1 – X) / X

M_solute = 2.30 (1 – 0.917) / 0.917 = 0.208 moles solute.

 But M_solute = Mass solute / Molar Mass, so Molar Mass = Mass solute / M_solute = 44.6g / 0.208 moles = 214 g/mole.