Question

22.0 grams of a nonvolatile solute are added to 60.0 grams of water. The vapor pressure above the solution is 21.2 Torr. Pure water has a vapor pressure of 23.8 at this temperature. What is the molecular weight of the solute?

Answer 1

According to Raoult’s law:

P = Po*X(solvent)

21.2 = 23.8*X(solvent)

X(solvent) = 0.8908

This is mole fraction of H2O

mass of H2O = 60 g

we have below equation to be used:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(60.0 g)/(18 g/mol)

= 3.33 mol

X(H2O) = n(H2O)/( n(H2O) + n(unknown))

0.8908 = 3.33 / ( 3.33+n(unknown))

2.97+0.8908*n(unknown) = 3.33

0.8908*n(unknown) = 0.36

n(unknown) = 0.404 mol

mass of solute = 22.0 g

we have below equation to be used:

number of mol = mass / molar mass

0.404 mol = (22.0 g)/molar mass

molar mass = 54.4 g/mol

Answer: 54.4 g/mol