Question

Calculate the mass of propylene glycol (C3H8O2) that must be added to 0.350 kg of water to reduce the vapor pressure by 2.87 torr at 40 ∘C (PH2O at 40 ∘C=55.3torr).

Answer 1

According to Raoult’s law:

P = Po*X(solvent)

2.87 = 55.3*X(solvent)

X(solvent) = 0.0519

This is mole fraction of H2O

mass(H2O)= 0.350 Kg

= 350.0 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(3.5*10^2 g)/(18 g/mol)

= 19.44 mol

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

5.19*10^-2 = 19.44 / ( 19.44+n(solute))

1.009+5.19*10^-2*n(solute) = 19.44

5.19*10^-2*n(solute) = 18.44

n(solute) = 3.552*10^2 mol

Molar mass of C3H8O2,

MM = 3*MM(C) + 8*MM(H) + 2*MM(O)

= 3*12.01 + 8*1.008 + 2*16.0

= 76.094 g/mol

use:

mass of C3H8O2,

m = number of mol * molar mass

= 3.552*10^2 mol * 76.09 g/mol

= 2.703*10^4 g

Answer: 2.70*10^4 g