Question

A 500.0 mL solution of NaNO3 in water has a vapor pressure of 21.445 mm Hg at 25°C. How many grams of NaNO3 (molar mass = 84.994 g/mol) were added to the water if the vapor pressure of pure water at 25°C is 23.76 mm Hg. Assume the volume of the water (d = 1.000 g/mL) is the same as the volume of the solution.

Answer 1

According to Raoult’s law:

P = Po*X(solvent)

21.445 = 23.76*X(solvent)

X(solvent) = 0.9026

This is mole fraction of H2O

mass(H2O)= 500 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(5*10^2 g)/(18 g/mol)

= 27.78 mol

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

0.9026 = 27.78 / ( 27.78+n(solute))

25.07+0.9026*n(solute) = 27.78

0.9026*n(solute) = 2.706

n(solute) = 2.999 mol

Molar mass of NaNO3,

MM = 1*MM(Na) + 1*MM(N) + 3*MM(O)

= 1*22.99 + 1*14.01 + 3*16.0

= 85 g/mol

use:

mass of NaNO3,

m = number of mol * molar mass

= 2.999 mol * 85 g/mol

= 255 g

Answer: 255 g