An aqueous CaCl2 solution has a vapor pressure of 82.2 mmHg at 50 degrees C. The vapor pressure of pure water at this temperature is 92.6 mmHg. What is the concentration of CaCl2 in mass percent?
From Raoult’s law:
Vapor pressure of solution = mole fraction of water x vapor pressure of water
mole fraction of water = 82.2/92.6 = 0.8877
mole fraction of Ca2+ and Cl– ions = 1 – 0.8877 = 0.1123
For every 0.8877 moles of water, there are 0.1123/3 = 0.03743 moles of CaCl2 since CaCl2 => Ca2+ + 2Cl– and each CaCl2 gives 3 solute ions.
Since mass = moles x molar mass,
mass% CaCl2 = mass of CaCl2/(mass of CaCl2 + mass of H2O) x 100
= (0.03743 x 110.98)/(0.03743 x 110.98 + 0.8877 x 18.02) x 100