If CaCl2 is completely ionized in water, what does that do to the vapor pressure of it?

Ok the actual question is:

A solution of 22.0 g of CaCl2 in 100.0 cm^3 of water has a vapor pressure of 22.8 torr at 26°C. Why does this result support the view that CaCl2 is completely ionized to Ca2+ and Cl- in aqueous solution? (At 26°C the vapor pressure of H2O is 25.2 torr and the density of H2O is 1.00 g/cm^3 ; CaCl2 is non-volatile)

Please help?? THANK YOU!!

This problem requires Raoult’s law, which states that the vapor pressure of a solution (at a particular temperature) is equal to the VP of the pure solvent, times the mole fraction of solvent in the solution. The mole fraction of solvent is given by Xsolv = moles solvent/(moles solvent + moles solute).

If you calculate the number of moles of CaCl2 and H2O in the solution, you get 0.2 moles of CaCl2 of solute and 55.6 moles of H2O. The mole fraction of H2O in the solution would be ~55.6/55.8 = 0.97.

Applying Raoult’s law gives Pvap (soln) = 0.97 x 25.2 torr = 24.33 torr, which doesn’t match the result given.

If you multiply the moles of solute by three, however, the math works out. What that means is that if you dissolve one mole of CaCl2, you actually get 3 moles of solute particles, and it’s the number of particles dissolved that controls colligative properties like vapor pressure changes. One CaCl2 gives three particles when it dissolves in solution.