A KCl solution is prepared by dissolving 40.0 g KCl in 250.0 g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?

Answer 1

Raoult's Law:

Vapor pressure of a solution containing a non volatile solute (KCl) is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent,

So, we will first find moles of KCl (and moles of K⁺ and Cl^-), add that to moles of H₂O to get total moles, and then find mole fraction of H₂O:

moles KCl = 40.0 g KCl x 1 mol KCl/74.6 g = 0.536 mol KCl x 2 = 1.07 moles of ions (K⁺ + Cl^-)

moles H₂O = 250.0 g H₂O x 1 mol H₂O/18 g = 13.9 moles H₂O

Total moles = 13.9 + 1.07 = 14.97 moles

mole fraction H₂O = 13.9 / 14.97 = 0.9285

Vapor pressure of solution = 0.9285 x 23.76 mm Hg = 22.06 mm Hg