A solution contains 0.159 mol K3PO4 and 0.941 molH2O. Calculate the vapor pressure of the solution at 55 ∘C. The vapor pressure of pure water at 55 ∘C is 118.1 torr. (Assume that the solute completely dissociates.)

49.8 Torr will be the vapor pressure of the solution

Explanation:

Vapor pressure lowerig is the colligative property that must be applied to solve this. The formula is: ΔP = P°. Xm. i

where ΔP = P° (vapor pressure of pure solvent) - Vapor pressure of solution (P')

Xm, the mole fraction of solute (moles of solute / total moles)

i = Van't Hoff factor (numbers of ions dissolved in the solution)

The solute, potassium phosphate dissociates like this:

K₃PO₄ → 3K⁺ + PO₄³⁻ so we have 3 moles of cation K⁺ and 1 mol of phosphate. Therefore, we have 4 moles of ions, is we assume that the solute completely dissociates → i = 4

Let's find out the mole fraction for solute (Xm)

Xm = mol of solute / Total moles

where total moles = mole of solute + moles of solvent

0.159 mol + 0.941 mol = 1.1 moles

Xm = 0.159 / 1.1 → 0.144

Let's replace the data in the formula

P° - P' = P°. Xm. i

118.1 Torr - P' = 118.1 Torr. 0.144. 4

P' = - (118.1 Torr. 0.144. 4 - 118.1 Torr)

Vapor pressure for solution is : 49.8 Torr