Answer :
The question is incomplete, the solute was not given.
Let the solute be K₂CrO₄ and the solvent be water
Complete Question should be like this:
The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.
Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.
Pvap = ________atm
Answer:
Pvap (of water above the solution) = 0.0306 atm
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
Explanation:
Given
volume of solution = 1 Litre = 1000 mL of the solution
density of the solution = 1.063 g/mL
concentration of the solution= 0.438M
temperature of the solution= 298 K
vapour pressure of pure water = 0.0313atm
Recall: density = mass/volume
∴mass of solution = volume x density
m = 1000 x 1.063 = 1063 g
To calculate the moles of K₂CrO₄ = volume x concentration
= 1 x 0.438 = 0.438 mol
Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g
Mass of water = mass of solution - mass of K₂CrO₄
= 1063 - 85.055 = 977.945 g
moles of water = mass/molar mass
∴ moles of water = 977.945/18.02 = 54.27 mol
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution
moles of ions = 3 x moles of K₂CrO₄
= 3 x 0.438 = 1.314 mol
Vapor pressure of solution = mole fraction of water x vapor pressure of water
= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm