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 of the solute. Express your answer in atmospheres to three significant figures. Pvap = atm

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

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