This question is incomplete, the complete question is;
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 = 0.0306 atm
Explanation:
we know that; 1 L = 1000 mL of the solution
Mass of solution = volume x density
= 1000 x 1.063 = 1063 g
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 = 1063 - 85.055 = 977.945 g
Moles of water = mass / molar mass
977.945 / 18.02 = 54.27 mol
K2CrO4 => 2 K+ + Cr2O42-
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
