The density of a 0.438 M solution of potassium chromate at 298 K is $1.063 g cm^{-3}$. Calculate the vapour pressure of water above this solution. Given: $P^0$ (water) = 23.79 mm Hg.

A solution of 0.438 M means 0.438 mol of $K_2 CrO_4$ is present in 1L of the solution.
Now, Mass of K2CrO4 dissolved per litre of the solution = $0.438 \times 194 = 84.972 g$
Mass of 1L of solution = $1000 \times1.063 = 1063$ g
Assuming $K_2CrO_4$ to be completely dissociated in the solution, we will have;
Amount of total solute species in the solution = 3 $\times$ 0.438 = 1.314 mol.
Finally, Vapour pressure of water above solution = 0.976 $\times$ 23.79 = 23.22 mm Hg