Question

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

• A. 23.22
• B. 43.23
• C. 54.2
• D. 12.1

Answer 1

The correct option is A 23.22

A solution of 0.438 M means 0.438 moles of $K_{2}Cr{O}_4$ is present in 1L of the solution. Now,
Mass of $K_{2}Cr{O}_4$ dissolved per liter of the solution = $0.438\times 194=84.972g$
Mass of 1L of solution = $1000\times 1.063=1063g$
Amount of water in 1L of solution = $\frac{978.028}{18}=54.255mol$
Assuming $K_{2}Cr{O}_4$ to be completely dissociated in the solution, we will have;
Amount of total solute species in the solution = $3\times 0.438=1.314mol$.
Mole fraction of water solution = $\frac{54.335}{(54.335+1.314)}=0.976$
Finally, Vapour pressure of water above solution = $0.976\times 23.79=23.22mmHg$