The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Explanation

We are given that
Vapour pressure of water = 12.3 k Pa
Temperatue = 300 K
We are to calculate the vapour pressure of 1 molar solution.
1 molar solution means 1 mol of the solution in 1 L of the solution = 1000 g of the solvent, `H_(2)O` (assuming the solution to be dilute).
Mole fraction of solute = `1/(1+55.5)=0.0177`
Now `(p^(@)-p_(s))/(p^(@))=x_(2)`, i.e., `(12.3-p_(s))/(12.3)=0.0177`
or `p_(s)=12.08" kPa"`.