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

Explanation

As solution is 1 molal, it means that 1 mole of solute is dissolved in 1000 g of solvent (water).
No. of moles of solute = 1 mol, `p_("solvent")^(@)=12.3kPa`
No. of moles of water `=(1000)/(18)=55.5` mol
Total moles = 56.55 mol
Mole fraction of water `=(chi_(1)("solvent"))/(chi_(1)+chi_(2)("solution"))=(55.55)/(56.55)`
`p_("solution")=p_("solution")^(@)xx chi_(H_(2)O)`
`p_("solution")=12.3xx (55.55)/(56.55)=12.08kPa`