A solution of a non-volatile solute in water has a boiling point of 375.3K. Calculate the vapour pressure of water above this solution at 338K. Given, `p_(0)` (water) = 0.2467 atm at 338K and `K_(b)` for water = 0.52.

Explanation

`DeltaT_(b)=(375.3-373.16)=2.15K`
We know that,
`DeltaT_(b)`= Molality `xx K_(b)`
2.15 = Molality `xx 0.52`
Molality `= (2.15)/(0.52) = 4.135`
i.e., 4.135 moles of the solute present in 1000g of water (55.5 moles)
Mole fraction of water `= (55.5)/(4.135+55.5)=(55.5)/(59.635)`
Vapour pressure of water above solution
= Mole fraction `xx p_(0)`
`=(55.5)/(59.635)xx0.2467=0.23`atm.