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

Explanation

1 molal solution implies one mole of the solute dissolved in 10 g (1 kg)( of solvent i.e., water.
`" No. of moles of solute"(n_(B))=1 mol`
`"No. of hmoles of water "(n_(B))=n_(B)/(n_(B)+n_(A))=((1 mol))/((1 mol +55.55 mol))`
`=1/(56.55)=0.0177 `
`" Vopour pressure of solution "=P_(A)^(@)x_(A)=P_(A)^(@)(1-x_(B))=12.3kPaxx(1-0.0177)`
`=12.3 kPa xx 0.9823=12.08kPa`