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 of the solute dissolved in 1000 g (1 kg) of solvent i.e. water.
`"No. of moles of solute" (n_(B)) 1 mol`
`"No. of moles of water" (n_(A))=("Mass of water")/("Molar mass")=((1000g))/((18 g mol^(-1)))=55.55 mol`
`"Mole fraction of solute" (x_(B))=n_(B)/(n_(B)+n_(A))=((1mol))/((1mol+55.55mol))`
`=1/(56.55)=0.0177`
`"Vapour pressure of solution" (P_(A))=P_(A)^(@)X_(A)=P_(A)^(@)(1-x_(B))=12.3kPaxx(1-0.0177)`
`=12.3 kaxx0.9823=12.08 kPa`