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 of solute means 1 mole of solute in 1000 g of the solvent.
Molar mass of water (solvent)`=18 " g mol"^(-1)`
`therefore " Molar of water"=(1000)/(18)=55.5` moles.
`therefore " Mole fraction of solute"=(1)/(1+55.5)=0.0177`
Now,`(P^(@)-P_(s))/(P^(@))=x_(2)`
`(12.3-P_(s))/(12.3)=0.0177`
`impliesP_(s)=12.08 kPa`