The vapour pressure of water is 12.3 kPa at 300 K. calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer

`P_(s)=12.08 kPa`

Explanation

`p^(@)=12.3kPa,p_(S)=?`
`M=1(1 mol e` of solute is dissolved in `1 L` of solution `)`
Molarity of water `=(1000)/(18)=55.56`
`(` or number of moles of water `=55.56)`
`(p^(@)-p_(S))/(p^(@))=CHMi_(2)=(n_(2))/(n_(1)+n_(2))=(n_(2))/(n_(1))(` for dilute solution `)`
`(12.3-p_(S))/(12.3)=(1)/(55.56)`
`12.30p_(S)=(12.3)/(55.56)=0.2213`
`p_(S)=12.3-0.2213=12.078~~12.08kPa`