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

Given Molality of solution `=1m`
vapour pressure of solvent `=12.3kPa`
Asked Vapour pressure of solution `=?`
Formulae: `(P_(A)^(0)-P_(A))/(P_(A)^(0))=x_(B)`
Explanation: `P_(A)^(0)=` vapour pressure of pure solvent
`P_(A)=` vapour pressure of solution
`x_(B)=` Mole fraction of solute
Calculation of `x_(B)`
`x_(B)=(m)/(m+(1000)/(M_(A)))=(1)/(1+(1000)/(18))=(18)/(1018)`
Substitution in formulae
`(P_(A)^(0)-P_(A))/(P_(A)^(0))rArr1-(P_(A))/(P_(A)^(0))=x_(B)=(18)/(1018)`
`1-(P_(A))/(12.3)=(18)/(1018)`
`rArr(P_(A))/(12.3)=1-(18)/(1018)=(1000)/(1018)`
`P_(A)=(12.3xx1000)/(1018)=(12300)/(1018)=12.08kPa`.