At `25^(@)`C temperature, vapour pressure of water is 23.55mm. What will be the vapour pressure of the solution when 6g of urea is mixed with 100g of water?

Explanation

No. of moles of urea present in solution`=6/(60)=0.1`
[`:.` molar mass of urea=60g]
No. of moles of water present in solution `=(100)/(18)=5.56`
[`:.` molar mass of water=18g]
Therefore, according to Raoult's law, we can write
`(p^0-p)/(p^0)=x^2`
`:." "(23.55-p)/(23.55)=0.0177`
or `p=23.55-23.55xx0.0177`
or `p=23.133mm`