Vapour presure of pure water at 298 K is 23.8 mm Hg. 50 g of urea `(NH_(2)CONH_(2))` is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Explanation

It is given that vapour pressure of water,
`p_(A)^(0)=23.8` mm of Hg
Weight of water taken, `w_(1)=850 g`
Weight of urea taken, `w_(2)=50 g`
Molecular weight of water, `M_(1)=18 g mol^(-1)`
Molecular weight of urea, `M_(2)=60 g mol^(-1)`
Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as `p_(1)`.
Now, from Raoult.s law, we have :
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=(n_(2))/(n_(2)+n_(2))`
`therefore (p_(1)^(0)-p_(1))/(p_(1)^(0))=((w_(2))/(M_(2)))/((w_(1))/(M_(1))+(w_(2))/(M_(2)))`
`therefore (23.8-p_(1))/(23.8)=((50)/(60))/((850)/(18))+(50)/(60)`
`therefore (23.8-p_(1))/(23.8)=(0.83)/(47.22+0.83)`
`therefore (23.8-p_(1))/(23.8)=0.0173`
`p_(1)=23.4` mm of Hg
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.