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.

Vapour presure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH_

1.	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.
Answer» Solution :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.