Vapour pressure 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 of this solution and its relative lowering.

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

1.	Vapour pressure 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 of this solution and its relative lowering.
Answer» <P> Solution :Here,`""p^(@)="23.8 mm"` `""w_(2)=50g,""M_(2)"(urea) = 60 G mol"^(-1),""w_(1)=850 g, M_(1)" (water) = 18 g mol"^(-1)` Our aim is to calculate `p_(s) and (p^(@)-p_(s))//p^(@)` `"Applying Raoult's law."=(p^(@)p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))= (w_(2)//M_(2))/(w_(1)//M_(1) +w_(2)//M_(2)) =(50//60)/(850//18+50//60)` `=(0.83)/(47.22+0.83)=(0.83)/(48.05)=0.17` THUS, RELATIVE lowering of vapour pressure = 0.017 SUBSTITUTING `p^(@)=23.8` mm, we get `(23.8-p_(s))/(p^(@))=0.017 or(23.8-p_(s))/(23.8)=0.017or23.8-p_(s)=0.017 xx23.8=0.40` `"or"p_(s)=23.8-0.40="23.4 mm"` Thus, vapour pressure of water in the solution = 23.4 mm.