Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH_(2) CONH_(2)) is dissolved is 850 g of water. Calculate the vapour pressure of water for 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 is 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Answer» Solution :Here `P _(1) ^(0) =23.8mm` `W _(2)= 50 g` `M_(2) `(urea) `=60g mol^(-1)` `W_(1)= 850G` `M _(1) `(Water) `= 18g mol ^(-1)` Here we have to CALCULATE `P _(S)` Applying Raoult.s law, `(P ^(0) - P_(S))/(P^(0))= (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)/(48.05) =0.017` Thus, relative lowering of VAPOUR pressure `=0.017` Substituting `P^(0) =23.8` mm Hg `(23.8 -P_(s))/(P _(S)) = 0.017` We get, `23.8 - P_(S) = 0.017 P _(S)` `P _(S) =23.4 mm Hg` Thus, vapour pressure of water in the solution `=23.4` mm Hg