Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH_2CONH_2)is dissolvedin 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_2CONH_2)is dissolvedin 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Answer» SOLUTION :The DATA provided is as GIVEN below : `p^0 = 23.8 mm` `w_2 = 50 g , M_2 ("urea") = 60 g "mol"^(-1)` ` w_1 = 850 g, M_1 ("water") = 18 g "mol"^(-1)` Applying Raoult.s law, `(p^0 - p_x)/(p^0) = (n_2)/(n_1+ n_2) = (w_2 // M_2)/(w_1 // M_1 + w_2 // M_2)` Substituting the values, we get `(p^0 - p_x)/(p^0) = (50 // 60)/(850//18 + 50//60) = (0.80)/(47.22 + 0.83) = 0.017` Thus, RELATIVE lowering of VAPOUR pressure = 0.017 Vapour pressure of water for this solution means `P_s` Substituting `p^0` = 23.8 mm, we get `(23.8 - p_0)/(p_s) = 0.017 " or " 23.8 -p_s = 0017 p_s` `1.017 p_s = 23.8 " or " p_s = 23.4 mm` Thus, vapour pressure of water in the solution = 23.4 mm Hg.