Question

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Answer 1

vapor pressure of water , $p_l^{\circ}$=23.8mm Hg
weight of water, W = 850g
weight of Urea, w = 50g
molar mass of water ,M = 18g/mol
molar mass of Urea, m = 60 g/mol

so, mole of water, N = W/M = 850/18 = 47.22mol
mole of Urea,n = w/m =50/60 = 0.83 mol
Let vapor pressure of water in the solution, $P_l$

now, use Rault's theorem,
$\frac{P_l^{\circ}-P_l}{P_l^{\circ}}=\frac{n}{n+N}$
=> (23.8 - Pl)/23.8 = 0.83/(0.83 + 47.22)
=> (23.8 - Pl)/23.8 = 0.83/48.05
=> (23.8 - Pl)/23.8 = 0.0173
=> Pl = 23.4 mm Hg

hence, The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.

Answer 2

Answer:

Explanation:

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.