Question

Vapour pressure of pure water is 23.8 mm of hg. 50gm of urea (Nh2conh2) is dissolved in 850gm of water. Calculate the vapour pressure of solution?

Answer 1

Explanation:

Given that

Vapour pressure of water,p1 = 23.8 mm of Hg

Weight of water = 850 g

Weight of urea = 50 g

Molecular weight of water(H2O) = 1 × 2 + 16 = 18 g mol−1

Molecular weight of urea(NH2CONH2)= 2N + 4H + C + O

= 2 × 14 + 4 × 1 + 12 + 16

= 60 g mol−1

Use formula

Number of moles of water n1 = 850 / 18 = 47.22

Number of mole of urea n2 = 50/60 = 0.83

Now, we have to calculate vapour pressure of water in the solution. We take vapour

pressure as p1.

Use the formula of Raoult’s law

Plug the values we get

Cross multiply

23.8 – p1 = 23.8 × 0.0173

Solve it we get

p1 = 23.4 mm Hg

Answer

Vapour pressure of water in the given solution = 23.4 mm of Hg

Relative lowering = 0.0173.