Question

30g of urea is dissolved in 846 g of water.calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.

Answer 1

Solution :-

Vapour pressure of water, p10 = 23.8 mm Hg Weight of water = 846 g

Weight of urea = 30 g Molecular weight of

water (H2O) = 1 × 2 + 16 = 18 g mol−1Molecular weight of urea (NH2CONH2)
= 2N + 4H + C + O

= 2 × 14 + 4 × 1 + 12 + 16

= 60 g mol−1

Number of moles of water n1 = 846 / 18 = 47

Number of mole of urea n2 = 30 / 60 = 0.5

We take vapour pressure as p1.

Raoult’s law =

(P10 - P1)  / P10  = n2 / (n1-n2)

(23.8 - p1) / 23.8 = 0.5 / (47+0.5)

(23.8 - p1) / 23.8 = 0.5106 { cross multiplay }

23.8 – p1 = 23.8 × 0.5106

p1 = 11.6 mm Hg

vapour pressure of water in the given solution= 11.6 mmHg

============

I hope it's help you....!!! :)

Answer 2

Hey !!

(P°a - Pa ) / P°a = ( wb × Ma ) / ( Mb × wa )

  = 23.8 - Pa / 23.8 = ( 30 × 18 ) / 60 × 846

   = 23.8 - Pa = 23.8 × [ ( 30 × 18 ) / 60 × 846 ]

    = 23.8 - Pa = 0.2532

    Pa = 23.55 mm Hg

Good luck !!