Question

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

Answer 1

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 
No. of moles = Mass /Molar Mass

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
(23.8 - p1) /23.8 = 0.83/(47.22+ 0.83)
Cross multiply

23.8 – p1 = 23.8 × 0.0173

Solve it we get

p1 = 23.4 mm Hg

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

Relative lowering = 0.0173.

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