Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea NH2CONH2)

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.

Solution

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.