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.

Calculate Vapor Pressure & its Relative Lowering

It is given that: Mass of water (solvent) = 850 g

Molar mass of water (H₂O) = 18 g/mol

Mass of urea (solute) = 50 g

Molar mass of urea (NH₂CONH₂) = 60 g/mol

Vapor pressure of pure water, p⁰₁= 23.8 mm Hg

Solution:

Moles of water (n_water) = mass of water / molar mass of water, n₁ = 850 g / 18 g/mol

Moles of urea (n_urea) = mass of urea / molar mass of urea, n₂ = 50 g / 60 g/mol

Total moles of solute and solvent (n_total) = n_water + n_urea

Let us assume pressure as p₁.

According to the Raoult’s law,

(p⁰₁-p₁)/p⁰₁ = n₂/(n₁+n₂)

(23.8-p₁)/23.8 = (50/60)/[(850/18)+(50/60)]

(23.8-p₁)/23.8 = 0.83/[47.22+0.83]

(23.8-p₁)/23.8 = 0.0173

p₁= 23.4 mm Hg

Summary:

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.

The vapor pressure of water in the given solution is determined to be 23.4 mm Hg, and the relative lowering of the vapor pressure is 0.0173.

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