# 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.

By BYJU'S Exam Prep

Updated on: September 25th, 2023

To calculate the vapor pressure of water for the solution and its relative lowering, we need to apply Raoult’s law. Raoult’s law states that the vapor pressure of a component in a solution is proportional to its mole fraction in the solution.

## 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 (nwater) = mass of water / molar mass of water, n1 = 850 g / 18 g/mol

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

Total moles of solute and solvent (ntotal) = nwater + nurea

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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