# Calculate Vapour Pressure

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Vapour pressure of the liquid is the equilibrium pressure applied by the vapour on the surface of liquid or solid. This is the pressure of the vapour resulting from the evaporation of that liquid in a closed container.

According to the Raoult’s law, the vapour pressure will be lowering when we add some solute in the liquid.

P = X P*
Where,

P is a vapour pressure of solution

X is a mole fraction

And P* is a vapour pressure of solvent

Unit of vapour pressure is atm or mm of Hg.

Example: A solution of 38.4g of glucose (C6H12O6) a non electrolyte in 180g of water is prepared at 25oC. What is the vapour pressure of water above the solution?

Solution: We can consider the solvent is pure water which has vapour pressure 23.6 mm of Hg.
And mole fraction of H2O, X = (180/18)/[(180/18)+(38.4/180)]
= 10/(10+0.21)
= 10/10.21
= 0.98
So that the vapour pressure of solution = 23.6 x 0.98
= 23.12 mm of Hg.

Example: The vapour pressure of pure water is 0.4atm at 300kelvin. Calculate the vapour pressure above a solution prepared from 410 grams of water and 120 grams of glucose (C6H12O6) at 300 kelvin.

Solution: Given,
Vapour pressure of solvent, V* = 0.4 atm
Mole fraction of H2O, X = (360/18)/[(360/18)+(410/180)]
= 20/(20+2.27)
= 20/22.27
= 0.89
Vapour pressure of solution = 0.89 x 0.4 = 0.356 atm.