Solved Examples on Vapour Pressure Formula
Example 1: At 25 °C, an aqueous solution’s vapour pressure is measured to be 23.80 mmHg. What fraction of the solute by a mole in this solution? At 25 °C, water has a vapour pressure of 25.756 mm Hg.
Solution:
Vapour pressure of the solution, P Solution = 23.80
Vapour pressure of solvent, P° Solvent= 25.756
Using vapour pressure formula:
Psolution = (Xsolvent) (P°solvent)
23.80= (XSolvent) 25.756
XSolvent = 0.92405
Mole fraction of Solvent = 0.92405
Mole fraction of Solute = 1 – 0.92405
= 0.07595
Example 2: At 25°C, an aqueous solution’s vapour pressure is measured at being 20mmHg. What is the mole fraction of the solvent in this solution? At 25 °C, water has a vapour pressure of 60 mm Hg.
Solution:
Vapour pressure of the solution, P Solution = 20
Vapour pressure of solvent, P° Solvent= 60
Using vapour pressure formula:
Psolution = (Xsolvent) (Posolvent)
20= (XSolvent) 60
XSolvent = 0.333
Mole fraction of solvent = 0.333
Example 3: What is the vapour pressure of the solution if the mole fraction of the solute is 0.3? Water has a 16.358 mmHg vapour pressure at 23 °C.
Solution:
Mole Fraction of solute = 0.3
Mole Fraction of solvent, XSolvent = 1 – 0.3
= 0.7
Now,
Vapour pressure of solvent, P°Solvent = 16.358
Mole Fraction of solvent = 0.7
Using vapour pressure formula:
Psolution = (Xsolvent) (P°solvent)
= (0.7)(16.358)
= 11.4506
Vapour pressure of the solution is 11.4506
Example 4: 200 grams of sucrose dissolved in 300 grams of water at 30 degrees. How much vapour pressure does this solution have? At 30.0 °C, the vapour pressure of water is 41.62 mmHg.
Solution:
First, we need to find the fraction of mole in a solvent.
Now, the amount of mole in solute
Moles of Sucrose = wt (gm)/ molecular mass (molecular weight of sucrose = 342.2948)
= 200 / 342.2948g
= 0.5843Moles of Water = 300 / 18.015 (molecular weight of water = 18.015)
= 16.6528Total mole of solution = 0.5843+16.6528 = 17.2371
Mole fraction of solvent (water) = Mole of water / Mole of solution
Mole fraction of solvent, X Solvent = 16.6528/(16.6528 + 0.5843)
= 16.6528/17.2371
= 0.9661
vapour pressure of solvent, P°Solvent = 41.62
Using vapour pressure formula:
Psolution = (Xsolvent) (P°solvent)
= (0.9661)(41.62)
= 40.2090
Vapour pressure of the solution is 40.2090.
Example 5: What is the vapour pressure of 400 g of propanol (molecular weight = 60 g/mol) and 130 g of acetone (molecular weight = 58 g/mol)? Propanol and acetone have vapour pressures of 11 and 20 mmHg, respectively, at 25 °C.
Solution:
Calculating Molar fractions:
Number of moles in acetone, nacetone = 130g / 58g/mol = 2.24 mol
Number of moles in propanol, npropanol = 400g / 60g/mol = 6.66 mol
ntotal = nacetone + npropanol = 2.24 + 6.66 = 8.9 mol
Molar fraction of acetone, xacetone= 2.24/8.9 = 0.2516
Molar fraction of propanol, xpropanol = 6.66/8.9 = 0.7483
Vapour pressure of acetone, Pacetone = 20
Vapour pressure of Propanol, P°Propanol= 11
Now, partial pressure can be calculated as:
Pacetone = (xacetone) ( P°acetone) = 0.2516 × 20 = 5.03 mmHg
PPropanol = (xPropanol) ( P°Propanol) = 0.7483 × 11 = 8.2313 mmHg
vapour Pressure of the Mixture
Pmixture = Pacetone + PPropanol
= 5.03 + 8.2313
= 13.2613 mmHg
Vapour Pressure
Vapour pressure is the force exerted by a liquid’s (or solid’s) vapour above the surface of the liquid. At a particular temperature and thermodynamic equilibrium, this pressure is formed in a closed container. The rate of liquid evaporation is controlled by the equilibrium vapour pressure. The vapour pressure increases with increasing temperature. When atmospheric pressure and vapour pressure are equal, a liquid is said to have reached its boiling point.