Form 5 Chemistry – PHYSICAL CHEMISTRY – Two Components Liquid System

1. LOWERING OF VAPOR PRESSURE

Vapor pressure is the pressure exerted by vapor against the atmospheric pressure.

Lowering of vapor pressure: Is the difference between the original pressure of liquids solvent (P^o) and the pressure of the solution.

Effect of solute on the vapor pressure of the solvent

When solute particles are added in the solvent, the vapor pressure of the solution is lowered.

When solute particles dissolve in a given solvent normally they collide with the solvent molecule and hence prevent / decrease the number of solvent molecules that escape from liquid phase to vapor phase. This causes the decrease amount of vapor above the solution and normally causes the decrease in vapor pressure.

Relative lowering of vapor pressure

Relative lowering of vapor pressure is the ratio lowering vapor pressure to the vapor pressure the solvent.

RAOULT’S LAW OF VAPOR PRESSURE

It states that “The relative lowering of vapor pressure is proportional to the mole fraction of the solute added”

Mathematically

Let P ^o be the vapor pressure of the solvent

      P be the vapor pressure of solution

      Xs be the mole fraction of solute

From  Raoult’s law of vapor pressure

For mole fraction of solute

Let n be number of mole solute

     N be number of mole solvent

The molar mass of solute can be determined

 Example 1.

What do you understand by the term ” Colligative property”?

Colligative property is the property of a liquid which change depending on the number of properties of solute added but not on the nature of the solute.

A solution is prepared from 90g of water and 10.6g of non – volatile solute. If the vapor pressure of the solution at 60^oC was found to be 0.1867 atm. Calculate the relative molecular mass of solute. Given that V. P of H₂O at 60^oC was 0.1966 atm

Solution

Mass of water (solvent) = 90g.

Mass of non – volatile solute = 10.6g.

Vapor pressure of solution (P) = 0.1867 atm.

Vapor pressure of solvent Po = 0.1966 atm.

Recall

Example 2.

When 114g of sucrose are dissolved in 1000g at water the vapor pressure was lowered 0.11mmHg . Calculate the relative molecular mass of sucrose if the vapor pressure of water at 20^oC was 17.54 mmHg.

Solution

Mass of solute = 114g

Mass of solvent = 1000g

Lower vapor pressure = 0.11mmHg (P^o – P)

Vapor pressure of solvent = 17.54mmHg

Recall

Example 3.

Calculate the vapor pressure lowering caused by the addition of 100g of sucrose of molar mass 342g/mol to 1000g of water if the vapor pressure of pure water at 25^oC is 23.8 mmHg.

Solution

Mass of solute = 100g

Mass of solvent = 1000g

Molar mass of solute = 342g/mol

Molar mass of solvent = 18g/mol

Vapor pressure of solvent = 23.8mmHg

Example 4.

The vapor pressure of ether (molar mass 74g/mol) is 442mmHg at 293k. If 3g of compound A are dissolved in 50g of ether at this temperature, the vapor pressure falls to 426mmHg. Calculate the molar mass of A assuming that the solution of A in ether is very dilute.

Solution

Mass of solute = 3g

Mass of solvent = 50g

Vapor pressure of solution P = 426mmHg

V.p of solvent P^o = 442mmHg

Mr of solvent = 74g/mol

Example 5.
18. 2g of urea is dissolved in 100g of water at 50^oC . The following of vapor pressure produced is 5mmHg. Calculate the molecular mass of urea if the vapor pressure of water at 50^oC is 92mmHg.

Solution

 Mass of solute = 18.2g

 Mass of solvent = 100g

Lowering vapor pressure (P^o  – P) = 5mmHg

V . p  of solvent (P^o) = 92mmHg

Required to find Mr of solute