Every pure liquid exerts a vapour pressure in the space above it. This is the vapour pressure of the solvent over it at that particular temperature. It depends upon the nature of the solvent and the temperature.
(a) If a volatile solute is dissolved, vapour pressure of the solvent is increased.
(b) However, if a non-volatile solute is dissolved in it, the vapour pressure of the solution is lowered. This is because, in a solution, the percentage of the volatile solvent molecules, which only contributes towards vapour pressure is diminished.
Fig. Decrease of vapour pressure when a non-volatile solute is added to the solvent.
Since, the solute molecules are non-volatile and show no measurable tendency to escape from the solution as vapour, consequently, the vapour pressure of a solution is always lower than that of its solvent.
Raoult’s gave a relation between the relative lowering of vapour pressure and the mole fraction of the solute. Mathematically:
(mole fraction of the solute)
Using the above equation, we can determine the molecular weight of the solute, when the lowering in v.p. is known, when a known weight of the solute w, dissolved in a known wt. of the solvent W.p0 is the vapour pressure of the pure solvent and m and M are the molecular weights of solute and solvent respectively.
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Volume of solution = 4.3 L
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.