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LOWERING IN THE VAPOUR PRESSURE
When a non-volatile solute is added to a solvent, the vapour pressure is lowered due to the following reasons:
(i) Percentage surface area occupied by the solvent decreases. Thus the rate of evaporation and vapour pressure decreases. The solute molecules occupy the surface, and so the per cent surface area occupied by the solvent decreases.
(ii) According to Graham’s law of evaporation,
Rate of evaporation ∝ 1/√density
When a non-volatile solute is dissolved in a liquid, its density increases. Thus both rate of evaporation and vapour pressure are lowered.
If p0 is the vapour pressure of pure solvent and ps is the vapour pressure of the solution, the difference (p0 – ps) is termed lowering in vapour pressure and the ratio [p0–ps/p0] is termed relative lowering in vapour pressure.
Raoult, in 1886, established a relationship between relative lowering in vapour pressure and composition of the solution after a series of experiments in various solvents. The relationship is known as Raoult’s law. It states that the relative lowering in vapour pressure of a dilute solution is equal to mole fraction of the solute present in the solution.
If n moles of solute be dissolved in N moles of the solvent, the mole fraction of the solute will be n/n+N.
According to Raoult’s law, [p0–ps/p0] = n/n+N.
This is the mathematical expression for Raoult’s law.
(Modified form of Raoult’s law: The above relationship can be written as,
p0/P0–ps = n+N/n = 1 + N/n
or p0/p0–ps – 1 = N/n or p0/p0–ps = N/n
or p0–ps/ps = n/N = wA/mA × mB/wB
This equation gives accurate results and easy to apply.]
Derivation of Raoult’s law for a dilute solution
When a non-volatile solute is dissolved in a volatile solvent, a fraction of the surface of solvent is blocked by solute molecules where no evaporation occurs. Thus, under similar conditions, the vapour pressure is decreased. The vapour pressure of the solution, thus, depends upon the number of molecules of the solvent present on the surface of the solution.
The number of such molecules is proportional to mole fraction of the solvent.
So, the vapour pressure of solution,
‘ps’ ∝ N/n+N
or ps = k.N/n+N ….(i)
(k is proportionality factor)
For pure solvent, n = 0
And hence, p0 = k N/0+N = k …(ii)
Putting the value of k in Eq. (i)
or ps = p0 N/n+N
or ps/p0 = N/n+N
or 1 – ps/p0 = 1 – N/n+N
or p0–ps/p0 = n/n+N …(iii)
This is Raoult’s equation.
If a solution is made by dissolving wA g of solute (molecular mass mA) in wB g of the solvent (molecular mass mB), the mole fraction of the solute will be
= wA/mA/wA/mA + wB/mB
If the solution is very dilute, wA/mA can be neglected in the denominator as compared to wB/mB. The Eq. (iii), thus, becomes
p0–ps/p0 = wA/mA × wB/mB …(iv)
This relationship is useful in the determination of the molecular mass of dissolved solute by measuring relative lowering of vapour pressure.