How is relative lowering of vapour pressure defined for a solution consisting a volatile solvent and non- volatile solute?
Answers
Colligative properties:
Colligative properties are defined as properties of the solution that depend only on the total number of solute particles in the solution and are independent of the chemical identity of the solute particles.
Colligative properties are properties that depend on the concentration of the solution and not on the nature of its contents.
Solutions containing non-volatile solutes exhibit the following Colligative properties:
Relative lowering of the vapour pressure of the solvent
Depression of its freezing point
Elevation of the boiling point
Osmotic pressure of the solution.
Lowering of the vapour pressure of the solvent:
When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases.
According to Raoult's Law, the vapour pressure of a solvent (P1) in a solution containing a non-volatile solute is given by
According to Raoult's Law,
Vapour pressure of the pure solvent = P1°
Vapour pressure of the solvent in solution = P1
P1 = x1P1°
ΔP1 = P1° - P1
= P1° - x1P1°
= P1° (1 - x1)
In a binary solution, 1 - x1 = x2
ΔP1 = P1° x2
ΔP1/P1° = (P1° - P1)/P1° = x2
The lowering of vapour pressure relative to the vapour pressure of pure solvent is called relative lowering of vapour pressure.
ΔP1/P1° → Relative lowering of Vapour pressure
Thus, the relative lowering in vapour pressure depends only on the concentration of solute particles and is independent of their identity.
If the solution contains more than one non-volatile solute, then the relative lowering in vapour pressure of a solvent is equal to the sum of the mole fractions of all the non-volatile solutes.
If n1 and n2 are respectively the number of moles of the solvent and solute in a binary solution, then the relative lowering in the vapour pressure of the solvent,
(P1° - P1)/P1° = x1 + x2 + x3 + ... + xn
if n1 and n2 are the number of moles of the solvent and solute,
(P1° - P1)/P1° = n2/(n1+n2)
For dilute solutions n2 << n1
(P1° - P1)/P1° = n2/n1
n1 = W1/M1 , n2 = W2/M2
(P1° - P1)/P1° = (W2xM1)/(W1xM2)
W1 = Mass of solvent
W2 = Mass of solute
M1 = Molar mass of solvent
M2 = Molar mass of solute
When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases.
Explanation:
Let xA = mole fraction of the solvent
xB = mole fraction of the solute
pAo = vapour pressure of the pure solvent
p = vapour pressure of the solution
Given that the solute is non-volatile, so there will be no contribution of the solute to the vapour pressure and the vapour pressure of the solution will only be due to the solvent. So, vapour pressure of the solution p will be equal to the vapour pressure of the solvent pA, over the solution.
Lowering in vapour pressure can be written as:
ΔΔpA = pAo - pA = pAo - pAo xA (from Raoult's Law)
as xA = 1 - xB
ΔΔpA = pAo - pAo (1-xB)
Rearranging them we get,
ΔpApAo = pAo − pApAo = xBΔpApAo = pAo - pApAo = xB
The relative lowering in vapour pressure of an ideal solution containing non-volatile solute is equal to the mole fraction of the solute at a given temperature.