Question

Derive an expression for the determination of molecular mass of a non-volatile solute on the basis of depression of freezing point.

Answer 1

An expression for the depression of the freezing point can be derived in the same way as used for the elevation of the boiling point. The derivation is as follows:

If the curves AB, AC, BD and CE shown in Fig. are assumed to be straight lines, triangles ABD and ACE may be regarded as similar triangles.

For similar triangles ABD and ACE, we have

`(BD)/(CE) = (AD)/(AE)`

or `(T_f - T_1)/(T_f - T_2) = (p^circ - p_1)/(p^circ - p_2)`    ...(i)

where, p° = Vapour pressure of liquid solvent at its freezing point T_f

p₁ = Vapour pressure of solution I at temperature T₁

p₂ = Vapour pressure of solution II at temperature T₂

Eq. (i) can be written as

`(Delta T_(f_1))/(Delta T_(f_2)) = (Delta p_1)/(Delta p_2)`    ...(ii)

It follows from eq. (ii) that in general,

ΔT_f ∝ Δp    ...(iii)

Thus, depression of freezing point is directly proportional to the lowering of vapour pressure.

According to Raoult’s law, for a dilute solution,

`(p^circ - p)/p^circ = (w M)/(W M')`

or `(Delta p)/p^circ = (w M)/(W M')`

or `Delta p = p^circ M xx w/(W M')`

Since p°M is a constant for a particular solvent, we have

`Delta p prop w/(W M')`    ....(iv)

Comparing eqs. (iii) and (iv), we have

`Delta T_f prop w/(W M')`    ...(v)

If W (mass of solvent) = 1 kg, `w/(W M')` represents the molality m of the solution. Therefore,

ΔT_f ∝ m

or ΔT_f ∝ k_f . m