Question

What is the relationship between molal elevation constant and the elevation of boiling point of a solution?

Answer 1

An expression for the elevation of boiling point can be derived as follows.

Triangles ABD and ACE in the figure may be assumed as similar triangles if curves BD and CE are taken to be straight lines.

For similar triangles ABD and ACE, we have

`(AB)/(AC) = (AD)/(AE)`

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

where, p° = Vapour pressure of pure solvent at its boiling point T_b (equal to 1 atm)

p₁ = Vapour pressure of solution I at temperature T_b

p₂ = Vapour pressure of solution II at temperature Tb.

Eq. (i) can be written as

`(Delta T_(b_1))/(Delta T_(b_2)) = (Delta p_1)/(Delta p_2)`    ...(ii)

where, `(Delta T_(b_1))` = Elevation of boiling point for solution I

`(Delta T_(b_2))` = Elevation of boiling point for solution II

Δp₁ = Lowering in vapour pressure for solution I

Δp₂ = Lowering in vapour pressure for solution II.

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

ΔT_b ∝ Δp    ...(iii)

i.e., the elevation of boiling point is directly proportional to the lowering in vapour pressure.

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

`(p^circ - p)/p^circ = (wM)/(WM')`

or `(Delta p)/p^circ = (wM)/(WM')`

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

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

`Delta p prop w/(WM')`     ...(iv)

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

`Delta T_b prop w/(WM')`    ...(v)

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

ΔT_b ∝ m

or ΔT_b = K_b . m

It represents a relationship between elevation of boiling point and molality of the solution. In this expression, K_b is a constant known as the molal elevation constant or ebullioscopic constant.