Question

If ΔT_b is the elevation of boiling point and m is the molality of solution, show that ΔT_b = K_bm.

Answer 1

In Fig. if curves BD and CE are assumed to be straight lines, triangles ABD and ACE may be regarded as similar triangles. 

For similar triangles ABD and ACE, we have

`(A B)/(A C) = (A D)/(A E)`

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 T_b.

Eq. (i) can be written as

`(Delta T_(b_1))/(Delta T_(b_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 = (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_b prop w/(W M^')`    ...(v)

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

ΔT_b ∝ m

or ΔT_b = K_b . m