Question

With the help of vapour pressure-temperature curves for solution and solvent, explain why boiling point of solvent is elevated when a nonvolatile solute is dissolved into it.

Why does the boiling point of a solvent get raised when a non-volatile solute is dissolved in it? Explain with vapour pressure-temperature graph.

Answer 1

1. The vapour pressures of the solution and of the pure solvent are plotted as a function of temperature in the given diagram.
   
2. At any temperature, the vapour pressure of the solution is lower than that of the pure solvent. Hence, the vapour pressure-temperature curve of solution (CD) lies below that of the solvent (AB).
3. The difference between the two vapour pressures increases as temperature and vapour pressure increase, as predicted by the equation.
   `Delta P = P_1^0 x_2`
4. The intersection of the curve CD with the line corresponding to 760 mm is the boiling point of the solution. The similar intersection of the curve AB is the boiling point of the pure solvent. It is clear from the diagram that the boiling point of the solution (Tb) is higher than that of pure solvent `(T_b^0)`.
5. At the boiling point of a liquid, its vapour pressure is equal to 1 atm.
6. In order to reach boiling point, the solution and solvent must be heated to a temperature at which their respective vapour pressures attain 1 atm.
7. At any given temperature the vapour pressure of the solution is lower than that of the pure solvent. Hence, the vapour pressure of the solution needs a higher temperature to reach 1 atm than that needed for the vapour pressure of the solvent.
8. In other words, the solution must be heated to a higher temperature to cause it to boil than the pure solvent. Thus, the solution containing nonvolatile solute boils at a temperature higher than the boiling point of the pure solvent, i.e., `T_b > T_b^0`.

Answer 2

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

`(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 T_b.

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