218. Show graphically how the vapour pressure of a solvent and a solution in it of a non-volatile solute change with temperature. Show on this graph the boiling points of the solvent and the solution. Which is higher and why?

Answer:

Elevation of boiling point with addition of non-volatile solute vapour pressure decrese and hence boiling point increase.

Let ${\mathrm{T}}_{\mathrm{b}}^0$ be the boiling point of pure solvent and
T_b be the boiling point of solution. The increase in
the boiling point  $∆{\mathrm{T}}_{\mathrm{b}} = {\mathrm{T}}_{\mathrm{b}} - {\mathrm{T}}_{\mathrm{b}}^0$  is known as
elevation of boiling point.

for dilute
solutions the elevation of boiling point (ΔTb) is
directly proportional to the molal concentration of
the solute in a solution. Thus

     ΔTb ∝ m 
or
   ΔTb = K_b m

Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant. The unit of K_b is K kg mol-1.If w₂ gram of solute of molar mass M₂ is dissolved in w₁ gram of solvent, then molality, m of the solution is given by the expression:

$$\begin{gathered} \mathrm{m} = \frac{1000 \times {\mathrm{w}}_2}{{\mathrm{M}}_2 \times {\mathrm{w}}_1} \\ \mathrm{putting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{molality} \mathrm{in} ∆{\mathrm{T}}_{\mathrm{b}} = {\mathrm{K}}_{\mathrm{b}}\mathrm{m} \\ \mathrm{we} \mathrm{get} \\ ∆{\mathrm{T}}_{\mathrm{b}} = \frac{{\mathrm{k}}_{\mathrm{b}} \times 1000\times {\mathrm{w}}_2}{{\mathrm{M}}_2\times {\mathrm{w}}_1} \end{gathered}$$

Thus, in order to determine M₂, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔT_b is
determined experimentally for a known solvent whose Kb value is known.