Question

Define molal elevation constant and molal depression constant?

Answer 1

Part 1: Molal elevation constant:

• Molal elevation constant can be defined as the elevation in boiling point produced when one mole of non-volatile solute is dissolved in 1 kg i.e. 1000 g of the solvent.
• It is also known as the ebullioscopic constant.
• The unit of molal elevation constant is $\mathrm{K}\mathrm{Kg}{\mathrm{mol}}^{-1}$.

If ${\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}$ is the boiling point of the pure solvent and ${\mathrm{T}}_{\mathrm{b}}$ is the boiling point of the solution then elevation in boiling point is given as

$\mathrm{Elevation}\mathrm{in}\mathrm{boiling}\mathrm{point}(∆{\mathrm{T}}_{\mathrm{b}})={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}$

Experimental results show that there is a relation between elevation in boiling point and molality ‘m’ of the solute present in solution.

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{b}}\propto \mathrm{m} \\ ∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\times \mathrm{m} \\ \mathrm{Where},{\mathrm{K}}_{\mathrm{b}}=\mathrm{molal}\mathrm{elevation}\mathrm{constant} \end{gathered}$$

Part 2: Molal depression constant:

• Molal depression constant can be defined as the depression in freezing point produced when one mole of non-volatile solute is dissolved in 1 kg i.e. 1000 g of the solvent.
• It is also known as the cryoscopic constant.
• The unit of molal depression constant is $\mathrm{K}\mathrm{Kg}{\mathrm{mol}}^{-1}$

If ${\mathrm{T}}_{\mathrm{f}}^{\mathrm{o}}$ is the boiling point of the pure solvent and ${\mathrm{T}}_{\mathrm{f}}$ is the boiling point of the solution then elevation in boiling point is given as

$\mathrm{Depression}\mathrm{in}\mathrm{freezing}\mathrm{point}(∆{\mathrm{T}}_{\mathrm{f}})={\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{f}}^{\mathrm{o}}$

Experimental results show that there is a relationship between depression in freezing point and molality ‘m’ of the solute present in the solution.

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}\propto \mathrm{m} \\ ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m} \\ \mathrm{Where},{\mathrm{K}}_{\mathrm{f}}=\mathrm{molal}\mathrm{depression}\mathrm{constant} \end{gathered}$$