Question

Derive the relationship between relative lowering of vapour pressure and molar mass of non-volatile solute?

Answer 1

Hint: Relative lowering of vapour pressure is the ratio of the vapour pressure lowering of the solvent from solution to the vapour pressure of the pure solvent. This lowering in vapour pressure has occurred when a solute is added to the solvent.

Complete step by step answer:
The expression for the relative lowering of vapour pressure can be written as;
$${\displaystyle \frac{\mathrm{\Delta }P}{P^{{}^{\circ }}}}={\displaystyle \frac{P^{{}^{\circ }}-P}{P^{{}^{\circ }}}}$$
$$\mathrm{\Delta }P=$$Lowering in vapour pressure
$$P^{{}^{\circ }}=$$Vapour pressure of the pure solvent
$$P=$$Vapour pressure of the solvent from the solution
Also, we have known that the relative lowering of vapour pressure is equal to the mole fraction of the solute($$x_2$$). Hence we can be written as;
$$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }P}{P^{{}^{\circ }}}}={\displaystyle \frac{P^{{}^{\circ }}-P}{P^{{}^{\circ }}}}=x_2$$
Hence, the mole fraction can be written as;
$$\Rightarrow x_2={\displaystyle \frac{n_2}{n_2+n_2}}={\displaystyle \frac{{\displaystyle \frac{W_2}{M_2}}}{{\displaystyle \frac{W_2}{M_2}}+{\displaystyle \frac{W_1}{M_1}}}}$$
$$n_1$$ and $$n_2$$ are the number of moles of solute and solvent, respectively.
$$W_1$$ and $$M_1$$ are the mass and molar mass of the solvent, respectively.
$$W_2$$ and $$M_2$$ are the mass and molar mass of the solute, respectively.

We have known that for dilute solutions, $$n_1>>n_2$$
Hence, $$n_1$$ is too small. So we can neglect the value of $$n_1$$ when compared with $$n_2$$
Hence , the equation for the mole fraction of solute becomes;
$$\Rightarrow x_2={\displaystyle \frac{n_2}{n_2+}}={\displaystyle \frac{{\displaystyle \frac{W_2}{M_2}}}{{\displaystyle \frac{W_1}{M_1}}}}$$
So, finally we got the equation for the mole fraction of solute in terms of molar mass of solute and solvent respectively. Also, we have known that the mole fraction of solute is equal to the relative lowering of vapour pressure. So we can equate both equations;
$$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }P}{P^{{}^{\circ }}}}={\displaystyle \frac{P^{{}^{\circ }}-P}{P^{{}^{\circ }}}}={\displaystyle \frac{{\displaystyle \frac{W_2}{M_2}}}{{\displaystyle \frac{W_1}{M_1}}}}$$
Hence, it derived the relationship between the relative lowering of vapour pressure and molar mass of the non-volatile solute.

Note: Relative lowering of vapour pressure is colligative property (which depends upon the number of particles only. Other than this, there are another three types of colligative properties such as boiling point elevation, depression in freezing point, and osmotic pressure.