7.1 Relative lowering of vapour pressure

2.1 $n$-factor

3.1 Solubility of gas in liquids

5.1 Vapour pressure of solids in liquid
5.2 Ideal Solution
5.3 Non Ideal Solution

7.1 Relative lowering of vapour pressure
7.2 Elevation in Boiling point
7.3 Depression of Freezing point
7.4 Osmosis and Osmotic pressure

As we know that vapour pressure decreases on addition of non volatiole solute. So, according to Raoult's law vapour pressure after mixing,

$${P_1} = {x_1}P_1^0$$

where,

$P_1^0$ = vapour pressure of pure solvent

${P_1}$ = vapour pressure of solvent with solute i.e solution

${x_1}$ = mole fraction of solvent

then change in vapour pressure after mixing,

$$\Delta {P_1} = P_1^0 - {P_1}$$

$$\begin{equation} \begin{aligned} \Rightarrow \Delta {P_1} = P_1^0(1 - {x_1}) \\ \Rightarrow \Delta {P_1} = P_1^0{x_2} \\\end{aligned} \end{equation} $$

where ${x_1}$ is mole fraction of solute

And relative lowering in vapour pressure is, basically change in vapour pressure divided by vapour pressure of pure solvent.

$ \Rightarrow \Delta P_1^{}/P_1^0$ = relative lowering of vapour pressure

$$ \Rightarrow \Delta P_1^{}/P_1^0 = (P_1^0 - {P_1})/P_1^0 = {x_2} = {n_2}/{n_2} + {n_1}\ ...(1)$$

Special case,

If ${n_1} > > {n_2}$

then $${n_1} + {n_2} \approx {n_1}$$

$$ \Rightarrow \Delta {P_1}/P_1^0 = {n_2}/{n_1} = {w_2}{M_1}/{M_2}{w_1}$$

So, from this equation we can get the molar mass of solute can be calculated.

Also, subtracting equation $(1)$ from one we get,

$${P_1}/P_1^0 = {n_1}/{n_1} + {n_2}$$

This can be used for easy calculations of vapour pressures. But, this equation do not give relative lowering of vapour pressure.

Question 6. The vapour pressure of pure benzene at a certain temperature is $0.850\ bar$. A non-volatile, non-electrolyte solid weighing $0.5\ g$ when added to $39.0\ g$ of benzene (molar mass $78\ \frac{g}{{mol}}$). Vapour pressure of the solution, then, is $0.845\ bar$. What is the molar mass of the solid substance?

Solution: The various quantities known to us are as follows:

$P_1^0$ = $0.850\ bar$; $p = 0.845\ bar$; ${M_1}$ = $78\frac{g}{{mol}}$; ${w_2}$ = $0.5\ g$; ${w_1}$ = $39\ g$

Substituting these values in following equation,

$$ \Rightarrow \Delta {P_1}/P_1^0 = {n_2}/{n_1} = {w_2}{M_1}/{M_2}{w_1}$$

we get,

$$\frac{{0.850 - 0.845}}{{0.850}} = \frac{{0.5 \times 78}}{{{M_2} \times 39}}$$

Therefore,

$${M_2} = 170 \frac{g}{{mol}}$$