##### Exercise | Q 3.8 | Page 46

##### Answer the following.

## Using Raoult’s law, how will you show that ΔP = $\text{P}}_{1}^{0}{\text{x}}_{2$ ? Where, x_{2} is the mole fraction of solute in the solution and $\text{P}}_{1}^{0$ vapour pressure of pure solvent.

Solution:

1. Raoult’s law expresses the quantitative relationship between vapour pressure of solution and vapour pressure of the solvent.

2. In solutions of nonvolatile solutes, the law is applicable only to the volatile solvent.

3. The law states that, “the vapour pressure of solvent over the solution is equal to the vapour pressure of pure solvent multiplied by its mole fraction in the solution.”

4. Suppose that for a binary solution containing solvent and one nonvolatile solute, P_{1} is the vapour pressure of solvent over the solution, x_{1} and x_{2} are the mole fractions of solvent and solute, respectively and $\text{P}}_{1}^{0$ is the vapour pressure of pure solvent, then, $\text{P}}_{1}={\text{P}}_{1}^{0}{\text{x}}_{1$ .

5. Since, x_{1} = 1 – x_{2},

$\text{P}}_{1}={\text{P}}_{1}^{0}{\text{x}}_{1}={\text{P}}_{1}^{0}(1-{\text{x}}_{2})={\text{P}}_{1}^{0}-{\text{P}}_{1}^{0}{\text{x}}_{2$

∴ $\text{P}}_{1}^{0}-{\text{P}}_{1}={\text{P}}_{1}^{0}{\text{x}}_{2$

∴ $\u25b3\text{P}={\text{P}}_{1}^{0}{\text{x}}_{2}$ (∵ Δ P is the lowering of vapour pressure)

**Note:** A plot of P_{1} versus x_{1} is a straight as shown below.

**Variation of vapour pressure of solution with mole fraction of solvent**

Solution:

1. Raoult’s law expresses the quantitative relationship between vapour pressure of solution and vapour pressure of the solvent.

2. In solutions of nonvolatile solutes, the law is applicable only to the volatile solvent.

3. The law states that, “the vapour pressure of solvent over the solution is equal to the vapour pressure of pure solvent multiplied by its mole fraction in the solution.”

4. Suppose that for a binary solution containing solvent and one nonvolatile solute, P_{1} is the vapour pressure of solvent over the solution, x_{1} and x_{2} are the mole fractions of solvent and solute, respectively and

5. Since, x_{1} = 1 – x_{2},

∴

∴

**Note:** A plot of P_{1} versus x_{1} is a straight as shown below.

**Variation of vapour pressure of solution with mole fraction of solvent**