Question

Using Raoult's law, show that ∆P = x₂ p° when P° is vapour pressure of a solvent and x₂ is the mole fraction of solute.​

Answer 1

Explanation:

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Answer 2

Given:

Vapour pressure of solvent= $P^{0}$

Mole fraction of solute= $x_{2$

To Find:

An equation for relation between the vapour pressure of solution (P) and the mole fraction of solution ($x_{1}$).

Solution:

Considering $P^{0}$ is vapour pressure of solvent and P is vapour pressure of solution, and $x_{1$ is mole fraction of solution while $x_{2$ is mole fraction of solute,

According to Raoult's law, we get the equation:

$\frac{P0 - P}{P0}$ = $\frac{x}{x-X}$

$\frac{P0 - P}{P0}$ = $x_{2}$

1 - $\frac{P}{P0}$ = $x_{2}$

$\frac{P}{P0}$ = 1 - $x_{2}$

$\frac{P}{P0}$ = $x_{1}$

$P$ = $P_{0}$ X $x_{1}$

The required equation is hence derived from Raoult's law.