Question

Vapour pressure of pure water is 28.4 torr. Cane sugar is dissolved in it and the vapour pressure of solution becomes 25 torr. The mole fraction is​

Answer 1

Given:

The vapour pressure of pure water is 28.4 torr. Cane sugar is dissolved in it and the vapour pressure of the solution becomes 25 torr.

To find:

The mole fraction

Solution:

To solve the given problem we will use the following Raoult's Law of relative lowering of vapour pressure:

$\boxed{\bold{P_1 = X_1P^o_1 }}\\\\\Rightarrow\boxed{\bold{ \frac{P^o_1 - P }{P^o_1} = X_1 }}$

where P₁⁰ = vapour pressure of pure solvent, P₁ = vapour pressure of solvent and X₁ = mole fraction of solvent

The vapour pressure of the pure solvent i.e., water, P⁰₁ = 28.4 torr

The vapour pressure of the solution, P = 25 torr

Now, on substituting the given values in Raoult's Law, we get

$X_1 = \frac{28.4\: -\: 25}{28.4}$

$\implies X_1 = \frac{3.4}{28.4}$

$\implies X_1 = 0.1197$

$\implies \bold{X_1 }$ ≈ $\bold{0.12}$

Thus, the mole fraction is → 0.12.

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