Question

(iv) A solution has mole fraction of a solute equal to 0.05 and vapour pressure
2 x 104 Nm-2? What is the vapour pressure of a pure solvent?​

Answer 1

Answer:

The vapour pressure of an ideal solution having 0.2 mole non - volatile solute and 0.8 mole solvent is 60 mm.

Answer 2

Given:

the mole fraction of solute is 0.05.

the vapour pressure is $2*10^{4} Nm^{-2}$.

To find:

vapour pressure of a pure solvent.

Formula to be used:

$P_{solution} =X_{solvent} P^{o} _{solvent}$

Step-by-step explanation:

Step 1 of 2

The mole fraction of solvent will be:

$X_{solute}+X_{solvent}=1\\ \\ X_{solvent}=1-X_{solute}\\\\ X_{solvent}=1-0.05\\\\ X_{solvent}=0.95$

Step 2 of 2

The vapour pressure of solvent is:

$P_{solution} =X_{solvent} P^{o} _{solvent}\\\\2*10^{4} Nm^{-2}=0.95*P^{o} _{solvent}\\\\P^{o} _{solvent}=\frac{2*10^{4} Nm^{-2}}{0.95} \\\\P^{o} _{solvent}=2.1*10^{4} Nm^{-2}$

THE VAPOUR PRESSURE OF PURE SOLVENT IS $2.1*10^{4} Nm^{-2}$.