Chemistry - Liquid Solution Question with Solution | TestHub

The problem involves Raoult's Law for relative lowering of vapor pressure.

According to Raoult's Law, the relative lowering of vapor pressure is equal to the mole fraction of the solute.

$$\frac{\Delta P}{P^{\circ }}=x_{solute}$$

where $\Delta P$ is the decrease in vapor pressure, and $P^{\circ }$ is the vapor pressure of the pure solvent.

From the first condition:

$\Delta P_1=10$ mm Hg

$x_{solute,1}=0.2$

So, $\frac{10}{P^{\circ }}=0.2⟹P^{\circ }=\frac{10}{0.2}=50$ mm Hg.

For the second condition:

$\Delta P_2=20$ mm Hg

We need to find $x_{solvent,2}$.

Using Raoult's Law:

$\frac{20}{P^{\circ }}=x_{solute,2}$

$\frac{20}{50}=x_{solute,2}$

$x_{solute,2}=0.4$

Since $x_{solute}+x_{solvent}=1$:

$x_{solvent,2}=1-x_{solute,2}=1-0.4=0.6$

The mole fraction of the solvent should be 0.6.

The final answer is $\boxed{0.6}$.