Question

When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $x \%$ of vapor pressure of pure solvent. What is $x \%$?

• A. 50%
• B. 60%
• C. 70%
• D. 80%

Hint:

Apply Raoult's Law: $P_s = X_{solvent} \cdot P^o$. Calculate the mole fraction of the solvent using the given moles of solute and solvent.

Answer 1

The Correct Option is D

We can also solve this by looking at the Relative Lowering of Vapor Pressure (RLVP).
According to the RLVP formula:
$$ \frac{P^o - P_s}{P^o} = X_{solute} $$
Where $P^o$ is the vapor pressure of pure solvent and $P_s$ is the vapor pressure of the solution.

Calculate the mole fraction of the solute ($X_{solute}$):
$$ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{0.25}{0.25 + 1} = \frac{0.25}{1.25} $$
$$ X_{solute} = \frac{25}{125} = \frac{1}{5} = 0.2 $$

Now, plug this into the RLVP equation:
$$ \frac{P^o - P_s}{P^o} = 0.2 $$
$$ 1 - \frac{P_s}{P^o} = 0.2 $$
$$ \frac{P_s}{P^o} = 1 - 0.2 = 0.8 $$

To find the percentage value $x$:
$$ P_s = 0.8 \cdot P^o = \frac{80}{100} \cdot P^o = 80 \% \text{ of } P^o $$
Therefore, $x = 80$.