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

This problem is based on Raoult's Law for a solution containing a non-volatile solute. Raoult's Law states that the vapor pressure of a solution ($P_s$) is equal to the product of the vapor pressure of the pure solvent ($P^o$) and the mole fraction of the solvent ($X_{solvent}$) in the solution.

The formula is given by:
$$ P_s = X_{solvent} \cdot P^o $$

First, let's calculate the mole fraction of the solvent ($X_{solvent}$):
The number of moles of solute ($n_{solute}$) = 0.25 mol.
The number of moles of solvent ($n_{solvent}$) = 1 mol.
Total moles in solution = $n_{solvent} + n_{solute} = 1 + 0.25 = 1.25$ mol.

$$ X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}} = \frac{1}{1.25} $$

To simplify the calculation:
$$ X_{solvent} = \frac{1}{1.25} = \frac{100}{125} = 0.8 $$

Now, substitute this back into Raoult's Law equation:
$$ P_s = 0.8 \cdot P^o $$

To find the percentage $x \%$, we express the ratio $\frac{P_s}{P^o}$ as a percentage:
$$ \frac{P_s}{P^o} \times 100 = 0.8 \times 100 = 80 \% $$

Thus, $x = 80$, and the vapor pressure of the solution is 80% of the vapor pressure of the pure solvent.