Question

The vapour pressure of a solvent decreases by 2.5 mm Hg by adding a solute. What is the mole fraction of solute? (Vapour pressure of pure solvent is 250 mm Hg)

• A. 0.88
• B. 0.01
• C. 0.1
• D. 0.99

Hint:

Relative lowering of vapor pressure is a colligative property that maps directly to the solute's presence. Just think of it as a percentage drop: a drop of $2.5$ out of $250$ is exactly a $1\%$ reduction, which written as a decimal fraction is $0.01$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem states that the addition of a non-volatile solute lowers the vapor pressure of a solvent by $2.5\text{ mm Hg}$. Given that the vapor pressure of the pure solvent is $250\text{ mm Hg}$, we need to determine the mole fraction of the solute in the solution.

Step 2: Key Formula or Approach:
According to Raoult's Law for a dilute solution containing a non-volatile solute, the relative lowering of vapor pressure is exactly equal to the mole fraction of the solute ($x_{\text{solute}}$):
$$\frac{\Delta P}{P^{\circ}} = x_{\text{solute}}$$ Where $\Delta P$ is the lowering of vapor pressure ($P^{\circ} - P_{\text{solution}}$) and $P^{\circ}$ is the vapor pressure of the pure solvent.

Step 3: Detailed Explanation:
Identify the values provided in the prompt:
Lowering of vapor pressure, $\Delta P = 2.5\text{ mm Hg}$
Vapor pressure of pure solvent, $P^{\circ} = 250\text{ mm Hg}$
Substitute these quantities directly into the Raoult's Law formula:
$$x_{\text{solute}} = \frac{2.5}{250}$$ To simplify the fraction, multiply both the numerator and denominator by 10:
$$x_{\text{solute}} = \frac{25}{2500}$$ $$x_{\text{solute}} = \frac{1}{100} = 0.01$$

Step 4: Final Answer:
The mole fraction of the solute is $0.01$, which corresponds perfectly to option (B).