Question

If vapour pressure of pure solvent and solution are 240 and 216 mm Hg respectively then mole fraction of solvent in solution is

• A. 0.9
• B. 0.1
• C. 0.6
• D. 0.4

Hint:

Raoult's law can be written simply as: $\text{Vapour Pressure ratio} = \text{Mole fraction of solvent}$. Just evaluate $\frac{216}{240}$. Since $\frac{216}{240}$ is slightly less than 1, 0.9 stands out instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the vapour pressure of a pure solvent ($P_1^\circ$) and the total vapour pressure of its solution ($P_1$). We need to determine the mole fraction of the solvent ($x_1$) in that solution.

Step 2: Key Formula or Approach:
According to Raoult's law for non-volatile solutes, the vapour pressure of a solvent over a solution is directly proportional to its mole fraction in the liquid phase: $$ P_1 = P_1^\circ \times x_1 $$ Rearranging this equation to solve directly for the mole fraction of the solvent ($x_1$): $$ x_1 = \frac{P_1}{P_1^\circ} $$

Step 3: Detailed Explanation:
Let's list the values provided in the problem statement:

• Vapour pressure of pure solvent, $P_1^\circ = 240\ \text{mm Hg}$

• Vapour pressure of the solution, $P_1 = 216\ \text{mm Hg}$
Substitute these figures directly into the rearranged Raoult's law expression: $$ x_1 = \frac{216\ \text{mm Hg}}{240\ \text{mm Hg}} $$ Let's simplify the fraction by dividing both the numerator and the denominator by 24: $$ x_1 = \frac{216 \div 24}{240 \div 24} = \frac{9}{10} = 0.9 $$ Alternatively, using the Relative Lowering of Vapour Pressure formula: $$ \frac{\Delta P}{P_1^\circ} = x_2 \implies \frac{240 - 216}{240} = \frac{24}{240} = 0.1 $$ where $x_2$ is the mole fraction of the solute. Knowing that the sum of mole fractions is always equal to 1: $$ x_1 = 1 - x_2 = 1 - 0.1 = 0.9 $$

Step 4: Final Answer: The mole fraction of the solvent in the solution is 0.9, which corresponds to option (A).