Question

Calculate the vapour pressure of solution if relative lowering of vapour pressure and vapour pressure of pure solvent are 0.018 and 18 mm Hg respectively at 300 K.

• A. 18.32 mm Hg
• B. 17.08 mm Hg
• C. 17.68 mm Hg
• D. 18.60 mm Hg

Hint:

Always apply a logic check to your final answer: the addition of a non-volatile solute strictly lowers the vapour pressure. Therefore, your calculated $P_s$ MUST be smaller than the given $P^\circ$. Options (a) and (d) are immediately impossible because they are higher than 18.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the Relative Lowering of Vapour Pressure (RLVP) and the vapour pressure of the pure solvent. We need to calculate the actual vapour pressure of the resulting solution.

Step 2: Key Formula or Approach:
The definition of Relative Lowering of Vapour Pressure is the actual drop in pressure divided by the original pure pressure:
$$RLVP = \frac{P^\circ - P_s}{P^\circ}$$
Where:
$P^\circ$ = Vapour pressure of the pure solvent
$P_s$ = Vapour pressure of the solution

Step 3: Detailed Explanation:
Given values:
$RLVP = 0.018$
$P^\circ = 18 \text{ mm Hg}$
Substitute these directly into the formula:
$$0.018 = \frac{18 - P_s}{18}$$
Multiply both sides by 18 to isolate the numerator:
$$18 \times 0.018 = 18 - P_s$$
$$0.324 = 18 - P_s$$
Rearrange to solve for $P_s$:
$$P_s = 18 - 0.324$$
$$P_s = 17.676 \text{ mm Hg}$$
Rounding this to two decimal places gives 17.68 mm Hg.

Step 4: Final Answer:
The vapour pressure of the solution is 17.68 mm Hg, which corresponds to option (c).