Question

Which of following is correct?

• A. the lowering of vapour pressure is equal to the mole fraction of solute
• B. the relative lowering of vapour pressure is equal to the mole fraction of solute
• C. the relative lowering of vapour pressure is proportional to the amount of solute in solution
• D. the vapour pressure of the solution is equal to the mole fraction of solvent

Hint:

Do not confuse "lowering of vapour pressure" (\( P^0 - P_s \)) with "relative lowering of vapour pressure" (\( \frac{P^0 - P_s}{P^0} \)). Only the relative lowering is equal to the mole fraction of the solute.

Answer 1

The Correct Option is B

Step 1: Recall Raoult's law for a non-volatile solute.
Adding a non-volatile solute lowers the solvent's vapour pressure from $P^0$ down to $P_s$.
Step 2: Define absolute lowering.
The drop itself is $\Delta P = P^0 - P_s$, an absolute quantity with units of pressure.
Step 3: Define relative lowering.
Dividing by the pure-solvent value gives the dimensionless ratio \[ \frac{P^0 - P_s}{P^0} \]
Step 4: Apply Raoult's law.
Raoult's law states this relative lowering equals the mole fraction of the solute: \[ \frac{P^0 - P_s}{P^0} = x_{\text{solute}} \]
Step 5: Test the wrong options.
The absolute lowering is not the mole fraction (option A confuses the two), and it is not simply proportional to amount unless expressed as mole fraction (option C is loose).
Step 6: Select the precise statement.
Only the relative lowering equals the solute mole fraction. \[ \boxed{\dfrac{P^0-P_s}{P^0}=x_{\text{solute}}} \]