Question

Statement-I: 30% (w/w) methanol in CCl₄ have mole fraction of solvent equal to 0.33.
Statement-II: CCl₄ and methanol mixture show positive deviation from Raoult's law.

• A. Both Statement I and Statement II are correct.
• B. Statement I is correct but Statement II is incorrect.
• C. Statement I is incorrect but Statement II is correct.
• D. Both Statement I and Statement II are incorrect.

Answer 1

The Correct Option is A

Step 1: Calculation of mole fraction.
Let weight of the solution = 100 g.
Weight of methanol = 30 g, and weight of CCl₄ = 100 - 30 = 70 g.
Moles of methanol = \( \frac{30}{32} = 0.9375 \, \text{mol} \)
Moles of CCl₄ = \( \frac{70}{154} = 0.4545 \, \text{mol} \)
Mole fraction of CCl₄ = \( \frac{n_{\text{CCl}_4}}{n_{\text{CCl}_4} + n_{\text{CH}_3\text{OH}}} = \frac{0.4545}{0.4545 + 0.9375} = 0.3267 \approx 0.33 \).
Step 2: Raoult's law and positive deviation.
CCl₄ and methanol show positive deviation from Raoult's law because the intermolecular forces between the two liquids are weaker than the forces between the molecules of each liquid. As a result, the vapor pressure of the solution is higher than predicted by Raoult’s law.
Step 3: Conclusion.
Both statements are correct. The mole fraction of CCl₄ is 0.33, and the mixture does exhibit positive deviation from Raoult's law. Final Answer: Both Statement I and Statement II are correct.