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: Understanding the Question:
This question tests the calculation of mole fraction in a weight-percent solution and the physical property of non-ideal liquid mixtures.
Step 2: Detailed Explanation:
Evaluating Statement-I:
30% (w/w) methanol ($CH_3OH$) means 30 g of methanol in 100 g of solution.
Mass of methanol = 30 g.
Mass of $CCl_4$ (solvent) = 100 - 30 = 70 g.
Molar mass of $CH_3OH = 12 + 4 + 16 = 32$ g/mol.
Molar mass of $CCl_4 = 12 + (4 \times 35.5) = 154$ g/mol.
Moles of $CH_3OH = 30 / 32 = 0.9375$ mol.
Moles of $CCl_4 = 70 / 154 = 0.4545$ mol.
Mole fraction of $CCl_4 = \frac{0.4545}{0.4545 + 0.9375} = \frac{0.4545}{1.392} \approx 0.3265 \approx 0.33$.
So, Statement-I is correct.

Evaluating Statement-II:
Methanol molecules are held together by strong hydrogen bonds. $CCl_4$ is a non-polar solvent. When mixed, $CCl_4$ molecules come between methanol molecules and break the H-bonds. This weakens the intermolecular forces compared to the pure components, leading to a positive deviation from Raoult's law.
So, Statement-II is correct.
Step 3: Final Answer:
Both Statement I and Statement II are correct.