Question

Given below are two statements:}
Statement I:} In 30% (w/w) solution of methanol in CCl$_4$ (at $T$ K), the mole fraction of CCl$_4$ is equal to 0.33.
Statement II:} Mixture of methanol and CCl$_4$ shows positive deviation from Raoult's law.

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We must calculate the mole fraction of a solvent in a binary mixture given its percentage by weight. Secondly, we evaluate the intermolecular interactions between polar and non-polar molecules to predict deviation from Raoult's Law.
Step 2: Key Formula or Approach:
1. Moles $n = \frac{\text{Mass}}{\text{Molar Mass}}$.
2. Mole fraction $\chi_A = \frac{n_A}{n_A + n_B}$.
3. Deviation from Raoult's Law: Positive deviation occurs when A-B interactions are weaker than A-A and B-B interactions.
Step 3: Detailed Explanation:
Statement I:
30% (w/w) solution of methanol ($CH_3OH$) in $CCl_4$ means 30 g of methanol is mixed with 70 g of $CCl_4$.
Molar mass of $CH_3OH = 12 + 3(1) + 16 + 1 = 32\text{ g/mol}$.
Moles of methanol $n_1 = \frac{30}{32} = 0.9375\text{ mol}$.
Molar mass of $CCl_4 = 12 + 4(35.5) = 12 + 142 = 154\text{ g/mol}$.
Moles of $CCl_4$ $n_2 = \frac{70}{154} \approx 0.4545\text{ mol}$.
Mole fraction of $CCl_4$ is $\chi_2 = \frac{n_2}{n_1 + n_2} = \frac{0.4545}{0.9375 + 0.4545} = \frac{0.4545}{1.392} \approx 0.326$.
Rounding to two decimal places gives 0.33. So, Statement I is true.
Statement II:
Methanol molecules exhibit strong intermolecular hydrogen bonding.
$CCl_4$ is a non-polar molecule. When $CCl_4$ is added to methanol, it inserts itself between the methanol molecules, effectively breaking their hydrogen bonds.
This results in the A-B intermolecular forces ($CH_3OH - CCl_4$) being significantly weaker than the pure A-A forces ($CH_3OH - CH_3OH$).
Weaker intermolecular forces make it easier for molecules to escape into the vapor phase, increasing vapor pressure above what Raoult's law predicts.
Thus, the mixture shows a positive deviation from Raoult's law. So, Statement II is true.
Step 4: Final Answer:
Both Statement I and Statement II are true.