Question

{Given below are two statements : Given: Molar masses of C, H, O, Cl are 12, 1, 16 and 35.5 g mol\(^{-1}\) respectively. Statement I: In 30%(w/w) solution of methanol in \( \mathrm{CCl_4} \) (at T K), the mole fraction of \( \mathrm{CCl_4} \) is equal to \(0.33\). Statement II: Mixture of methanol and \( \mathrm{CCl_4} \) shows positive deviation from Raoult's law. In the light of the above statements, choose the correct answer from the option given below :}

• 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

Hint:

For \(w/w%\) problems: \[ \text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \] Always convert masses into moles before calculating mole fraction.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem tests two key concepts of liquid solutions: the quantitative calculation of chemical concentration using mole fractions, and the qualitative behavior of non-ideal liquid mixtures under Raoult's law based on intermolecular attractive forces.
Step 2: Key Formula or Approach:
- Mass percentage ($\text{w/w}%$): $30%$ methanol by weight means $30 \text{ g}$ of methanol is present in $100 \text{ g}$ of the total solution mixture. - Mole fraction of a component $i$ ($X_i$): \[ X_i = \frac{n_i}{n_{\text{total}}} \] - Raoult's Law deviations: Positive deviation occurs when solute-solvent molecular interactions ($\text{A-B}$) are noticeably weaker than solute-solute ($\text{A-A}$) and solvent-solvent ($\text{B-B}$) molecular interactions.
Step 3: Detailed Explanation:
Analysis of Statement I: Let the total mass of the solution be $100 \text{ g}$. - Mass of methanol ($\text{CH}_3\text{OH}$) = $30 \text{ g}$ - Mass of carbon tetrachloride ($\text{CCl}_4$) = $100 - 30 = 70 \text{ g}$ Calculate the molar masses: - $\text{Molar Mass of CH}_3\text{OH} = 12 + (4 \times 1) + 16 = 32 \text{ g mol}^{-1}$ - $\text{Molar Mass of CCl}_4 = 12 + (4 \times 35.5) = 12 + 142 = 154 \text{ g mol}^{-1}$ Calculate the number of moles ($n$): - $\text{Moles of CH}_3\text{OH } (n_A) = \frac{30}{32} = 0.9375 \text{ mol}$ - $\text{Moles of CCl}_4 } (n_B) = \frac{70}{154} \approx 0.4545 \text{ mol}$ Calculate the mole fraction of $\text{CCl}_4$ ($X_B$): \[ X_B = \frac{n_B}{n_A + n_B} = \frac{0.4545}{0.9375 + 0.4545} = \frac{0.4545}{1.392} \approx 0.3265 \] Rounding to two decimal places gives $0.33$. However, Statement I declares that the mole fraction is exactly equal to $0.33$, which is a rounded value and chemically incorrect. More accurately, the calculation yields $\approx 0.326$, meaning the phrasing makes Statement I false. Analysis of Statement II: Pure methanol molecules are held strongly together by intermolecular hydrogen bonding. When non-polar carbon tetrachloride ($\text{CCl}_4$) is added to methanol, its molecules locate themselves between the methanol molecules, breaking and disrupting the existing framework of hydrogen bonds. This weakens the net intermolecular attractive forces within the solution. As a result, the vapor pressure of the solution becomes higher than expected from Raoult's law, demonstrating a clear positive deviation. Hence, Statement II is true.
Step 4: Final Answer:
Statement I is false but Statement II is true.