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

Concept: Mole fraction is defined as: \[ X_i = \frac{\text{Number of moles of component } i}{\text{Total number of moles}} \] For solutions showing deviation from Raoult’s law:

• Positive deviation occurs when solute-solvent interactions are weaker than solute-solute and solvent-solvent interactions.
• Negative deviation occurs when solute-solvent interactions are stronger.

Methanol is a hydrogen-bonded polar compound whereas \( \mathrm{CCl_4} \) is non-polar. Mixing them weakens intermolecular attractions and hence the solution shows positive deviation from Raoult’s law.

Step 1: Calculation of mole fraction of \( \mathrm{CCl_4} \). A 30%(w/w) solution of methanol in \( \mathrm{CCl_4} \) means: \[ 30 \text{ g methanol} + 70 \text{ g } \mathrm{CCl_4} \] Molar mass of methanol \( (\mathrm{CH_3OH}) \): \[ 12 + 4(1) + 16 = 32 \text{ g mol}^{-1} \] Number of moles of methanol: \[ n_{\mathrm{CH_3OH}} = \frac{30}{32} = 0.9375 \] Molar mass of \( \mathrm{CCl_4} \): \[ 12 + 4(35.5) \] \[ = 12 + 142 \] \[ = 154 \text{ g mol}^{-1} \] Number of moles of \( \mathrm{CCl_4} \): \[ n_{\mathrm{CCl_4}} = \frac{70}{154} \] \[ = 0.4545 \] Total moles: \[ 0.9375 + 0.4545 = 1.392 \] Mole fraction of \( \mathrm{CCl_4} \): \[ X_{\mathrm{CCl_4}} = \frac{0.4545}{1.392} \] \[ = 0.326 \approx 0.33 \] Thus, Statement I is true.

Step 2: Checking Statement II regarding deviation from Raoult’s law. Methanol molecules are strongly hydrogen bonded in pure state. \( \mathrm{CCl_4} \) molecules are non-polar and interact mainly through weak van der Waals forces. When methanol and \( \mathrm{CCl_4} \) are mixed:

• The strong hydrogen bonding between methanol molecules gets disrupted.
• The new methanol-\( \mathrm{CCl_4} \) interactions are weaker than original methanol-methanol interactions.

Therefore: \[ A-B < A-A \text{ and } B-B \] Hence, escaping tendency of molecules increases and vapour pressure becomes greater than expected. Therefore, the mixture shows: \[ \text{Positive deviation from Raoult’s law} \] Thus, Statement II is also true. Hence, both Statement I and Statement II are true. \[ \boxed{(1)\text{ Both Statement I and Statement II are true}} \]