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ChemMedMain 2026 • 8th Apr - Shift 2

Mole Fraction and Raoult Law Deviation in Methanol Carbon Tetrachloride Mixture

Given below are two statements :

Given: Molar mass of C\text{C}, H\text{H}, O\text{O}, Cl\text{Cl} are 1212, 11, 1616 and 35.5 g mol−135.5\text{ g mol}^{-1}, respectively

Statement I : In 30%30\% (w/w) solution of methanol in CCl4\text{CCl}_4 (at T KT\text{ K}), the mole fraction of CCl4\text{CCl}_4 is equal to 0.330.33.

Statement II : Mixture of methanol and CCl4\text{CCl}_4 shows positive deviation from Raoult's law.

In the light of the above statements, choose the correct answer from the options given below :

Options

A

Both Statement I and Statement II are true

Correct
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

Topics & Concepts

SolutionsConcentration Terms

Step-by-Step Solution

To determine the correctness of the two statements, let us analyze them step-by-step:

Analysis of Statement I:

  1. Calculate the molar masses:

    • Molar mass of methanol (CH3OH\text{CH}_3\text{OH}): Mmethanol=12+4(1)+16=32 g mol−1M_{\text{methanol}} = 12 + 4(1) + 16 = 32 \text{ g mol}^{-1}
    • Molar mass of carbon tetrachloride (CCl4\text{CCl}_4): MCCl4=12+4(35.5)=12+142=154 g mol−1M_{\text{CCl}_4} = 12 + 4(35.5) = 12 + 142 = 154 \text{ g mol}^{-1}

  2. Calculate the mass of each component: A 30%30\% (w/w) solution of methanol in CCl4\text{CCl}_4 means that in 100 g100\text{ g} of the solution:

    • Mass of methanol (mmethanolm_{\text{methanol}}) = 30 g30\text{ g}
    • Mass of CCl4\text{CCl}_4 (mCCl4m_{\text{CCl}_4}) = 100 g−30 g=70 g100\text{ g} - 30\text{ g} = 70\text{ g}

  3. Calculate the number of moles:

    • Moles of methanol (nmethanoln_{\text{methanol}}): nmethanol=30 g32 g mol−1=0.9375 moln_{\text{methanol}} = \frac{30\text{ g}}{32\text{ g mol}^{-1}} = 0.9375\text{ mol}
    • Moles of CCl4\text{CCl}_4 (nCCl4n_{\text{CCl}_4}): nCCl4=70 g154 g mol−1=511 mol≈0.4545 moln_{\text{CCl}_4} = \frac{70\text{ g}}{154\text{ g mol}^{-1}} = \frac{5}{11}\text{ mol} \approx 0.4545\text{ mol}

  4. Calculate the mole fraction of CCl4\text{CCl}_4 (xCCl4x_{\text{CCl}_4}): ntotal=nmethanol+nCCl4=0.9375+0.4545=1.3920 moln_{\text{total}} = n_{\text{methanol}} + n_{\text{CCl}_4} = 0.9375 + 0.4545 = 1.3920\text{ mol} xCCl4=nCCl4ntotal=511245176=1649≈0.3265≈0.33x_{\text{CCl}_4} = \frac{n_{\text{CCl}_4}}{n_{\text{total}}} = \frac{\frac{5}{11}}{\frac{245}{176}} = \frac{16}{49} \approx 0.3265 \approx 0.33

Hence, Statement I is true.

Analysis of Statement II:

In pure methanol, molecules are held together by strong intermolecular hydrogen bonding. When non-polar carbon tetrachloride (CCl4\text{CCl}_4) is added to methanol, CCl4\text{CCl}_4 molecules occupy spaces between methanol molecules and disrupt/break these hydrogen bonds. This weakens the overall solute-solvent intermolecular attractive forces compared to the solute-solute and solvent-solvent interactions, increasing the escaping tendency of the components into the vapor phase.

As a result, the vapor pressure of the solution is greater than predicted by Raoult's law, indicating a positive deviation from Raoult's law.

Hence, Statement II is true.

Conclusion:

Both Statement I and Statement II are true.

Correct Answer: A (Both Statement I and Statement II are true)

Mole Fraction and Raoult Law Deviation in Methanol Carbon Tetrachloride Mixture | Chemistry PYQ Solution - JEE Challenger