Given below are two statements: Given:…

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Given below are two statements:

Given: Molar mass 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 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.

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

Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true

Solution

Statement I. Take 100 g of solution: 30 g methanol and 70 g CCl $_4$ .

$n(\text{CH}_3\text{OH}) = \frac{30}{32} = 0.9375$ mol
$n(\text{CCl}_4) = \frac{70}{154} = 0.4545$ mol

$x(\text{CCl}_4) = \frac{0.4545}{0.9375 + 0.4545} = 0.327 \approx 0.33$ . True.

Statement II. CCl $_4$ breaks up the hydrogen bonding between methanol molecules, and the new methanol–CCl $_4$ attractions are weaker. Molecules escape more easily than Raoult's law predicts: positive deviation. True.

Answer: both statements are true.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Solutions

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