Chemistry Q52 – JEE-MAIN 2026

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 :

Clue 1 of 3: Solution Composition

Understand how to calculate the mass of solute and solvent from a given percentage (w/w) concentration.

🧠 Full Solution Path

Step 1: Evaluate Statement I: Calculate Mole Fraction of CCl

_{4}

✦ Active

Assume 100 g of the $30 %$ (w/w) solution. This means 30 g of methanol ( $C H_{3} O H$ ) and 70 g of carbon tetrachloride ( $C C l_{4}$ ). Given molar masses: C = 12, H = 1, O = 16, Cl = 35.5 g/mol. Calculate molar masses of components: Molar mass of $C H_{3} O H = 12 + (3 \times 1) + 16 + 1 = 32$ g/mol. Molar mass of $C C l_{4} = 12 + (4 \times 35.5) = 12 + 142 = 154$ g/mol. Calculate moles of each component: Moles of $C H_{3} O H (n_{C H_{3} O H}) = \frac{30 g}{32 g/mol} = 0.9375$ mol. Moles of $C C l_{4} (n_{C C l_{4}}) = \frac{70 g}{154 g/mol} \approx 0.4545$ mol. Total moles = $n_{C H_{3} O H} + n_{C C l_{4}} = 0.9375 + 0.4545 = 1.392$ mol. Calculate mole fraction of $C C l_{4}$ : $X_{C C l_{4}} = \frac{n_{C C l_{4}}}{Total moles} = \frac{0.4545}{1.392} \approx 0.3265 \approx 0.33$ . Thus, Statement I is TRUE.

Step 2: Evaluate Statement II: Deviation from Raoult's Law○ Expand

Methanol ( $C H_{3} O H$ ) is a polar molecule that forms strong intermolecular hydrogen bonds. Carbon tetrachloride ( $C C l_{4}$ ) is a non-polar molecule. When methanol and CCl $_{4}$ are mixed, the strong hydrogen bonds between methanol molecules are disrupted by the introduction of non-polar CCl $_{4}$ molecules. The new interactions between methanol and CCl $_{4}$ are weaker than the original methanol-methanol hydrogen bonds. Weaker intermolecular forces in the solution lead to a higher tendency for molecules to escape into the vapor phase, resulting in a higher vapor pressure than predicted by Raoult's law. This phenomenon is known as positive deviation from Raoult's law. Thus, Statement II is TRUE.

💡 Teacher's Secret Hint

Remember that positive deviation occurs when A-B interactions are weaker than A-A and B-B interactions, leading to higher vapor pressure.

Step 3: Conclusion○ Expand

Since both Statement I and Statement II are true, the correct option is 'Both Statement I and Statement II are true'.

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Chemistry Question 52 – JEE-MAIN 2026

Clue 1 of 3: Solution Composition

🧠 Full Solution Path