JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Solutions

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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 CCl4 (at T K), the mole fraction of CCl4 is equal to 0.33. Statement II: Mixture of methanol and CCl4 shows positive deviation from Raoult's law. In the light of the above statements, choose the correct answer:

(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

Answer: (A)

Solution

Take

100 g

of the solution:

30 g

methanol is

\frac{30}{32} = 0.9375

mol and

70 g

CC l_{4}

\frac{70}{154} = 0.4545

mol. The mole fraction of carbon tetrachloride is

\frac{0.4545}{0.9375 + 0.4545} = 0.33

, so the first statement holds. Methanol molecules are associated through hydrogen bonding; mixing with non-polar

CC l_{4}

disrupts this hydrogen bonding, raising the escaping tendency of the components, so the mixture shows positive deviation from Raoult's law and the second statement also holds. Both statements are true.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Solutions

Solution

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