JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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:

Official previous-year question

Held on 8 Apr 2026 · Verified 6 Jul 2026.

Options

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

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Solution

Statement I:

$30\%$ (w/w) solution of methanol in CCl $_4$ means $30$ g of methanol is present in $70$ g of CCl $_4$ .

Molar mass of methanol (CH $_3$ OH) $= 12 + 4 \times 1 + 16 = 32$ g mol $^{-1}$

Molar mass of CCl $_4 = 12 + 4 \times 35.5 = 154$ g mol $^{-1}$

Moles of methanol $= \dfrac{30}{32} = 0.9375$ mol

Moles of CCl $_4 = \dfrac{70}{154} = 0.4545$ mol

Mole fraction of CCl $_4 = \dfrac{0.4545}{0.9375 + 0.4545} = \dfrac{0.4545}{1.392} \approx 0.326 \approx 0.33$

Thus, Statement I is true.

Statement II:

In pure methanol, molecules are held together by strong intermolecular hydrogen bonding. On adding CCl $_4$ , its molecules come between methanol molecules and break the hydrogen bonds. This decreases the intermolecular attractive forces, leading to an increase in vapour pressure. Hence, the mixture shows a positive deviation from Raoult's law.

Thus, Statement II is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true

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Solution

Statement I:

$30\%$ (w/w) solution of methanol in CCl $_4$ means $30$ g of methanol is present in $70$ g of CCl $_4$ .

Molar mass of methanol (CH $_3$ OH) $= 12 + 4 \times 1 + 16 = 32$ g mol $^{-1}$

Molar mass of CCl $_4 = 12 + 4 \times 35.5 = 154$ g mol $^{-1}$

Moles of methanol $= \dfrac{30}{32} = 0.9375$ mol

Moles of CCl $_4 = \dfrac{70}{154} = 0.4545$ mol

Mole fraction of CCl $_4 = \dfrac{0.4545}{0.9375 + 0.4545} = \dfrac{0.4545}{1.392} \approx 0.326 \approx 0.33$

Thus, Statement I is true.

Statement II:

Thus, Statement II is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true

JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry