JEE Main 2026 — Solutions Question with 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 $=\frac{30}{32}=0.9375$ mol
Moles of CCl${}_4=\frac{70}{154}=0.4545$ mol

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