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 :

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.

30% (w/w) methanol in $$\mathrm{CCl_4}$$​ means:

• 30 g methanol
• 70 g $$\mathrm{CCl_4}$$
  

No. of Moles :

For Methanol: $$M=12+4\left(1\right)+16=32$$

$$n_{CH_3OH}=\frac{30}{32}=0.9375$$

For $$CCl_4$$: $$M=12+4(35.5)=154$$

$$n_{CCl_4}=\frac{70}{154}=0.4545$$

Mole Fraction of $$CCl_4$$:

$$X_{CCl_4}=\ \frac{0.4545}{0.4545+0.9375}=\frac{0.4545}{1.392}\approx\ 0.326\approx\ 0.33$$

So Statement I is correct.

Statement II : Mixture of methanol and CCl$$_4$$ shows positive deviation from Raoult's law.

Methanol has strong hydrogen bonding, while $$\mathrm{CCl_4}$$​ is non-polar.

When mixed:

• methanol-methanol interactions decrease,
• new methanol-$$\mathrm{CCl_4}$$ interactions are weaker.

Therefore vapour pressure becomes higher than expected.

Hence the mixture shows:

Positive Deviation from Raoult's Law

Statement II is Correct,

Hence, Option A is correct.