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:

Step-by-step 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 1 + 16 = 32 g mol ⁻¹ Molar mass of CCl ₄ = 12 + 4 35.5 = 154 g mol ⁻¹ Moles of methanol = 30 32 = 0.9375 mol Moles of CCl ₄ = 70 154 = 0.4545 mol Mole fraction of CCl ₄ = 0.4545 0.9375 + 0.4545 = 0.4545 1.392 0.326 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