Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to :

Core Logic

(A) Molality (m$m$) = Moles of soluteMass of solvent (kg) = (2.5 / 60)/(75 / 1000) = (2.5)/(60) × (1000)/(75) = 0.5555 ≈ 0.556 m$\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{2.5 / 60}{75 / 1000} = \frac{2.5}{60} \times \frac{1000}{75} = 0.5555 \approx 0.556 \text{ m}$. (Correct) (B) Molarity (M$M$) = Moles of soluteVolume of solution (L) = (5 / 40)/(450 / 1000) = (5)/(40) × (1000)/(450) = 0.2777 ≈ 0.278 M$\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{5 / 40}{450 / 1000} = \frac{5}{40} \times \frac{1000}{450} = 0.2777 \approx 0.278 \text{ M}$. (Correct) (C) Aquatic species are more comfortable in cold water because the solubility of dissolved oxygen is higher at lower temperatures (Henry's Law constant KH$K_H$ increases with temperature, implying solubility decreases as T increases). (Correct) (D) According to Henry's Law (P = KH χ$P = K_H \chi$), solubility of a gas is directly proportional to partial pressure. So, solubility decreases with a decrease in pressure. (Incorrect) (E) Mole fraction of B (xB$x_B$) should be (nB)/(nA + nB)$\frac{n_B}{n_A + n_B}$. The given formula has nA$n_A$ in the numerator, which is for xA$x_A$. (Incorrect)

Step 1: Final Conclusion

Only statements A, B, and C are correct.

Pattern Recognition

Henry's Law facts are staples: Higher T = Lower solubility (fishes prefer cold). Higher P = Higher solubility (soda fizzes when pressure drops).

Chapter Mix

Class 12 Chemistry: Solutions