Chemistry · Determination of molecular mass using colligative properties, abnormal value of molar mass, van't Hoff factor and its significance
The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benze
The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by \( 0.45^{\circ} \mathrm{C} \). The degree of association of acetic acid in benzene is: [Assume acetic acid dimerizes in benzene and \( K_{f} \) for benzene \( =\mathbf{5 . 1 2} \mathbf{K} \) \( \left.\mathrm{kg} \mathrm{mol}^{-1}\right] \)
- A. \( 94.5 \% \)
- B. \( 54.9 \% \)
- C. \( 78.2 \% \)
- D. 100\%
Step-by-step solution
Given ΔTf = 0.45°C, Kf = 5.12 K kg mol⁻¹, mass of acetic acid = 0.2 g, mass of benzene = 20.0 g. Molar mass of acetic acid = 60 g/mol. Molality (if no association) = (0.2/60) / (20.0/1000) = 0.1667 mol/kg. Observed van't Hoff factor i = ΔTf / (Kf × m) = 0.45 / (5.12 × 0.1667) = 0.5273. For dimerization, i = 1 - α/2, so α = 2(1 - i) = 2(1 - 0.5273) = 0.9454 = 94.54% ≈ 94.5%.