To 500 cm3 of water, 3.0 × 10−3 kg of acetic acid is added. If 23% of the acetic acid is dissociated, what will be the depression in the freezing point? Kf and the density of water are 1.86 K kg mol−1 and 0.997 g cm−3 respectively.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.229 K
Acetic acid is a weak electrolyte: it dissociates only partly, so the solution holds more particles than the moles of acid added. The van’t Hoff factor i accounts for this:
\( \displaystyle \Delta T_f=i\,K_f\,m \)
van’t Hoff factor: \( \ce{CH3COOH <=> CH3COO^- + H^+} \). With α = 0.23, each mole of acid gives (1 − α) mol of undissociated acid + α mol of each ion, so i = 1 + α = 1.23.
Moles of acid: 3.0 × 10−3 kg = 3.0 g; molar mass of \( \ce{CH3COOH} \) = 60 g mol−1, so n = 3.0/60 = 0.05 mol.
Mass of water: 500 cm3 × 0.997 g cm−3 = 498.5 g = 0.4985 kg.
\( \displaystyle m=\frac{0.05\,\mathrm{mol}}{0.4985\,\mathrm{kg}}=0.1003\,\mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_f=1.23\times1.86\,\mathrm{K\,kg\,mol^{-1}}\times0.1003\,\mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_f=0.229\,\mathrm{K} \)
The depression in the freezing point is 0.229 K.