What will be the change in freezing point of water, when 0.05 mol of calcium phosphate is dissolved in 1000 g of water? Assume calcium phosphate completely dissociates into its ions in aqueous solution. (Kf = 1.86 K kg mol−1)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.465 K
Depression in freezing point is a colligative property, so it depends on the number of solute particles. For an electrolyte, the molality is multiplied by the van’t Hoff factor i, the number of ions each formula unit gives:
\( \displaystyle \Delta T_f=i\,K_f\,m \)
Calcium phosphate dissociates completely as:
\( \displaystyle \ce{Ca3(PO4)2 -> 3Ca^2+ + 2PO4^3-} \)
This gives 3 + 2 = 5 ions, so i = 5.
\( \displaystyle m=\frac{0.05\ \mathrm{mol}}{1\ \mathrm{kg}}=0.05\ \mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_f=5\times1.86\times0.05 \)
\( \displaystyle \Delta T_f=0.465\ \mathrm{K} \)
So the freezing point of water falls by 0.465 K.