KBr is 80% dissociated in an aqueous solution of molality 0.5 mol kg−1. (Kf for water = 1.86 K kg mol−1; freezing point of water = 273 K) The solution freezes at:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 271.326 K
Depression of freezing point is a colligative property: it depends on the number of solute particles. For an electrolyte the van’t Hoff factor i counts the particles formed from each formula unit.
Dissociation: \( \ce{KBr -> K^+ + Br^-} \); each formula unit that dissociates gives 2 ions.
Particles from 1 mol of KBr: 0.8 mol dissociates into 1.6 mol of ions and 0.2 mol stays undissociated, so i = 1.6 + 0.2 = 1.8.
\( \displaystyle \Delta T_f=i\,K_f\,m \)
\( \displaystyle \Delta T_f=1.8\times 1.86\ \mathrm{K\,kg\,mol^{-1}}\times 0.5\ \mathrm{mol\,kg^{-1}}=1.674\ \mathrm{K} \)
\( \displaystyle T_f=273\ \mathrm{K}-1.674\ \mathrm{K}=271.326\ \mathrm{K} \)
The solution freezes at 271.326 K.