If 1 mol of KCl is dissolved in 1 kg of water and full dissociation is assumed (i = 2), the expected elevation of boiling point is (Kb for water = 0.52 K kg/mol):

1. With the van't Hoff factor, $\Delta T_b = i K_b m$. 2. Here $i = 2$, $K_b = 0.52$ K kg/mol, $m = 1$ mol/kg. 3. $\Delta T_b = 2 \times 0.52 \times 1 = 1.04$ K. (Option A omits the factor i.) _Source: syllabus Chemistry Ch 1 "Solutions", p.22_