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):
A2.08 K
B0.52 K
C0.26 K
D1.04 K
Answer & Solution
Correct answer: D. 1.04 K
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: NCERT Class 12 Chemistry Ch 1 "Solutions", p.22_
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