18 g of glucose (molar mass 180 g/mol) is dissolved in 1 kg of water. Given Kb for water is 0.52 K kg/mol and water boils at 373.15 K, the boiling point of the solution is:
A373.150 K
B373.202 K
C372.630 K
D373.670 K
Answer & Solution
Correct answer: B. 373.202 K
1. Moles of glucose = $18 / 180 = 0.1$ mol; molality = $0.1 / 1 = 0.1$ m.
2. $\Delta T_b = K_b m = 0.52 \times 0.1 = 0.052$ K.
3. Boiling point = $373.15 + 0.052 = 373.202$ K. (Option A ignores the elevation.)
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.16_
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