The molality of a 1 M aqueous NaCl solution (density = 1.04 g/mL, NaCl M = 58.5 g/mol) is closest to
A0.98 m
B1.02 m
C1.10 m
D0.94 m
Answer & Solution
Correct answer: A. 0.98 m
1. 1 L of 1 M NaCl contains 1 mol = 58.5 g of NaCl.
2. Mass of the solution = 1000 mL × 1.04 g/mL = 1040 g.
3. Mass of water = 1040 − 58.5 = 981.5 g = 0.9815 kg.
4. Molality = 1 / 0.9815 = 1.019 m ≈ 1.02 m.
_Source: syllabus Chemistry, Unit 1 "Solutions", §1.2_
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