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The molality of a 1 M aqueous NaCl solution (density = 1.04 g/mL, NaCl M = 58.5 g/mol) is closest to

A{'text': '0.98 m', 'label': 'A'}
B{'text': '1.02 m', 'label': 'B'}
C{'text': '1.10 m', 'label': 'C'}
D{'text': '0.94 m', 'label': 'D'}
Answer & Solution
Correct answer: A. {'text': '0.98 m', 'label': 'A'}
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: NCERT Class 12 Chemistry, Unit 1 "Solutions", §1.2_
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