Molality of a solution containing 2.5 g of ethanoic acid (CH₃COOH, M =…

Question

Molality of a solution containing 2.5 g of ethanoic acid (CH₃COOH, M = 60 g/mol) in 75 g of benzene is closest to

Accepted Answer

D. {'text': '0.556 m', 'label': 'D'}

1. Moles of CH₃COOH = 2.5 / 60 = 0.0417 mol.
2. Mass of benzene (solvent) in kg = 75 / 1000 = 0.075 kg.
3. Molality = 0.0417 / 0.075 = 0.556 mol/kg.
4. So the solution is 0.556 m in ethanoic acid.
_Source: NCERT Class 12 Chemistry, Unit 1 "Solutions", Example 1.3_

Answer

{'text': '0.278 m', 'label': 'A'}

Answer

{'text': '0.333 m', 'label': 'B'}

Answer

{'text': '1.000 m', 'label': 'C'}