alleryCalculate the molality of a solution prepared by dissolving 2.5 g of ethanoic acid (CH₃COOH, M = 60 g mol⁻¹) in 75 g of benzene.
A0.556 mol kg⁻¹
B1.20 mol kg⁻¹
C0.42 mol kg⁻¹
D0.083 mol kg⁻¹
Answer & Solution
Correct answer: A. 0.556 mol kg⁻¹
Moles solute = 2.5 / 60 = 0.0417 mol.
Mass solvent = 0.075 kg.
Molality = 0.0417 / 0.075 = **0.556 mol kg⁻¹**.
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