alleryCalculate the molality of a solution of 2.5 g of ethanoic acid (CH3COOH, molar mass 60 g/mol) in 75 g of benzene.
A0.0417 m
B0.556 m
C0.750 m
D0.333 m
Answer & Solution
Correct answer: B. 0.556 m
1. Molality = moles of solute / mass of solvent in kg.
2. Moles of ethanoic acid = $2.5 / 60 = 0.0417$ mol.
3. Mass of benzene = $75 / 1000 = 0.075$ kg.
4. Molality = $0.0417 / 0.075 = 0.556$ m. (Option A is the mole count of solute.)
_Source: syllabus Chemistry Ch 1 "Solutions", p.4_
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