A 15 percent by mass glucose solution is given. What is its approximate molality?
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Answer and explanation
Correct answer: 0.98 mol kg−1
Assume 100 g solution. It contains 15 g glucose and 85 g water. Moles of glucose = 15/180 = 0.0833 mol, and solvent mass = 0.085 kg. Hence molality = 0.0833/0.085 ≈ 0.980 mol kg−1. The denominator is the mass of water only, which is why using 100 g would produce a wrong result.
Frequently asked questions
What is the correct answer to this question?
0.98 mol kg−1
Why is this the correct answer?
Assume 100 g solution. It contains 15 g glucose and 85 g water. Moles of glucose = 15/180 = 0.0833 mol, and solvent mass = 0.085 kg. Hence molality = 0.0833/0.085 ≈ 0.980 mol kg−1. The denominator is the mass of water only, which is why using 100 g would produce a wrong result.
Which subject and chapter does this question cover?
This is a Class 11 Chemistry question. Chapter: Some Basic Concepts of Chemistry. Topic: Mole concept and molar mass.
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