A 20% by mass glucose solution is prepared in water. What is its approximate molality?
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Answer and explanation
Correct answer: 1.39 mol kg−1
Take 100 g of solution. It contains 20 g glucose and 80 g water, or 0.080 kg solvent. Glucose moles are 20/180 = 0.1111 mol. Therefore molality = 0.1111/0.080 = 1.3889 mol kg−1, approximately 1.39 mol kg−1. Option B is correct; using total solution mass instead of solvent mass would give a wrong value.
Frequently asked questions
What is the correct answer to this question?
1.39 mol kg−1
Why is this the correct answer?
Take 100 g of solution. It contains 20 g glucose and 80 g water, or 0.080 kg solvent. Glucose moles are 20/180 = 0.1111 mol. Therefore molality = 0.1111/0.080 = 1.3889 mol kg−1, approximately 1.39 mol kg−1. Option B is correct; using total solution mass instead of solvent mass would give a wrong value.
Which subject and chapter does this question cover?
This is a Class 11 Chemistry question. Chapter: Chapter 01: Solutions.
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