100 mL of 1 M hydrochloric acid is mixed with 200 mL of 0.5 M same acid. If final volume is 300 mL what is the molarity?
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Answer and explanation
Correct answer: 0.67 mol L−1
Calculate solute moles in each portion: n1 = 1.0 × 0.100 = 0.100 mol and n2 = 0.5 × 0.200 = 0.100 mol. Total moles are 0.200 mol. With final volume 0.300 L, M = 0.200/0.300 = 0.6667 M, approximately 0.67 M. Thus C is correct; concentrations cannot be averaged without weighting by volumes.
Frequently asked questions
What is the correct answer to this question?
0.67 mol L−1
Why is this the correct answer?
Calculate solute moles in each portion: n1 = 1.0 × 0.100 = 0.100 mol and n2 = 0.5 × 0.200 = 0.100 mol. Total moles are 0.200 mol. With final volume 0.300 L, M = 0.200/0.300 = 0.6667 M, approximately 0.67 M. Thus C is correct; concentrations cannot be averaged without weighting by volumes.
Which subject and chapter does this question cover?
This is a Class 11 Chemistry question. Chapter: Chapter 01: Solutions.
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