How much pure solute should be added to 200 g of 10 percent solution to make the mass percent 20?
Answer and explanation
Correct answer: 25 g
The initial solution contains 10% of 200 g = 20 g solute. Let x g pure solute be added. The final solute mass is 20 + x, while the final solution mass is 200 + x because the added material is also part of the solution. Set the required mass percentage: (20 + x)/(200 + x) = 0.20. Solving gives 20 + x = 40 + 0.2x, so 0.8x = 20 and x = 25 g. Option C is correct.
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What is the correct answer to this question?
25 g
Why is this the correct answer?
The initial solution contains 10% of 200 g = 20 g solute. Let x g pure solute be added. The final solute mass is 20 + x, while the final solution mass is 200 + x because the added material is also part of the solution. Set the required mass percentage: (20 + x)/(200 + x) = 0.20. Solving gives 20 + x = 40 + 0.2x, so 0.8x = 20 and x = 25 g. Option C is correct.
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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