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Class 11ChemistryChapter 01: SolutionsMediumLevel 5

What mass of sodium carbonate is present in 250 mL of 0.5 M sodium carbonate solution?

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Answer and explanation

Correct answer: 13.25 g

Convert 250 mL to 0.250 L. The amount of sodium carbonate is n = M × V = 0.5 × 0.250 = 0.125 mol. Anhydrous Na2CO3 has molar mass 106 g mol−1. Hence mass = 0.125 × 106 = 13.25 g. The result would differ if a hydrated salt were specified, but the question names sodium carbonate without hydration.

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solutionsmolarity massChapter 01: Solutionschapter 01 solutionsChemistryClass 11 MCQ

Frequently asked questions

What is the correct answer to this question?

13.25 g

Why is this the correct answer?

Convert 250 mL to 0.250 L. The amount of sodium carbonate is n = M × V = 0.5 × 0.250 = 0.125 mol. Anhydrous Na2CO3 has molar mass 106 g mol−1. Hence mass = 0.125 × 106 = 13.25 g. The result would differ if a hydrated salt were specified, but the question names sodium carbonate without hydration.

Which subject and chapter does this question cover?

This is a Class 11 Chemistry question. Chapter: Chapter 01: Solutions.

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