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3.4 Using concentrations of solutions in mol/dm3

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Question 17
H

In a titration, 25.0 cm3 of sodium carbonate solution was exactly neutralised by 20.0 cm3 of 0.100 mol/dm3 hydrochloric acid. The equation is Na2CO3+2HCl→2NaCl+H2O+CO2\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2Na2​CO3​+2HCl→2NaCl+H2​O+CO2​. What is the concentration of the sodium carbonate solution?

[1]
A

0.0200 mol/dm30.0200\ \text{mol/dm}^30.0200 mol/dm3

B

0.0400 mol/dm30.0400\ \text{mol/dm}^30.0400 mol/dm3

C

0.0800 mol/dm30.0800\ \text{mol/dm}^30.0800 mol/dm3

D

0.160 mol/dm30.160\ \text{mol/dm}^30.160 mol/dm3

Markscheme

3.4 Using concentrations of solutions in mol/dm3 Questions

  1. GCSE
  2. /Chemistry
  3. /3.4 Using concentrations of solutions in mol/dm3

40 exam-style questions on AQA GCSE Chemistry 3.4 Using concentrations of solutions in mol/dm3, covering 3.4.1 Concentration in mol/dm3. Each one has a worked solution and a mark scheme showing where the marks go.

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