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10.0 mL of a Na2CO3 solution is titrated against 0.2 M HCl. Five successive titre readings are recorded: 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL, and 5.0 mL. Using the standard convention of discarding outliers and taking the mean of concordant readings, find the concentration of the Na2CO3 solution in mM. (Round off to the nearest integer.)

1. 50
2. 25
3. 100
4. 10

Correct answer: 50

Solution

The concordant volume is 5.0 mL. Moles of HCl = 0.2 * 5.0/1000 = 0.001 mol; moles of Na2CO3 = 0.0005 mol; concentration = 0.0005/0.010 = 0.05 M = 50 mM.

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