How much 12 mol/L acid is needed to prepare 500 mL of 0.25 mol/L acid by dilution?
Answer and explanation
Correct answer: D. 10.417 mL
- Step 1: C1 V1 = C2 V2.
- Step 2: V1 = 0.25 x 500/12 = 10.417 mL.
Answer
10.417 mL.
Why not the others: 24000 mL, 489.583 mL and 20.833 mL do not fit.
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