A vessel contains 2 litres (2000 mL) of pure acid. 200 mL of the mixture is…
2019
A vessel contains 2 litres (2000 mL) of pure acid. 200 mL of the mixture is taken out and replaced with 200 mL of water; this removal-and-replacement is carried out two more times, so it is done a total of 3 times in all. What is the amount of pure acid left in the vessel?
Answer: C. 1458 mL — When a solution of volume V is repeatedly diluted by removing a fixed volume v and replacing it with the same volume of a different liquid, the fraction of…
- A.
1446 mL
- B.
1400 mL
- C.
1458 mL
- D.
1428 mL
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Correct answer: C
When a solution of volume V is repeatedly diluted by removing a fixed volume v and replacing it with the same volume of a different liquid, the fraction of the original substance retained after one such step is (1 − v/V). Repeating the process n times leaves V × (1 − v/V)n of the substance, because each step removes a FRACTION of whatever is currently present rather than a fixed amount — so the quantity shrinks geometrically, not linearly.
Convert the volume to millilitres: V = 2 litres = 2000 mL.
Each replacement removes v = 200 mL from the vessel, so the fraction of acid retained per step = 1 − 200/2000 = 0.9.
The removal-and-replacement is carried out a total of n = 3 times (the first time, plus two more).
Apply the formula: remaining acid = 2000 × (0.9)3 = 2000 × 0.729 = 1458 mL.
Cross-check by tracking each step individually: after the 1st replacement, acid = 2000 × 0.9 = 1800 mL; after the 2nd, acid = 1800 × 0.9 = 1620 mL; after the 3rd, acid = 1620 × 0.9 = 1458 mL — the same total as the formula gives.
So 1458 mL of pure acid remains in the vessel.