A 360 ml aqueous solution contains 40% alcohol. How much will be the approximate percentage of alcohol if 3600 ml of water is added to the solution?
Part ACSIR NET September 2022solute-fixed-solvent-added
Solute fixed solvent added
- A.2.6
- B.3.6✓ Correct
- C.4.0
- D.1.0
The idea, in plain words
Pouring in water adds no alcohol and removes none. The alcohol is frozen at whatever it was; only the total volume grows, so the percentage drops by exactly the factor the volume grew by.
Solution
The alcohol is 40% of 360 = 144 ml and does not change.
The volume becomes 360 + 3600 = 3960 ml
so the concentration is 144/3960 = 40/11 ≈ 3.64%.
Why each option is right or wrong
Checked against 1 symbolic computationFalse.
2.6 would follow from dividing by a total that omits the original solution — 144/5400 or similar.
The new volume is 360 + 3600 = 3960, not 3600 alone.
- B.Correct
True.
The alcohol is fixed at 40% of 360 = 144 ml; only the water changes.
The new concentration is 144/3960 = 40/11 ≈ 3.636% (verified)
which rounds to 3.6.
The trigger: in every dilution problem, identify the quantity that does not change and put it in the numerator.
False.
4.0 comes from dividing 144 by 3600 — the added water alone — giving exactly 4%.
Forgetting that the original 360 ml is still in the beaker is the single most common slip here.
False.
1.0 is roughly 40 divided by 40, that is treating the tenfold dilution as if it acted on the percentage rather than on the volume.
Dilution by a factor of 11 takes 40% to 3.64%, not to 1%.
3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.
The trap it tests
Execution slipThe idea was right. The computation was not.
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