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✓
- 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 computation- A.WrongExecution slip
False. 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.
- C.WrongExecution slip
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.
- D.WrongExecution slip
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.