Question:
A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is
- 52%
- 50%
- 55%
- 48%
Let the volume of the first and the second Solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture
= (20 + 300S)
Let this Solution be mixed with equal volume i.e. 400 of third Solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final Solution
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = \(\frac{1}{2}\) = 50%
Hence, 50 is the correct answer.
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