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 31.25% ethanol solution, then the unknown concentration of S is

We have a 20 per cent ethanol solution
So, in 100 ml of this solution, we have 20 ml of alcohol
Now, there is another solution with S % concentration of alcohol
So, in 100 ml of this solution there will be S ml of alcohol
Now, the 20 per cent ethanol and S% ethanol solutions are mixed in 1:3
So, let's mix 100 ml of 20 per cent ethanol solution with 300 ml of S% ethanol solution
So, in this 400 ml, the total amount of ethanol will be: 1(20)+3(S)=20+3S ml
Now, another 400 ml of 20 per cent ethanol solution is mixed in this earlier mixture
So, in the final 800 ml mixture of ethanol solution, the total amount of ethanol will be =20+3S+4(20)=100+3S ml
It was given to us that the ethanol concentration in the final mixture is 31.25 per cent of our final mixture volume, which is 800 ml
So, [( 100+3S)*100]/800=31.25
or, 100+3S=31.25*8
or, 100+3S=250
or, 3S=150
or, S=50
Hence, the percentage concentration of S% ethanol solution is =50%

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