How much pure alcohol has to be added to 400 ml of a solution containing 15% alcohol to change the concentration
of alcohol in the mixture to 32% ?
Quantity of alcohol in 400 ml solution = `(15/100 xx 400)`ml = 60 ml
Quantity of water = (400 - 60)ml = 340 ml
Let `x` ml of alcohol be added .
Then , `(60 + x)/(400 + x) = 32/100 hArr 6000 + 100x = 12800 + 32x hArr 68x = 6800 hArr x = 100`
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