Question

How much pure alcohol must be added to 400 mL of a 15% solution to make its strength 32%.

Answer 1

Let the volume of the pure alcohol be x ml.
Initial concentration=15%
So, initial amount of alcohol in the solution will be=$\frac{15}{100}\times 400=60$ ml.
To make the strength of the solution 32%,
we will keep the amount of water constant and add x volume of pure alcohol.
On adding pure alcohol, the volume of the solution increases to 400 + according to the question, we have:$\frac{x+60}{400+x}=\frac{32}{100}$
$100x+6000)=12800+32x$
$\Rightarrow 100x-32x=12800-6000$
$\Rightarrow 68x=6800$
$\therefore x=100$
So, amount of pure alcohol to be added$=100$ ml