question_answer

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

Answer:

Quantity of pure alcohol in 400 ml. of 15% \[=\text{ }400\times \frac{15}{100}\] = 60 ml. Now, we add x ml of pure alcohol to the sample. So, total pure alcohol = (60 + x) ml. But volume of new sample = (400 + x) ml. ∴ Percentage of pure alcohol in new sample =\[\frac{\left( 60+x \right)}{\left( 400+x \right)}\times 100\] which is equal to 32% \[\frac{60+x}{400+x}\times 100=32\] \[\frac{60+x}{400+x}=\frac{32}{100}\] 100(60 + x) = 32(400 + x) 100x + 6000 = 32x + 12800 100x ? 32x = 12800 ? 6000 68x = 6800 \[x=\frac{6800}{68}=100\,\,ml\].