Question

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

Answer 1

MARK IT AS BRAINLIEST....

Answer:

100mL

Step-by-step explanation:

Initial amount of alcohol is

400 × 15 = 60

100

Let x ml of pure alcohol is to be added.

Then,

Amount of alcohol becomes 60 + x

Total amount of solution becomes 400 + x

60 + x/400 + x 100 = 32

60 + x = 128 + 0.32x

0.68x = 68

x = 100

Answer 2

Answer:

Initial amount of alcohol is

400 \times \frac{15}{100} = 60 \\

Let x ml of pure alcohol is to be added.

Then,

Amount of alcohol becomes 60 + x

Total amount of solution becomes 400 + x

\frac{60 + x}{400 + x} \times 100 = 32 \\ 60 + x = 128 + 0.32x \\ 0.68x = 68 \\ x = 100 \: ml......

hope it's helps........