Question

1. How much pure alcohol is added
to 400 mL of 15% solution to make
its strength 32%​

Answer 1

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

Answer 2

Hi mate,

Solution :

=> Volume of alcohol in 400 ml solution

➾ 0.15 × 400 ml

=> Volume of alcohol in 400 ml solution

➾ 60 ml

=> Volume of others liquid

➾ 340 ml

=> Let x ml of alcohol is added to make it 32% strong

=> Then, volume of alcohol

➾ 60+x ml

=> Volume of other liquid

➾ 340 ml

=> Volume of solution

➾ 400 + x ml

➾ 0.32 × (400 + x) = 60 + x

➾ 128 + 0.32x = 60 + x

➾ 0.68x = 68

➾ 68x = 6800

➾ x = 100 ml ...

Answer : 100 ml pure alcohol must be added to 400ml of a 15% solution to make it's strength 32%

i hope it helps you