Question

You need a 70% alcohol solution. On hand, you have a 420 mL of a 85% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

Answer 1

Answer:

Pure water =74.1 mL

Step-by-step explanation:

You need 420 mL of a 70% alcohol solution.

420 mL of solution and 70% of it should be alcohol.

So,

420 * 70/100 mL

= 420 * 0.7

= 294 mL of alcohol and the rest should be water.

On hand, you have a 85% alcohol mixture.

The alcohol can only come from the 85% mixture. So, how much of 85% alcohol do we need to get 294 mL of alcohol:

Let

y = alcohol solution needed

85% of y = 294

85/100 * y = 294

y = 294 ÷ 85/100

y = 294 × 100/85

y = 29,400/85

y = 345.88235294117

Approximately, 345.9mL

The rest of 420 mL solution has to be pure water

Pure water = 420 mL - 345.9mL

= 74.1 mL

Pure water =74.1 mL