Question

You need 420 mL of a 80% alcohol solution. On hand, you have a 30% alcohol mixture. How much of the 30%alcohol mixture and pure alcohol will you need to obtain the desired solution?

Answer 1

Answer:

We need 300 mL of 80% pure alcohol solution and 120 mL of 30% alcohol mixture

Explanation:

Let x represent 80% alcohol solution, the quantity is 0.8 mL

Let y represent a 30% alcohol mixture, the quantity is 0.3 mL

We need to obtain the amount of 30% alcohol mixture and pure alcohol that it will take to obtain a 420 mL of a 80% alcohol solution.

x + y = 420 ............................................(1)

x = 420 - y

Since there is 80% alcohol solution in the final result, we have:

0.8 * 420 = 336 mL

0.3y + x = 336

0.3y + (420 - y) = 336

0.3y - y = 336 - 420

0.7y = 84

y = 84/0.7

= 120 mL

x = 420 - 120

= 300 mL

Therefore, we need 300 mL of 80% pure alcohol solution and 120 mL of 30% alcohol mixture