Question

You need 320 mL of a 65% alcohol solution. On hand, you have a 60% alcohol mixture. How much of the 60% alcohol mixture and pure alcohol will you need to obtain the desired solution?
You will need
____ mL of the 60% solution
and
_____ mL of pure alcohol.

Answer 1

Answer:

You will need

280 mL of the 60% solution

and  40 mL of pure alcohol.

Step-by-step explanation:

Let 'x' be the amount of 60% alcohol solution and y the amount of pure alcohol.

Therefore:

(0.6x + y)/320 = 0.65 ⇒ 0.6x + y = 208

x + y = 320

Solving the sistem of equations:

x = 280 and y = 40

Therefore, You will need

280 mL of the 60% solution

and  40 mL of pure alcohol.