Question

How many mL of 75% alcohol should be mixed with 10% of 1000 cc alcohol to prepare 30% of 500 mL alcohol solution? a. 346.16 mL b. 234.43 mL c. 153.84 mL d. 121.12 mL e.

Answer 1

Answer:

C. 153.84 mL

Step-by-step explanation:

Let's say x is the volume of 75% solution and y is the volume of 10% solution.

Sum of the volumes:

x + y = 500

Sum of the alcohol amounts:

0.75x + 0.10y = 0.30(500)

0.75x + 0.10y = 150

Solve the system of equations using either substitution or elimination.  I'll use substitution.

y = 500 - x

0.75x + 0.10 (500 - x) = 150

0.75x + 50 - 0.10x = 150

0.65x = 100

x = 153.84

You need 153.84 mL of 75% solution.