Question

Three solutions contain a certain acid. The first contains 10% acid, the second 30%, and the third 50%. A chemist wishes to use all three solutions to obtain a 50-liter mixture containing 32% acid. If the chemist wants to use twice as much of the 50% solution as of the 30% solution, how many liters of each solution should be used?

Answer 1

Answer:

To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.

Explanation:

A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:

2x*50% + x*30% + y*10% = 50L*32%

130x + 10y = 1600 (1)

-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-

Also, it is possible to write a formula using the total volume (50L), thus:

2x + x +y = 50L

3x + y = 50L (2)

If you replace (2) in (1):

130x + 10(50-3x) = 1600

100x + 500 = 1600

100x = 1100

x = 11L -Volume of 30% solution-

2x = 22L -Volume of 50% solution-

50L - 22L - 11L = 17 L -Volume of 10% solution-

I hope it helps!