Question

A chemist has some 40% acid solution, some 60% acid solution, and a wholebunch of free time. How many liters of each should be used to get 40 liters of a 45%solution?

Answer 1

Answer:

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

Explanation:

Let volume of the 40% acid solution be x.

Let volume of the 60% acid solution be y.

Volume of solution formed after mixing both solution = 40 L

x + y = 40 L..[1]

Volume of acid 40% solution = 40% of x= 0.4x

Volume of acid 60% solution = 60% of y= 0.6y

Volume of acid formed = 45% of 40 L = $\frac{45}{100}\times 40=18L$

$0.4x+0.6y=18 L$..[2]

Solving [1] and [2]

x = 30 L  ,   y = 10 L

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.