can not answer is been over35 years i have been out of school

Question

you want to create a mixture of 51% alcohol from 5L of 22% solution .How many liters of 92% alcohol should be added

Answer 1

Danny, here's how to tackle this kind of problem:

The essential point is that the volumes of the desired component (i.e., alcohol) must balance out.

We start with 5L of a solution containing 22% of alcohol => solution contains 22/100 x 5 = 1.1 L of alcohol.

We want a solution with 51% of alcohol.

Suppose we add n L of the 92% solution => we are adding 92/100 x n = 0.92n L of pure alcohol.

Therefore, we now have a total of 1.1 + 0.92n L of alcohol in a solution of (n + 5) L.

This must be a 51% mixture => (1.1 + 0.92n) / (n + 5) = 0.51 . . . . . . . . . [51% = 0.51

Now, we can solve for n: 1.1 + 0.92n = 0.51((n + 5) . . . . [cross-multiplying

1.1 + 0.92n = 0.51n + 2.55

0.92n - 0.51n = 2.55 - 1.1

0.41n = 1.45

n = 1.45 / 0.41

n = 3.54 L of 92% solution was added to get the desired 51% solution