Question

in chemistry class, 9 Liters of a 40% silver iodine solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed?

Answer 1

9 L of 40% silver iodine solution contains 0.40 (9 L) = 3.6 L of silver iodine.

x L of a 10% solution contains 0.10 (x L) = 0.1x L of silver iodine.

Mixing these two solutions results in another solution of volume (x + 9) L and containing (0.1x + 3.6) L of silver iodine, which gives it a concentration of 6%.

But this is impossible! Let a < b. You cannot mix two solutions at concentrations of a and b and magically end up with a mixture whose concentration is smaller than a or larger than b, it has to fall somewhere in the middle.

We can still try solving for x to demonstrate this:

(0.1x + 3.6)/(x + 9) = 0.06

0.1x + 3.6 = 0.06 (x + 9)

0.1x + 3.6 = 0.06x + 0.54

0.04x = -3.06

x = -76.5

Nonsense.

Perhaps you meant the target concentration to be 16%? In that case, there is a valid solution:

(0.1x + 3.6)/(x + 9) = 0.16

0.1x + 3.6 = 0.16 (x + 9)

0.1x + 3.6 = 0.16x + 1.44

0.06x = 2.16

x = 36