Question

How many mL of a 15% acid solution and how many mL of a 3% acid solution must be mixed to get 45 mL of a 7% acid solution?

Answer 1

Answer:

15 mL of 15% acid solution and 30 mL of 3% acid solution

Explanation:

The total number of moles of the mixture $(n_{3})$ is equivalent to the addition of the number of moles of the first solution $(n_{1})$ and the number of moles of the second solution $(n_{2})$. Mathematically,

$n_{1} +n_{2} = n_{3}$

$n_{1} = V_{1}*C_{1}$

$n_{2} = V_{2}*C_{2}$

$n_{3} = V_{3}*C_{3}$

The volume of the mixture = 45 mL

The volume of first solution = x

The volume of second solution = 45 - x

Therefore:

0.15x + 0.03(45-x) = 0.07*45

0.15x + 1.35-0.03x = 3.15

0.12x = 1.8

x = 15

Thus, the volume of the first solution is 15 mL while the volume of the second solution is 30 mL.