Question

For a college chemistry experiment, students need to prepare a solution containing ethylene glycol. They are each given 20 fluid ounces of a solution containing 2% ethylene glycol, to which they will add another solution that contains 12% ethylene glycol. How much of the 12% solution should they add to obtain a 5% ethylene glycol solution?
Write your answer as a whole number or as a decimal rounded to the nearest tenth.
fluid o

Answer 1

To solve this problem, we need to determine how much of the 12% ethylene glycol solution should be added to achieve a final concentration of 5% ethylene glycol after mixing it with the 20 fluid ounces of 2% ethylene glycol solution.

Let's denote:
- \( V_2 = \) volume of the 2% solution = 20 fluid ounces
- \( C_2 = \) concentration of the 2% solution = 2% = 0.02
- \( V_1 = \) volume of the 12% solution that is to be added
- \( C_1 = \) concentration of the 12% solution = 12% = 0.12
- \( C_f = \) final concentration we want = 5% = 0.05

The total volume of the resulting solution will be:
\[ V = V_2 + V_1 \]

The total amount of ethylene glycol in the 2% solution is:
\[ \text{Amount from } V_2 = C_2 \times V_2 = 0.02 \times 20 = 0.4 \text{ ounces of ethylene glycol} \]

The total amount of ethylene glycol in the 12% solution is:
\[ \text{Amount from } V_1 = C_1 \times V_1 = 0.12 \times V_1 \]

We want the final amount of ethylene glycol to equal the final concentration of 5% times the total volume of the solution:
\[ C_f \times V = 0.05 \times (20 + V_1) \]

Now, setting up the equation for the amount of ethylene glycol:
\[
0.4 + 0.12 V_1 = 0.05 \times (20 + V_1)
\]

Now expanding the right-hand side:
\[
0.4 + 0.12 V_1 = 1 + 0.05 V_1
\]

Now we move terms involving \( V_1 \) to one side and constant terms to the other side:
\[
0.12 V_1 - 0.05 V_1 = 1 - 0.4
\]
\[
0.07 V_1 = 0.6
\]

Now we solve for \( V_1 \):
\[
V_1 = \frac{0.6}{0.07} \approx 8.5714
\]

Rounding to the nearest tenth,
\[
V_1 \approx 8.6 \text{ fluid ounces}
\]

Thus, the amount of 12% ethylene glycol solution that should be added is approximately **8.6 fluid ounces**.