Question 21
1 pts
How many milliliters of a $5 \%$ acid solution and how many milliliters of a $17 \%$ acid solution must be mixed to obtain $60 \mathrm{~mL}$ of a $13 \%$ acid solution?
$40 \mathrm{~mL}$ of $5 \%$ solution and $20 \mathrm{~mL}$ of $17 \%$ solution
$15 \mathrm{~mL}$ of $5 \%$ solution and $45 \mathrm{~mL}$ of $17 \%$ solution
$20 \mathrm{~mL}$ of $5 \%$ solution and $40 \mathrm{~mL}$ of $17 \%$ solution
$30 \mathrm{~mL}$ of $5 \%$ solution and $30 \mathrm{~mL}$ of $17 \%$ solution
- Previous
Next

Step 1 ：Let's denote the volume of the 5% solution as \(x\) (in mL) and the volume of the 17% solution as \(y\) (in mL).

Step 2 ：We know that the total volume of the solution is 60 mL, so we can write the first equation as: \(x + y = 60\).

Step 3 ：The total amount of acid in the final solution is 13% of the total volume, or \(0.13 \times 60 = 7.8\) mL. The amount of acid in the 5% solution is \(0.05x\) and in the 17% solution is \(0.17y\). So we can write the second equation as: \(0.05x + 0.17y = 7.8\).

Step 4 ：Let's multiply the first equation by 0.05: \(0.05x + 0.05y = 3\).

Step 5 ：Subtract this equation from the second equation: \(0.12y = 4.8\).

Step 6 ：Solving for \(y\) gives: \(y = \frac{4.8}{0.12} = 40\) mL.

Step 7 ：Substituting \(y = 40\) into the first equation gives: \(x + 40 = 60\).

Step 8 ：Solving for \(x\) gives: \(x = 60 - 40 = 20\) mL.

Step 9 ：\(\boxed{20}\) mL of a 5% acid solution and \(\boxed{40}\) mL of a 17% acid solution must be mixed to obtain 60 mL of a 13% acid solution.