Question

A scientist needs 35 mL of a 60% acid solution for an experiment. The lab has available 83% solution and 80% solution. How many millimeters of the 30% solution and how many millimeters of the 80% solution should the scientist mix to make a 60% solution. Set up a system of equations, state both variables, and solve.

A) Let x represent the volume of 30% solution and y represent the volume of 80% solution. The system of equations is:
0.30x + 0.80y = 0.60(35)
x + y = 35
Solve for x and y. Answer: x = 20 mL, y = 15 mL.

Answer 1

To create 35 mL of a 60% acid solution, the scientist needs to mix 14 mL of the 30% acid solution with 21 mL of the 80% acid solution.

Step-by-step explanation:

To solve the problem of mixing two acid solutions to achieve a specific concentration, we will use a system of linear equations. We have two variables: x representing the volume of 30% acid solution, and y representing the volume of 80% acid solution. The goal is to obtain 35 mL of a 60% acid solution.

The first equation comes from the concentration requirement:

0.30x + 0.80y = 0.60(35)

The second equation represents the total volume constraint:

x + y = 35

To solve, we can use substitution or elimination. Let's multiply the second equation by 0.30, which gives us:

0.30x + 0.30y = 10.50

Subtract this from the first equation to eliminate x:

0.80y - 0.30y = 21 - 10.50

This simplifies to:

0.50y = 10.5

Dividing both sides by 0.50 gives:

y = 21 mL

Now, substituting y = 21 back into the second equation:

x + 21 = 35

Which leads to:

x = 14 mL

Therefore, the scientist should mix 14 mL of the 30% acid solution and 21 mL of the 80% acid solution to obtain the desired 60% acid solution.