Question

There is a beaker of 3.5% acid solution and a beaker of 6% acid solution in the science lab. Mr. Larson needs 200 milliliters of a 4.5% acid solution for an experiment. How many milliliters of each solution should he combine?

Answer 1

To obtain 200 mL of a 4.5% acid solution, Mr. Larson needs 120 mL of the 3.5% acid solution and 80 mL of the 6% acid solution.

Step-by-step explanation:

To find the volumes of each of the two solutions needed, we can set up a system of equations based on the principle of concentration:

Let x be the volume of the 3.5% acid solution needed, and y be the volume of the 6% acid solution needed.

The equation for the total volume is x + y = 200 mL.

The equation for the concentration is (0.035x + 0.06y) / 200 = 0.045.

Solving the system of equations, we can find the values of x and y.

x = 200 - y

(0.035(200 - y) + 0.06y) / 200 = 0.045

Simplifying the equation, we get:

7 - 0.035y + 0.06y = 9

0.025y = 2

y = 80 mL

Substituting the value of y into the equation x = 200 - y, we get:

x = 200 - 80

x = 120 mL

Therefore, Mr. Larson needs 120 mL of the 3.5% acid solution and 80 mL of the 6% acid solution to obtain 200 mL of a 4.5% acid solution.

Answer 2

et x = the volume of the 3.5% solution
Let y = the volume 6% solution

(3.5%)x + (6%)y = (4.5%)(200)
0.035)x + 0.06y = 0.045(200)
0.035x + 0.06y = 9 ( Equation 1 )

x + y = 200 ( Equation 2 )
x = 200 - y

Substitute x = 200 - y into equation 1:
0.035x + 0.06y = 9
0.035(200 - y) + 0.06y = 9
7 - 0.035y + 0.06y = 9
0.06y - 0.035y = 9 - 7
0.025y = 2
y = 2/(0.025)
y = 80 mL (the volume of 6% solution.)

Substitute y = 80 into equation 2:
x + y = 200
x + 80 = 200
x = 200 - 80
x = 120 mL (the volume of the 3.5% solution.)

Therefore, Mr. Larson should combine 120 mL of the 3.5% solution
and 80 mL of the 6% solution to make 200 mL of 4.5% solution.

Answer is 120 mL of the 3.5% solution & 80 mL of the 6% solution

Hope I helped :)