How much of a 6% solution and a 11% solution would you use to make 250 mL of a 8%-electrolyte solution?
A) 130 mL of the 11% solution,
120 mL of the 6% solution.
B) 90 mL of the 11% solution,
160 mL of the 6% solution.
C) 100 mL of the 11% solution,
150 mL of the 6% solution.
D) 85 mL of the 11% solution,
165 mL of the 6% solution.
E) 150 mL of the 11% solution,
100 mL of the 6% solution.

Question

Quizplus · Accepted Answer

Let x be the amount of the 6% solution and y be the amount of the 11% solution needed.
We can set up two equations using the given information:

x + y = 250 (we need a total of 250 mL)
0.06x + 0.11y = 0.08(250) (the amount of electrolyte in the final solution must be 8%)

Simplifying the second equation, we get:

0.06x + 0.11y = 20

Multiplying both sides of the first equation by 0.06 and subtracting it from the second equation, we can solve for y:

0.11y - 0.06x = 20 - 0.06(250)
0.11y - 0.06x = 5
11y - 6x = 500

Multiplying the first equation by 6 and subtracting it from this equation, we can solve for x:

11y - 6x = 500
-6x - 6y = -1500
-6x = -1000
x = 166.67

Substituting this value of x into the first equation, we get:

166.67 + y = 250
y = 83.33

Therefore, we need 150 mL of the 6% solution and 100 mL of the 11% solution to make 250 mL of an 8%-electrolyte solution.

How Much of a 6% Solution and a 11% Solution