Final answer:
To prepare 8 oz of 30% solution, you should combine 140 mL of a 10% solution with 100 mL of a 50% solution. The correct answer is a. 100 mL of 50% solution; 140 mL of 10% solution.
Step-by-step explanation:
To solve this problem, we can use the concept of mixtures. Let x be the volume (in mL) of the 10% solution and y be the volume (in mL) of the 50% solution.
We can set up two equations based on the information given:
x + y = 8 (since we need a total of 8 oz of solution)
0.10x + 0.50y = 0.30(8)
Solving these equations, we find that 140 mL of the 10% solution and 100 mL of the 50% solution need to be combined to prepare 8 oz of 30% solution.
To answer the student's question regarding how many mL of 10% and 50% solutions must be combined to prepare 8 oz of a 30% solution, we need to use the concept of mixtures in algebra. First, we must convert the volume from ounces to milliliters, since the rest of the problem uses the metric system.
There are approximately 29.5735 mL in 1 oz, so 8 oz is approximately 236.588 mL (8 x 29.5735 = 236.588). Let's call the volume of the 10% solution 'V10' and the volume of the 50% solution 'V50'. The equation is then:
0.10V10 + 0.50V50 = 0.30(236.588)
We must also take into account that the total volume of the mixed solution is the sum of V10 and V50:
V10 + V50 = 236.588