Question

You have a 3 M stock solution of sulfuric acid and you make the following dilutions 12 mL of the 3 M stock solution is diluted to 150 mL to make solution 25 mL of solution is diluted to 350 mL to make solution

what is the sulfuric acid concentration of solution (2)
a.) .017 M
b.) .023 M
c.) .0058 M
d.) .35 M"

Answer 1

The sulfuric acid concentration of solution (2) is 0.24 M.

Step-by-step explanation:

To determine the sulfuric acid concentration of the solution, we can use the dilution formula: C1V1 = C2V2 where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. Let's calculate the concentration of the final solution (2):

1. C1 = 3 M, V1 = 12 mL, C2 = ?, V2 = 150 mL
2. 3 M * 12 mL = C2 * 150 mL
3. C2 = (3 M * 12 mL) / 150 mL = 0.24 M

Therefore, the sulfuric acid concentration of solution (2) is 0.24 M. None of the provided options match this concentration, so the correct answer is not given. Please double-check your options or consult your teacher for clarification.