Question

22.0 mL of stock solution is used

to produce a .458 M solution after
dilution with 50.0 mL of water. What
is the molarity of the stock solution?

Answer 1

0.202M

Step-by-step explanation:

you will want to use C₁V₁=C₂V₂ to solve for this.

C₁ is the initial concentration, in this problem it is .458 mol/L or M

V₁ is the initial volume, so it will be the 22.0 mL or 0.0220L

C₂ is the new concentration but here it is unknown.

V₂ is the new volume which is 50.0 mL or 0.0500L.

1. So now we plug this information in to the equation:

`(.458 mol/l)(0.0220L)=(C_(2) )(0.0500L)`

2. We then do algebra to get C₂ alone:

`C_(2) =((.458 mol/L)(0.0220L))/((0.0500L)) =0.20152 mol/L or M`

The liters cancel out and take note of the significant figures, there are 3 digits in all numbers in the question, so that means your answer must contain 3 sig figs. The answer 0.0202 rounded.

** instead of converting the mL to L you can also just keep them mL since they will be crossed out anyway. You will still get the correct answer.

Hope this help you! good luck :)