Question

22.0 mL of stock solution is used

Answer 1

Answer:

0.202M

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:

$C_{2} =\frac{(.458 mol/L)(0.0220L)}{(0.0500L)} =0.20152 mol/L or M$

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 :)