Question

To what final volume should you dilute 51.5mL of a 5.55M KI solution so that 21.0mL of the diluted solution contains 2.70g of KI?

Answer 1

you can use \(M_1V_1=M_2V_2\)

and sub in for one: \(M=\dfrac{n_{solute}}{L_{solution}}\)

then \(n=\dfrac{m}{M}\)

Answer 2

sorry that's kinda confusing. you can do it in steps first find the moles 2.7 g is equivalent to, then plug into the molarity formula.

Answer 3

Yes the first part is a bit confusing. 2.7g of KI is .016265 mol KI. Now from there where would I convert to? Will I find the Molarity of the second part? .016265 mol KI\[\frac{.016265 mol KI }{ .0210 L } \]. If that is right, then I got .774524 M KI. where would I go from there? Thank you for the help

Answer 4

indeed, that's good thinking. Now just use \(M_1V_1=M_2V_2\) with this info and 51.5mL of a 5.55M KI.

you'll find the final volume, the water you need to add though needs to be subtracted from the original.

Answer 5

I will be left with .0515 L and 5.55 M KI. Do I plug that back into \[M _{1}V _{1}=M _{2}V _{2}\]? (.0515L)(5.55 mol/L) = V2 (.774524mol/L) and solve for V2? V2 will equal .369022L. What do you mean subtracting the water I need from the original?

Answer 6

Or a better question would be, how do I figure out the amount of water needed to subtract from the original?

Answer 7

oh wait, sorry i read the question wrong. Ignore that last part.

0.369022L is your final answer because the want the final volume, not the volume of water added (as i initially thought).

Answer 8

Ohh ok awesome. Thank you. It took me awhile to understand that problem for some reason. I knew to use the dilution equation, but I just didn't know which part to convert. Overall, Thank you for your help

Answer 9

yeah, sometimes these get tricky. But you were pretty good at catching on. No problem!