Asked by leo
What volume of 2.30 M potassiom iodide, KI, solution is needed to prepare 250.0 mL of .220 M potassium iodide solution? Find volume in mL.
Answers
Answered by
DrBob222
mL1 x M1 = mL2 x M2
?mL x 2.30M = 250 mL x 0.220 M
solve for ?mL.
?mL x 2.30M = 250 mL x 0.220 M
solve for ?mL.
Answered by
leo
23.91ml?
Answered by
DrBob222
You're allowed three places so I would round the answer to 23.9 mL.
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