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.
mL1 x M1 = mL2 x M2
?mL x 2.30M = 250 mL x 0.220 M
solve for ?mL.
23.91ml?
You're allowed three places so I would round the answer to 23.9 mL.
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To find the volume of the 2.30 M potassium iodide (KI) solution needed to prepare 250.0 mL of a 0.220 M potassium iodide solution, you can use the formula for dilution.
The equation for dilution is:
M1V1 = M2V2
Where:
M1 = initial concentration of the solution (2.30 M)
V1 = initial volume of the solution
M2 = final concentration of the solution (0.220 M)
V2 = final volume of the solution (250.0 mL)
Since we are trying to find the volume, V1, we can rearrange the formula to solve for V1:
V1 = (M2 * V2) / M1
Now we can substitute the given values into the equation:
V1 = (0.220 M * 250.0 mL) / 2.30 M
V1 = 0.220 * 250.0 / 2.30
V1 = 24.0869565217... mL
Therefore, the volume of the 2.30 M potassium iodide (KI) solution needed to prepare 250.0 mL of a 0.220 M potassium iodide solution is approximately 24.1 mL.