The mol mass of KI is 166, so 3.5g = 0.021 moles KI
.021mol/.0275L = 0.767 M
So, you need to dilute the original solution from 5.95M to 0.767M
That's a dilution by a factor of 7.757
So, the volume must increase by that factor, so you need .047*7.757 = 365mL
To what volume should you dilute 47.0 mL of a 5.95 M KI solution so that 27.5 mL of the diluted solution contains 3.50 g of KI?
2 answers
How many mols is 3.50 g KI? That's mols = grams/molar mass = 3.5/approx 366 = approx 0.021 mols.
You want to dilute the 5.95 to what volume? The new concentration of the 5.95M solution will be 5.95 x (0.047/y) = ? where y is the volume to which it is diluted but you don't know what y is yet.
Then 0.0275 L of that should contain that 0.021 mols KI.
0.0275 x [5.95 x (0.047/y) = 0.021
(Remember M x L = mols. The 0.0275 is the L and the 5.95 x 0.047/y is the M term.)
Solve for y = final volume.
You want to dilute the 5.95 to what volume? The new concentration of the 5.95M solution will be 5.95 x (0.047/y) = ? where y is the volume to which it is diluted but you don't know what y is yet.
Then 0.0275 L of that should contain that 0.021 mols KI.
0.0275 x [5.95 x (0.047/y) = 0.021
(Remember M x L = mols. The 0.0275 is the L and the 5.95 x 0.047/y is the M term.)
Solve for y = final volume.
3 answers