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?
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
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.
To find the volume of the diluted solution, we need to use the equation for dilution:
C1 × V1 = C2 × V2
where:
C1 = initial concentration of the solution = 5.95 M
V1 = initial volume of the solution = 47.0 mL
C2 = final concentration of the solution = unknown
V2 = final volume of the solution = 27.5 mL
First, we need to convert the given mass of KI (3.50 g) to moles. We can use the molecular weight of KI to do this calculation. The molecular weight of KI can be found by adding the atomic weights of potassium (K) and iodine (I).
Atomic weight of K = 39.10 g/mol
Atomic weight of I = 126.90 g/mol
So, the molecular weight of KI = (39.10 g/mol) + (126.90 g/mol) = 166.00 g/mol
Now, we can calculate the number of moles of KI in 3.50 g:
Moles of KI = Mass of KI / Molecular weight of KI
= 3.50 g / 166.00 g/mol
= 0.0211 mol
Next, we can substitute the known values into the equation for dilution to solve for C2:
C1 × V1 = C2 × V2
(5.95 M) × (47.0 mL) = C2 × (27.5 mL)
Solve for C2:
C2 = (5.95 M × 47.0 mL) / (27.5 mL)
= 10.17 M
Now we know the final concentration of the solution (C2 = 10.17 M). We can use this value to calculate the final volume of the solution (V2) that contains 0.0211 mol of KI:
V2 = Moles of KI / C2
= 0.0211 mol / 10.17 M
= 0.00207 L = 2.07 mL
Therefore, you need to dilute 47.0 mL of a 5.95 M KI solution to a final volume of 2.07 mL to obtain a diluted solution where 27.5 mL of it contains 3.50 g of KI.