To what volume should you dilute 43.5mL of a 4.20M KI solution so that 24.0mL of the diluted solution contains 2.80g of KI? Please if you can explain it to me I am so confused.

How many mols KI do you want? That's mols KI = grams KI/molar mass KI.

2.80/about 166 = about 0.0169 mols KI.
You want that to be contained in 24.0 mL; that's the equivalent of 0.0169/0.024L = about 0.703 M

So 4.2 M soln is to be diluted so that it will be 0.0703 M

4.2M x 43.5 mL/X mL = 0.703 M
x mL = about 260 mL.

Let's check it to see if that works.
Dilute 43.5 mL of 4.2M KI to 260 mL.\
4.2 x (43.5 mL/260 mL) = 0.703M
If we take 24.0 mL of that solution, how many mols will it be?
That will be M x L = 0.703 x 0.024 = 0.0169 mols KI. How many g is that? That's g = mols x molar mass = 0.0169 mols x 166 g/mol = 2.80 g KI which is what we wanted.

thank you so very much

To solve this problem, we need to use the concept of molarity and dilution formula.

Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for calculating molarity is:

Molarity (M) = moles of solute / volume of solution in liters

The dilution formula can be written as:

C1V1 = C2V2

Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume

Now let's break down the given information step-by-step:

Step 1: Calculate the number of moles of KI in the 24.0 mL of diluted solution.

To calculate the moles of KI, we can use the formula:

Moles (n) = Mass (m) / Molar Mass (Mw)

The molar mass of KI is 166 g/mol.

Given: mass of KI = 2.80 g

Moles of KI = 2.80 g / 166 g/mol = 0.016867 moles

Step 2: Calculate the final concentration (C2) of KI in the diluted solution.

Using the dilution formula:

C1V1 = C2V2

Given:
C1 = 4.20 M (initial concentration of KI solution)
V1 = 43.5 mL (initial volume of KI solution)
V2 = 24.0 mL (final volume of diluted solution)

Substituting the values into the formula:

(4.20 M)(43.5 mL) = C2(24.0 mL)

C2 = (4.20 M)(43.5 mL) / 24.0 mL
C2 = 7.6125 M

Step 3: Calculate the final volume (V2) of the diluted solution.

Using the dilution formula:

C1V1 = C2V2

Given:
C1 = 4.20 M (initial concentration of KI solution)
V1 = 43.5 mL (initial volume of KI solution)
C2 = 7.6125 M (final concentration of KI in diluted solution)

Substituting the values into the formula:

(4.20 M)(43.5 mL) = (7.6125 M)(V2)

V2 = (4.20 M)(43.5 mL) / 7.6125 M

V2 = 168 mL

So, you need to dilute the 43.5 mL of the 4.20 M KI solution to a final volume of 168 mL to obtain a diluted solution where 24.0 mL of the solution contains 2.80 g of KI.

To determine the required volume to dilute the 43.5mL of a 4.20M KI solution, we need to use the concept of dilution and the formula:

CiVi = CfVf

Where:
Ci = initial concentration of the solution before dilution
Vi = initial volume of the solution before dilution
Cf = final concentration of the solution after dilution
Vf = final volume of the solution after dilution

Given:
Ci = 4.20M
Vi = 43.5mL

We need to solve for Vf, the final volume of the solution after dilution.

Step 1: Convert mL to L
To maintain consistent units, we need to convert mL to L:
Vi = 43.5mL * (1L / 1000mL)
Vi = 0.0435L

Step 2: Calculate the final concentration (Cf) after dilution.
We have the final concentration (Cf) as 2.80g/L, and we need to find the final volume, Vf.

Step 3: Substitute the values into the formula and solve for Vf.
CiVi = CfVf
(4.20M)(0.0435L) = (2.80g/L)(Vf)

Step 4: Convert grams to moles
Since we are given the mass in grams, we need to convert it to moles using the molar mass.

Molar mass of KI:
K (potassium) = 39.10 g/mol
I (iodine) = 126.90 g/mol

Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol

Step 5: Calculate the number of moles of KI required.
Number of moles = mass / molar mass
Number of moles = 2.80g / 166.00 g/mol

Step 6: Calculate the final volume (Vf).
(4.20M)(0.0435L) = (2.80g / 166.00 g/mol)(Vf)

Vf = (4.20M)(0.0435L)(166.00 g/mol) / (2.80g)

Vf ≈ 35.359 L

Therefore, you need to dilute the 43.5mL of a 4.20M KI solution to a volume of approximately 35.359 L so that 24.0mL of the diluted solution contains 2.80g of KI.