to what volume should you dilute 125mL of an 8.00M CuCl2 solution so that 50.0mL of the diluted solution contains 5.9g CuCl2

I have a hard time determing the
M1V1=M2V2 is

Ah, the famous equation M1V1 = M2V2! Don't worry, my friend, I'll help you make sense of it, with a touch of humor, of course!

So, let's break it down. M1V1 represents the initial concentration and volume, while M2V2 represents the final concentration and volume. In your case, you're given the initial volume (125mL) and concentration (8.00M), and you need to find the final volume.

To solve this, we first need to find the final concentration using the given information. We know that a 50.0mL sample of the diluted solution contains 5.9g of CuCl2. Now, let's convert that 5.9g to moles by using the molar mass of CuCl2. Just to spice things up, I'll summon my imaginary calculator...

*poof!*

The molar mass of CuCl2 is approximately 134.45 g/mol. So, 5.9g CuCl2 is roughly 5.9 g / 134.45 g/mol = 0.0438 mol CuCl2. (Let's not forget to put on our imaginary safety goggles when dealing with chemicals.)

Now that we have the moles of CuCl2, we can calculate the final concentration. Molarity is defined as moles of solute divided by liters of solution. In this case, we have 0.0438 mol CuCl2 in 0.0500L (since 50.0mL = 0.0500L). So, the final concentration (M2) is:

M2 = 0.0438 mol / 0.0500 L โ‰ˆ 0.876 M

Now that we have the final concentration (0.876 M) and initial concentration (8.00 M), we can plug everything into the equation M1V1 = M2V2 and solve for V2:

(8.00 M)(125 mL) = (0.876 M)(V2)

Now, let's let the magic happen...

*abracadabra!*

V2 โ‰ˆ (8.00 M)(125 mL) / 0.876 M โ‰ˆ 1143 mL

So, you'll need to dilute your 125mL of 8.00M CuCl2 solution to a final volume of approximately 1143 mL to achieve a concentration of 0.876M. Have fun diluting, my chemically inclined friend!

M1V1 = M2V2 is the dilution equation, also known as the formula for dilution. It relates the initial concentration and volume of a stock solution (M1 and V1) to the final concentration and volume of the diluted solution (M2 and V2).

In this case, you want to know the volume (V2) to which you need to dilute a 125 mL 8.00M CuCl2 solution in order to obtain a 50.0 mL diluted solution containing 5.9 g of CuCl2.

Let's use the dilution equation to solve for V2:

M1V1 = M2V2

Given:
M1 = 8.00 M (concentration of the stock solution)
V1 = 125 mL (volume of the stock solution)
M2 = unknown (concentration of the diluted solution; CuCl2 should be 5.9g/50.0mL)
V2 = 50.0 mL (volume of the diluted solution)

Substituting the given values into the equation:

8.00 M * 125 mL = M2 * 50.0 mL

Now, let's solve for M2:

M2 = (8.00 M * 125 mL) / 50.0 mL

M2 = 20

Therefore, you would need to dilute 125 mL of an 8.00M CuCl2 solution to a final volume of 50.0 mL to obtain a diluted solution with a concentration of 20 M.

To solve this problem, you can use the formula M1V1 = M2V2, which is known as the dilution formula. M1 and V1 represent the initial concentration and volume of the concentrated solution, while M2 and V2 represent the final concentration and volume of the diluted solution.

Let's break down the given information:

M1 = 8.00M (initial concentration of the CuCl2 solution)
V1 = 125mL (initial volume of the CuCl2 solution)
M2 = ? (final concentration of the diluted CuCl2 solution)
V2 = 50.0mL (final volume of the diluted CuCl2 solution)
5.9g is the amount of CuCl2 that needs to be present in the diluted solution.

Now, let's find M2 by rearranging the dilution formula:

M1V1 = M2V2

Substituting the known values:

8.00M x 125mL = M2 x 50.0mL

Simplifying:

M2 = (8.00M x 125mL) / 50.0mL

M2 = 20M

Therefore, the final concentration of the diluted CuCl2 solution should be 20M.

To determine the final volume of the diluted solution, we need to find V2. Since the initial volume (V1) is being diluted, we can set up the equation as follows:

V1 x M1 = V2 x M2

Substituting the known values:

125mL x 8.00M = V2 x 20M

Simplifying:

(125mL x 8.00M) / 20M = V2

V2 = 50.0mL

Therefore, the final volume of the diluted CuCl2 solution should also be 50.0mL.

To summarize, you need to dilute 125mL of an 8.00M CuCl2 solution to a final volume of 50.0mL in order to obtain a final concentration of 20M.

molar mass CuCl2 = 134.45

8.00M x 0.125L = 1.00 mole CuCl2 in 125 mL. This is 134.45g/125 mL = 1.0756 g/mL in hand.
We want 5.9g/50 mL or 0.118 g/mL.
We want to dilute by a factor of
1.0756/0.118 = 9.11625 (I know that's too many significant figures).
Then 9.11525 x 125 mL = total volume of 1139.407. Diluted, that soln now is
1.0756g/mL x (125/1139.407) = 0.118 g/mL.
If we take 50 mL of that soln we will have 0.118g/mL x 50 mL = 5.90grams CuCl2.
So the answer is to dilute the 125 mL of 8.00M CuCl2 to 1139 mL. Check my work; it's late here.