What volume of 1.30 M SrCl2 is needed to prepare 525 mL of 5.00 mM SrCl2?
Well, let's calculate it step by step, shall we? First, convert the desired concentration of SrCl2 to moles per liter (M).
5.00 mM SrCl2 = 5.00 x 10^-3 moles/L SrCl2
Now, find the number of moles of SrCl2 required in the 525 mL of the solution.
(5.00 x 10^-3 moles/L SrCl2) x (0.525 L) = 2.625 x 10^-3 moles SrCl2
Finally, let's calculate the volume of the 1.30 M SrCl2 solution needed to obtain 2.625 x 10^-3 moles.
(2.625 x 10^-3 moles SrCl2) / (1.30 moles/L SrCl2) = 2.02 mL
Therefore, you will need approximately 2.02 mL of 1.30 M SrCl2 to prepare 525 mL of 5.00 mM SrCl2. Just a tiny amount! It's like a drop in the ocean. Or in this case, a drop in the beaker.
To determine the volume of 1.30 M SrCl2 needed to prepare 525 mL of 5.00 mM SrCl2, we can use the concept of dilution.
The formula for dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration of the stock solution
V1 = initial volume of the stock solution
C2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
We can rearrange the formula to solve for V1:
V1 = (C2 * V2) / C1
In this case:
C1 = 1.30 M (concentration of the stock solution)
V2 = 525 mL (final volume of the diluted solution)
C2 = 5.00 mM (final concentration of the diluted solution in mol/L)
First, let's convert the final concentration (C2) from millimolar (mM) to molar (M):
5.00 mM = 5.00 * 10^-3 M
Now we can substitute the values into the formula:
V1 = ((5.00 * 10^-3 M) * 525 mL) / 1.30 M
V1 = (0.002625 mol) / (1.30 mol/L)
V1 ≈ 0.00202 L ≈ 2.02 mL
Therefore, approximately 2.02 mL of the 1.30 M SrCl2 solution is needed to prepare 525 mL of 5.00 mM SrCl2.
To find the volume of 1.30 M SrCl2 needed to prepare 525 mL of 5.00 mM SrCl2, we need to use the formula:
M1V1 = M2V2
where:
M1 = molarity of the first solution
V1 = volume of the first solution
M2 = molarity of the second solution
V2 = volume of the second solution
Let's substitute the values we have:
M1 = 1.30 M
V1 = unknown (what we need to find)
M2 = 5.00 mM = 0.0050 M
V2 = 525 mL
Now we can plug in the values and solve for V1:
(1.30 M)(V1) = (0.0050 M)(525 mL)
First, we need to convert milliliters (mL) to liters (L) since our molarity is in moles per liter:
V2 = 525 mL = 525 mL * (1 L / 1000 mL) = 0.525 L
Now let's substitute the values and solve for V1:
(1.30 M)(V1) = (0.0050 M)(0.525 L)
V1 = (0.0050 M)(0.525 L) / 1.30 M
V1 = 0.002625 L or 2.625 mL
Therefore, you would need 2.625 mL of 1.30 M SrCl2 to prepare 525 mL of 5.00 mM SrCl2.
Sometimes mM can be confusing and one can avoid the confusion by converting to M.
5.00 mM = 0.005 M
M = moles/L
You know M and you know L, solve for moles = M x L = 0.005 x 0.525 = ??
Then moles = grams/molar mass
You know moles and molar mass, solve for grams.