I have to use the given concentration of 7.03 micromoles/liters to determine the concentration for the following dilutions.
1. mL stock solution-2
mL distilled water-8
2. mL stock solution-4
mL distilled water-6
3. mL stock solution-6
mL distilled water-4
You need to explain this a little better; I don't know what you're asking. I understand the 7.03 umol/L is what you have but I don't know what you're to do with it. What's the 2,8,4,6,6,4 mean?
I need to use those numbers to find the final concentrations for each individual solution.
You didn't explain anything. My assumption is that for #1 you are taking 2 mL of the stock solution and adding 8 mL distilled H2O and you want to know the concentration of the final solution when you start with that 2 mL of 7.03 umol/L. Again, I assume you want it in umol/L.
For #1 you have
7.03 umol/L x (2/10) = ?
For #2 you have
7.03 umol/L x (4/10) = ?
etc.
BTW, this works ONLY if the solution volumes are additive. Technically they are not but for such small concentrations you can assume they are.
To determine the concentration for the given dilutions, you need to use the concept of dilution. Dilution is a process of reducing the concentration of a solute in a solution by adding more solvent. It can be calculated using the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
Let's apply this formula to each of the given dilutions:
1. mL stock solution-2, mL distilled water-8:
Here, the initial volume (V1) is 2 mL, and the final volume (V2) is 10 mL (2 mL stock solution + 8 mL distilled water). The initial concentration (C1) is 7.03 micromoles/liter, and we need to find the final concentration (C2).
Let's plug these values into the formula:
C1V1 = C2V2
(7.03 micromoles/L)(2 mL) = C2(10 mL)
Now, solve for C2:
(7.03 micromoles/L)(2 mL) / 10 mL = C2
C2 ≈ 1.406 micromoles/liter
Therefore, the concentration for this dilution is approximately 1.406 micromoles/liter.
2. mL stock solution-4, mL distilled water-6:
Following the same procedure as above:
V1 = 4 mL, V2 = 10 mL (4 mL stock solution + 6 mL distilled water), C1 = 7.03 micromoles/liter (given), and we need to find C2.
C1V1 = C2V2
(7.03 micromoles/L)(4 mL) = C2(10 mL)
Solve for C2:
(7.03 micromoles/L)(4 mL) / 10 mL = C2
C2 ≈ 2.812 micromoles/liter
Therefore, the concentration for this dilution is approximately 2.812 micromoles/liter.
3. mL stock solution-6, mL distilled water-4:
Similar to the previous examples:
V1 = 6 mL, V2 = 10 mL (6 mL stock solution + 4 mL distilled water), C1 = 7.03 micromoles/liter (given), and we need to find C2.
C1V1 = C2V2
(7.03 micromoles/L)(6 mL) = C2(10 mL)
Solve for C2:
(7.03 micromoles/L)(6 mL) / 10 mL = C2
C2 ≈ 4.218 micromoles/liter
Therefore, the concentration for this dilution is approximately 4.218 micromoles/liter.
Remember to always double-check your calculations and units to ensure accuracy.