What is the final concentration of a solution if 15-mL of a 2.4-M solution was diluted to 100-mL and this process was repeated two more times?
Please help me how to work out the problem. I am not just looking for an answer. I really want to learn.
If you want a formula for dilution it is mL1 x M1 = mL2 x M2
For example 15 x 2.4 = 100 x M2 and you solve for M2. Then you repeat that two more times to arrive at the final answer. For problems like this; however, I think it is easier to run each dilution like this. You know the new concn will be 2.4M x a factor.
2.4 x (15/100 = ? but all of this can be run together like this
2.4 x (15/100) x (15/100) x (15/100) = ? Voila
To calculate the final concentration of a solution after multiple dilutions, you need to understand the concept of dilution and the equation used to calculate it.
Dilution is the process of reducing the concentration of a solute in a solution by adding more solvent. In this case, you are diluting a 2.4 M solution by adding more solvent.
The equation used to calculate dilution is:
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
Now, let's solve the problem step by step.
Step 1:
The initial concentration (C1) is given as 2.4 M.
The initial volume (V1) is given as 15 mL.
Step 2:
The final volume (V2) after dilution is given as 100 mL.
The final concentration (C2) is what we need to find.
Step 3:
Using the dilution equation (C1V1 = C2V2), we can rearrange it to solve for C2:
C2 = (C1V1) / V2
Step 4:
Substitute the values into the equation:
C2 = (2.4 M * 15 mL) / 100 mL
Step 5:
Simplify the equation:
C2 = 0.36 M
So, the final concentration of the solution after the first dilution is 0.36 M.
You mentioned that this process was repeated two more times, so let's continue with the next dilution.
Step 6:
For the second dilution, the initial concentration (C1) is now 0.36 M (the final concentration from the previous step).
The initial volume (V1) is still 15 mL.
Step 7:
The final volume (V2) after the second dilution is 100 mL.
The final concentration (C2) is what we need to find.
Step 8:
Using the dilution equation (C1V1 = C2V2), we can rearrange it to solve for C2:
C2 = (C1V1) / V2
Step 9:
Substitute the values into the equation:
C2 = (0.36 M * 15 mL) / 100 mL
Step 10:
Simplify the equation:
C2 = 0.054 M
So, the final concentration of the solution after the second dilution is 0.054 M.
Now, let's repeat the process one more time.
Step 11:
For the third dilution, the initial concentration (C1) is now 0.054 M (the final concentration from the previous step).
The initial volume (V1) is still 15 mL.
Step 12:
The final volume (V2) after the third dilution is 100 mL.
The final concentration (C2) is what we need to find.
Step 13:
Using the dilution equation (C1V1 = C2V2), we can rearrange it to solve for C2:
C2 = (C1V1) / V2
Step 14:
Substitute the values into the equation:
C2 = (0.054 M * 15 mL) / 100 mL
Step 15:
Simplify the equation:
C2 = 0.0081 M
So, the final concentration of the solution after the third dilution is 0.0081 M.
Therefore, the final concentration of the solution after being diluted three times is 0.0081 M.