75.0 mL of a 1.20 M solution is diluted to a total volume of 248 mL. A 124-mL portion of that solution is diluted by adding 111 mL of water. What is the final concentration? Assume the volumes are additive.
Look before you leap. I worked this an hour or more ago.
To find the final concentration, we need to use the formula for dilution:
C1V1 = C2V2
Where:
C1 is the initial concentration
V1 is the initial volume
C2 is the final concentration
V2 is the final volume
Let's break down the given information:
Initial solution:
C1 = 1.20 M (concentration)
V1 = 75.0 mL (volume)
Diluted solution:
V2 = 248 mL (final volume)
Now, let's calculate the final concentration:
C1V1 = C2V2
(1.20 M)(75.0 mL) = C2(248 mL)
90.0 M mL = C2(248 mL)
To find C2, we need to rearrange the equation:
C2 = (90.0 M mL) / (248 mL)
Now, let's move on to the second part of the question, where a 124-mL portion of the solution is diluted by adding 111 mL of water. We need to find the final concentration for this new solution.
For this calculation, we will use the same formula:
C1V1 = C2V2
Where:
C1 is the initial concentration
V1 is the initial volume
C2 is the final concentration
V2 is the final volume
In this case, the initial concentration (C1) is the concentration calculated from the previous step, which is 90.0 M mL.
Initial solution:
C1 = 90.0 M mL (concentration)
V1 = 124 mL (volume)
Diluted solution:
V2 = 124 mL + 111 mL = 235 mL (final volume)
Now, let's calculate the final concentration:
C1V1 = C2V2
(90.0 M mL)(124 mL) = C2(235 mL)
11,160 M mL = C2(235 mL)
Rearranging the equation:
C2 = (11,160 M mL) / (235 mL)
Now, we can calculate C2 to find the final concentration.