67.0 mL of a 1.50 M solution is diluted to a volume of 258 mL. A 129-mL portion of that solution is diluted using 163 mL of water. What is the final concentration?
Assuming that the 129 mL plus the 163 mL added water adds to a total volume of 129+163. (That's the same thing as saying that I'm assuming the volumes are additive.
1.50 x (67.0 mL/258 mL) x [129 mL/(129+163)]
To find the final concentration, we need to calculate the number of moles of solute in each step and then use the formula for concentration.
Step 1:
We have 67.0 mL of a 1.50 M solution that is diluted to a volume of 258 mL.
To find the number of moles in the initial solution, we need to use the formula:
moles = concentration × volume
moles = 1.50 M × (67.0 mL / 1000 mL/L)
moles = 0.1005 moles
Step 2:
We take a 129 mL portion of the initial solution and dilute it with 163 mL of water.
To find the number of moles in this solution, we can use the same formula:
moles = concentration × volume
moles = C × (V/1000 mL/L)
moles = C × (129 mL / 1000 mL/L)
moles = C × 0.129
Step 3:
The diluted solution from Step 2 is the final solution, and its volume is 129 mL + 163 mL = 292 mL.
To find the final concentration, we need to calculate the moles in this solution and divide by the final volume:
final concentration = moles / (V/1000 mL/L)
final concentration = (C × 0.129) / (292 mL / 1000 mL/L)
final concentration = (C × 0.129) / 0.292
Now, let's substitute the values we know:
1.50 M = (C × 0.129) / 0.292
To find the final concentration (C), we can rearrange the equation:
C = (1.50 M × 0.292) / 0.129
C ≈ 3.38 M
Therefore, the final concentration is approximately 3.38 M.