77.0 mL of a 1.20 M solution is diluted to a total volume of 288 mL. A 144-mL portion of that solution is diluted by adding 155 mL of water. What is the final concentration? Assume the volumes are additive.
1.20 x (77.0/288) x [144/(144+155)]
To find the final concentration of the solution, we need to calculate the moles of solute before and after dilution and use the formula:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Given information:
Initial concentration, C1 = 1.20 M
Initial volume, V1 = 77.0 mL
Final volume after dilution = 288 mL
New volume after second dilution = 144 mL + 155 mL = 299 mL
First, let's calculate the moles of solute before dilution:
n1 = C1 * V1
n1 = 1.20 M * 0.0770 L (since 1 mL = 0.001 L)
n1 = 0.0924 moles
Since the volumes are additive, we can apply the formula to find the final concentration after the initial dilution.
C1V1 = C2V2
(1.20 M)(0.0770 L) = C2(0.288 L)
C2 = (1.20 M * 0.0770 L) / 0.288 L
C2 = 0.3225 M
The concentration after the initial dilution is 0.3225 M.
Now, for the second dilution, we need to calculate the moles of solute after the initial dilution (n2).
n2 = C2 * V2
n2 = 0.3225 M * 0.144 L (since 1 mL = 0.001 L)
n2 = 0.0465 moles
Finally, we can find the final concentration after the second dilution:
C2V2 = C3V3
(0.3225 M)(0.144 L) = C3(0.299 L)
C3 = (0.3225 M * 0.144 L) / 0.299 L
C3 = 0.1561 M
Hence, the final concentration of the solution is approximately 0.1561 M.