Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required
Answer:
We need 31.5 mL of the 2.0 M FeCl2 solution
Explanation:
Step 1: Data given
Molarity of a FeCl2 solution = 2.0 M
Initial volume of FeCl2 = 100 mL
Initial molarity of FeCl2 = 0.630 M
Step 2: Calculate volume of the stock solution
C1V1 = C2V2
⇒with C1 = the initial molarity FeCl2 = 0.630 M
⇒with V1 = the initial volume = 100 mL = 0.100 L
⇒with C2 = the new molarity FeCl2 = 2.0 M
⇒with V2 = the new volume = TO BE DETERMINED
0.630M * 0.100 L = 2.0 M * V2
V2 = (0.630 * 0.100) / 2.0
V2 = 0.0315 L = 31.5 mL
We need 31.5 mL of the 2.0 M FeCl2 solution
