Question

A stock solution of FeCl2 is available to prepare solutions that are more dilute. Calculate the volume, in mL, of a 2.0-M solution of FeCl2 required to prepare exactly 100 mL of a 0.630-M solution of FeCl2.

Answer 1

We need 31.5 mL of the 2.0 M FeCl2 solution

Step-by-step 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

Answer 2

31.5mL

Step-by-step 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