Question

To what volume should you dilute 35 mL of a 13 M stock HCl solution to obtain a 0.580 M HCl solution?

Answer 1

From relation: C1V1 = C2V2

given, C1 = 13M, V1 = 35mL, C2 = 0.580M and V2 = ?

Hence, from above relation, V2 = C1V1 / C2

V2 = 13 x 35 / 0.580

V2 = 784.5 mL

Now Added water = 784.5 - 35

= 749.5mL

So, take 35mL of original solution and dilute with 749.5mL of water to get total volume of 784.5mL of solution which will provide strength of 0.580M HCl.