Question

15. What volume of water must be added to 300 mL of 0.75 M HCl to dilute the solution to 0.25 M?

Answer 1

Known :

V1 = 300 mL

M1 = 0.75 M

M2 = 0.25 M

Solution :

M1 • V1 = M2 • V2

(0.75 M) • (300 mL) = (0.25 M) V2

V2 = 900 mL

Water add to this solution is :

∆V = V2 - V1

∆V = 900 - 300

∆V = 600 mL