300 cm3 of water is added to a solution of 200 cm3 of 0.50 mol dm–3 sodium chloride. Calculate the concentration of sodium chloride in the new solution.

Answer 1

We can use V1M1 = V2M2 to solve this, where V1 and M1 are the initial volume and molarity, respectively, and V2 and M2 are the final volume and molarity, respectively.

Initial volume = 200 cm3 and initial molarity = 0.50 M (mol/dm3)

Final volume = 200 cm3 + 300 cm3 = 500 cm3

Final molarity = ?

(200 cm3)(0.50 M) = (500 cm3)(x m)

x = M2 = 0.20 M