If 20 cm3 of distilled water is added to 80 cm3 of 0.50 mol dm-3 hydrochloric acid, the concentration of the acid will change to
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
From M1V1 = M 2
V2; m 1 = 0.50 mol dm-3; m2 = ?; v1 = 80cm 3 = 100cm3
:. m2 = m1v2/v2 = 0.50 x 80/100 = 0.40 mol dm-3
From M1V1 = M 2
V2; m 1 = 0.50 mol dm-3; m2 = ?; v1 = 80cm 3 = 100cm3
:. m2 = m1v2/v2 = 0.50 x 80/100 = 0.40 mol dm-3