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
- 20 mol dm-3
-
0.40 mol dm-3
- 2.00 mol dm-3
- 5.00 mol dm -3
The correct answer is: B
Explanation
From C1V1 = C2V2
C1 = 0.50 mol dm-3; C2 = ?; V1 = 80cm3; V2 = 80 + 20 = 100cm3
:. C2 = C1V2/V2 = 0.50 x 80/100 = 0.40 mol dm-3 - Option B