If 20 cm3 of distilled water is added to 80cm 3 of 0. 50 mol dm-3 hydrochloric acid, the new concentration of the acid will be
-
A.
0.10mol dm-3 -
B.
0.20mol dm-3 -
C.
0.40mol dm-3 -
D.
2.00mol dm-3
Correct Answer: Option D
Explanation
C1 x 100 = 0.5 x 80
C1 = (frac{0.5 x 80}{100}/) = (frac{4}{10}/) = 0 . 40mol/dm-3