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
- 0.10mol dm-3
- 0.20mol dm-3
-
0.40mol dm-3
- 2.00mol dm-3
The correct answer is: C
Explanation
\(V_1\) = 80cm3, \(C_1\) = 0.5mol/dm3
\(C_2\) = ? \(V_2) = 20cm3 + 80cm3 = 100cm3
using the dilution formula, \(C_1 V_1 = C_2 V_2\)
0.5 x 80 = \(C_2\) x 100
\(C_2 = \frac{ 40}{100}\) = 0.4mol\(dm^{-3}\)