What volume of 11.0 M hydrochloric acid must be dilute to obtain 1 dm3 of 0.05 M acid?
-
0.05 dm3
- 0.10 dm3
- 0.55 dm3
- 11.0 dm3
The correct answer is: A
Explanation
using \(C_1 V_1\) = \(C_2 V_2\)
\(C_1\) = 11.0M, \(V_1\) = ?, \(C_2\) = 0.05M and \(V_2\) = 1\(dm^3\)
\(V_1\) = \(\frac{(C_2 V_2)}{(C_2)}\) = \(\frac{(0.05 \times 1)}{(11)}\) = 0.005\(dm^3\)
OPTION A is the closest value.