JAMB - Chemistry (1990 - No. 40)

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

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.