If 20 cm3 of distilled water is added to 80cm 3 of 0. 50 ...
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
Correct answer is C
V1 = 80cm3, C1 = 0.5mol/dm3
C2 = ? \(V_2) = 20cm3 + 80cm3 = 100cm3
using the dilution formula, C1V1=C2V2
0.5 x 80 = C2 x 100
C2=40100 = 0.4moldm−3
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