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
waec 2006chemistry

The correct answer is D. 2.00mol dm-3

C1 x 100 = 0.5 x 80

C1 = \(\frac{0.5 x 80}{100}/) = \(\frac{4}{10}/) = 0 . 40mol/dm-3

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