If 20 cm3 of distilled water is added to 80 cm3 of 0.50 mol dm-3…

If 20 cm3 of distilled water is added to 80 cm3 of 0.50 mol dm-3 hydrochloric acid, the concentration of the acid will change to
  • A.
    20 mol dm-3
  • B.
    0.40 mol dm-3
  • C.
    2.00 mol dm-3
  • D.
    5.00 mol dm -3
Correct Answer: Option B
Explanation

From M1V1 = M 2
V2; m 1 = 0.50 mol dm-3; m2 = ?; v1 = 80cm 3 = 100cm3
:. m2 = m1v2/v2 = 0.50 x 80/100 = 0.40 mol dm-3

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