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If 20 cm3 of distilled water is added to 80 cm

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

View answer & explanation

Correct answer is B

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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