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

Answer Details

When 20 cm³ of distilled water is added to 80 cm³ of 0.50 mol dm^-3 hydrochloric acid, the total volume of the solution becomes 100 cm³. However, the amount of HCl molecules in the solution remains the same before and after the dilution. Therefore, we can use the formula: C1V1 = C2V2 where C1 and V1 are the initial concentration and volume of the acid, and C2 and V2 are the final concentration and volume of the acid after dilution. Substituting the values, we get: C1V1 = C2V2 (0.50 mol dm^-3) x (80 cm³) = C2 x (100 cm³) C2 = (0.50 mol dm^-3) x (80 cm³) / (100 cm³) C2 = 0.40 mol dm^-3 Therefore, the concentration of the acid will change to 0.40 mol dm^-3 after the dilution. Thus, the correct option is (B) 0.40 mol dm^-3.