Final answer:
To dilute 200 cm³ of 0.5 mol/dm³ HNO3 to 0.25 mol/dm³, 200 cm³ of water needs to be added. The calculation is based on the dilution formula M1V1 = M2V2.
Step-by-step explanation:
The question involves calculating the volume of water needed to dilute a solution of nitric acid to a specific concentration. This is a common type of problem in chemistry where the concept of molarity and dilution is applied.
To calculate the volume of water to be added to dilute 200 cm³ of 0.5 mol/dm³ HNO3 to 0.25 mol/dm³, we can use the dilution formula M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
By rearranging the formula, we find that V2 = (M1V1)/M2. Substituting our values in gives us V2 = (0.5 mol/dm³ × 200 cm³)/(0.25 mol/dm³). When we calculate this, we find that V2 = 400 cm³. This means the final volume of the solution must be 400 cm³. To find the volume of water needed, we subtract the initial volume from the final volume: 400 cm³ - 200 cm³ = 200 cm³. Therefore, 200 cm³ of water must be added to achieve the desired concentration.