129k views
5 votes
Calculate the volume of water that would be added to 200 cm³ of 0.5 mol/dm³ HNO3 to 0.25 mol/dm³.

200 cm³
100 cm³
300 cm³
400 cm³

User Weisj
by
8.2k points

1 Answer

0 votes

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.

User Farhood ET
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.