How much 5.0 M nitric acid (HNO3) in mL, is needed to make 225 mL of 1.0 M HNO3?

The answer to your question is 45 ml

Explanation:

Data

Molarity 1 = 5 M

volume 1 = ?

Molarity 2 = 1 M

volume 2 = 225 ml

Formula

             Molarity 1 x volume 1 = Molarity 2 x volume 2

-Solve for volume 1

             volume 1 = Molarity 2 x volume 2 / Molarity 1

-Substitution

             volume 1 = (1 x 225) / 5

-Simplification

             Volume 1 = 225/5

-Result

             Volume 1 = 45 ml

-Conclusion

45 ml are needed to prepare 225 ml of HNO₃ 1.0 M

1 a

2 d

3 a

4 c

explanation: