Question

When 71.43 mL of 0.855 M HNO3 is diluted with 20.57 mL of water, the molarity of the nitric acid becomes

Answer 1

The dilution is

c1v2 = c2v2 but I prefer to reason it saying M has been diluted by how much?
0.855M x (71.43 mL/(71.43+20.57) = ?
By the way, this assumes that the volumes are additive which is not always the case.