Asked by Anonymous
When 71.43 mL of 0.855 M HNO3 is diluted with 20.57 mL of water, the molarity of the nitric acid becomes
Answers
Answered by
DrBob222
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.
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.
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