[JohnRed]

A reagent has a density of 1.41 g/ml and contains 71% nitric acid by weight and 29% H2O. What volume of reagent would you need to get 12.5 g of nitric acid for the reaction?

The answer in the key is 12.5,  but I'm not sure how to get there?

[Sophia7X]

First, determine how much volume is needed if it was 100% nitric acid.

So 12.5 g HNO3 * (1 mL/1.41 g) = 8.87 mL.

But since the solution is 71% HNO3, you must divide 8.87 mL by 0.71.


This will give you 12.5 mL. 71% of 12.5 is 8.87 mL, which is the volume of HNO3 needed. That's why you divide 8.87 by 0.71.

[Borek]

Many ways to skin that cat. Sophia gave one method, but you can also use slightly different approach - you need 12.5g of nitric acid, as it is 71% of the mass of the solution, you can easily calculate mass of teh solution needed - 12.5g/0.71. Next step is just to convert the mass to volume using density given.