How many moles of `HNO_3` are needed to prepare 5.0 liters of a 2.0 M solution of `HNO_3`?

All you have to do here is to use the molarity of the solution as a conversion factor to determine the number of moles of solute that must be dissolved in `"5.0 L"` of solution in order to have a concentration of `"2.0 M"`.

As you know, molarity tells you the number of moles of solute present in `"1 L"` of solution. In your case, a `"2.0 M"` solution will contain `2.0` moles of solute for every `"1 L"` of solution.

Now, because solutions are homogeneous mixtures, i.e. they have the same composition throughout, you can use the molarity of the solution as a conversion factor to get the moles of solute needed for your solution

`5.0 color(red)(cancel(color(black)("L solution"))) * overbrace("2.0 moles HNO"_3/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 2.0 M")) = color(darkgreen)(ul(color(black)("10. moles HNO"_3)))`

The answer is rounded to two sig figs.

This tells you that dissolving `10.` moles of nitric acid in enough water to get the total volume of the solution to `"5.0 L"` will give you the same molarity as dissolving `2.0` moles of nitric acid in enough water to get the total volume of the solution to `"1.0 L"`.