9.86 x 10²⁸ o-atoms require what volume (l) n₂o₂ at stp?

First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:

9.86 x 10²⁸ O-atoms ( 1 mol / 6.022x10^23 atoms O) ( 1 mol N2O2 / 2 mol O ) ( 22.4 L / 1 mol ) = 1833809.37 L needed

The correct answer is "atomic weight".