Question

How many oxygen atoms are present in a `120*g` mass of `"nitric acid"`?

Answer 1

Approx. `3.5xx10^24` `"oxygen atoms......"`

Explanation:

`"Moles of nitric acid"-=(120*g)/(63.01*g*mol^-1)=1.90*mol`.

And thus, given a formula of `HNO_3`, there are `3xx1.90*mol` `"oxygen atoms"` in the given mass of nitric acid.

And since `1*mol` specifies a quantity of `6.022xx10^23*mol^-1`, there are:

`3xx1.90*molxx6.022xx10^23*mol^-1` `"oxygen atoms"`

`=??` `"oxygen atoms............"`

Answer 2

Depends on the concentration.....

Explanation:

If you are thinking of concentrated nitric acid, then it is typically 68% `HNO_3`, 32% water. Fuming nitric acid will have `HNO_3` content of 86%, and only 14% water.

Oxygen atoms are present in both components, so you'd need to work out the number of moles of both.......

If you take standard conc. nitric acid, then 120 g contains 81.6 g of `HNO_3` and 38.4 g of water. This gives you (81.6 / 63.01) = 1.295 moles of `HNO_3` and (38.4 / 18.02) = 2.131 moles of water.

Each mole of `HNO_3` contains 6.02 x `10^23` molecules each containing an oxygen atom. The same applies to water.

So you have (6.02 x `10^23` x 1.295) + (6.02 x `10^23` x 2.131) = 2.062 x `10^24` oxygen atoms in total.