Question #322e6

In your case, I think that the number of gram-atoms is actually equal to the number of moles. A gram-atom is simply the quantity of a substance that contains Avogadro's number of atoms.

When you're dealing with an element that's made up of individual atoms rather than molecules, the gram-atom is equal to the mole.

In other words, one mole of helium will contain `6.022 * 10^(23` atoms of helium and will be equal to one gram-atom of helium.

On the other hand, one mole of oxygen gas, `"O"_2`, contains `6.022 * 10^(23)` molecules of oxygen. Since you get 2 atoms of oxygen per molecule of oxygen, one mole of oxygen gas would contain `2 * 6.022 * 10^(23)` atoms of oxygen.

This means that one mole of oxygen molecules is equal to 2 gram-atoms of oxygen.

In your case, you have element `"X"`. This tells you that element `"X"` is made up of atoms, not molecules.

So, use Avogadro's number to determine the mass of one mole of `"X"` atoms

`6.022 * 10^(23)color(red)(cancel(color(black)("atoms"))) * (6.664 * 10^(-23)"g")/(1color(red)(cancel(color(black)("atom")))) = "40.13 g"`

So one mole of `"X"` has a mass of `"40.13 g"`. Now all you have to do is figure out how many moles of `"X"` you get in 49 kg

`49color(red)(cancel(color(black)("kg"))) * "100 g"/(1color(red)(cancel(color(black)("kg")))) = 49 * 10^3"g"`

This means that you have

`49 * 10^(3)color(red)(cancel(color(black)("g"))) * "1 mole of X"/(40.13color(red)(cancel(color(black)("g")))) = color(green)("1221 moles of X"`

Therefore, 49 kg of element `"X"` will contain 1221 gram-atoms of `"X"`.