Answer: [tex]4.69(10)^{23} atoms[/tex]
Explanation:
Firstly, we have to find the Molecular mass of potassium oxide ([tex]K_{2}0[/tex]):
[tex]K[/tex] atomic mass: 39 u
[tex]O[/tex] atomic mass: 16 u
[tex]K_{2}O[/tex] molecular mass: [tex]39(2) g/mol+16g/mol=94 g/mol[/tex]
This means that in 1 mole of [tex]K_{2}O[/tex] there are [tex]94 g[/tex] and we need to find how many moles there are in [tex]73.9 g[/tex] [tex]K_{2}O[/tex]:
1 mole of [tex]K_{2}O[/tex]-----[tex]94 g[/tex] of [tex]K_{2}O[/tex]
[tex]X[/tex]-----[tex]73.9 g[/tex] of [tex]K_{2}O[/tex]
[tex]X=\frac{(73.9 g)(1 mole)}{94 g}[/tex]
[tex]X=0.78 mole[/tex] This is the quantity of moles in 73.9 g of potassium oxide
Now we can calculate the number of atoms in 73.9 g of potassium oxide by the following relation:
[tex]N_{atoms}=(X)(N_{A})[/tex]
Where:
[tex]N_{atoms}[/tex] is the number of atoms in 73.9g of potassium oxide
[tex]N_{A}=6.0221(10)^{23}/mol[/tex] is the Avogadro's number, which is determined by the number of particles (or atoms) in a mole.
Then:
[tex]N_{atoms}=(0.78 mole)(6.0221(10)^{23}/mol)[/tex]
[tex]N_{atoms}=4.69(10)^{23} atoms[/tex] This is the quantity of atoms in 73.9g of potassium oxide
