Step I : No. of the oxygen atoms in 22 g of `CO_(2)`
Molar mass of `CO_(2)+12 + 2 xx 16 = 44 g`
44g of `CO_(2)` represent = 1 mol
22g of `CO_(2)` represent `= (1mol) xx((22g))/((44g))=0.5 mol`
Now, 1 mole of `CO_(2)` contain oxygen atoms `= 2xx 6.022 xx 10^(23)`
`0.5` moles of `CO_(2)` contain oxygen atoms `= 2xx 6.022 xx 10^(23) xx 0.5 = 6.022 xx 10^(23)` atoms
Step II. Weight of carbon monoxide
Molar mass of `CO=12+16 = 28` g By definitioon, `6.022 xx 10^(23)` atoms (or 1 mole oxygen atoms) are present in CO = 28 g.
