Calculate no. of each atom present in `106.5 g` of`NaCIO_(3^(.)`

Answer

`6.023 xx 10^(23)"atom" Na, 6.023 xx 10^(23)"atom"Cl, 18.06 xx 10^(23)"atom" O`

Explanation

`n_(NaClO_(3))=(106.5)/(106.5)=1` mole
`NO`. of atom of
`{:(Na=1xxN_(A)),(Cl=1xxN_(A)),(O=1xxN_(A)):}`