1.

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

Answer» Correct Answer - `6.023 xx 10^(23)"atom" Na, 6.023 xx 10^(23)"atom"Cl, 18.06 xx 10^(23)"atom" O`
`n_(NaClO_(3))=(106.5)/(106.5)=1` mole
`NO`. of atom of
`{:(Na=1xxN_(A)),(Cl=1xxN_(A)),(O=1xxN_(A)):}`


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