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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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