The number of silver atoms present in a 90% pure silver wire weighing

1.	The number of silver atoms present in a 90% pure silver wire weighing 10 g is (Ag = 108)
Answer» `5.57xx10^(22)` `0.62xx10^(23)` `5.0 xx10^(22)` `6.2 xx10^(29)` SOLUTION :Amount of PURE silver in 10 g of`90%` sample = 9 g 108 g of Ag CONTAINS `6XX10^(23)` ATOMS of Ag 9 gm of Ag contains `(6xx10^(23)xx9)/(108)` `=0.5xx10^(23)=5xx10^(22)` atoms of Ag

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