The number of silver atoms present in a `90%` pure silver wire weighing 10 g is (Ag = 108)

Options

A. `5.57xx10^(22)`
B. `0.62xx10^(23)`
C. `5.0 xx10^(22)`
D. `6.2 xx10^(29)`

Answer

C

Explanation

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