The number of silver atoms present in a 90% pure silver wire weighing 10 g is :

Options

A. `5.57 xx 10^22`
B. `0.62 xx 10^23`
C. `5.0 xx 10^22`
D. `6.2 xx 10^29`

Answer

C

Explanation

Amount of pure silver in 10g wire
` = 10 xx 90/100 = 9g `
108 g of Ag contain atoms =`6 xx 10^23`
9g of Ag contain atoms
` = (6 xx 10^23)/(108) xx 9 = 5 xx 10^22`