0.005 cm thick coating of silver is deposited on a plate of `0.5 m^2` area. The number of silver atoms deposited on the plate is (atomic mass of silver = 108 and density `= 7.9 g cm^(-3)` )

Options

A. `1.1xx10^(10)`
B. `1.1xx10^(15)`
C. `1.1xx10^(24)`
D. `1.1xx10^(30)`

Answer

C

Explanation

Volume of Ag deposited `= 0.5 xx 10^4 cm^2 xx0.005 cm = 25 cm^3`
Mass of Ag deposited = volume `xx` density
So, number of Ag atoms deposited `=( "volume" xx "density")/(108)xxN_A`
`=(25xx 7.9xx6.023xx10^(23))/(108)=1.1xx10^(24)`