The number of atoms in `100 g an fcc` crystal with density `d = 10 g//cm^(3)` and the edge equal to 100 pm is equal to

Options

A. `4 xx 10^(25)`
B. `3 xx 10^(25)`
C. `2 xx 10^(25)`
D. `1 xx 10^(25)`

Answer

a

Explanation

`M= ( rho xx a^(3) xx N_(0) xx 10^(-30))/(z)`
`= (10 xx (100)^(2) (6.02 xx 10^(23))xx 10^(-30))/(4) = 15.05`
No of atoms in `100 g = (6.02 xx 10^(23))/(15.05) xx 100 = 4 xx 10^(23)`