The number of atoms in `100 g` of an fcc crystal with density `= 10.0g cm^(-3)` and cell edge equal to `200 pm` is equal to

Options

A. `6xx10^(24)`
B. `4.8xx10^(25)`
C. `1.2xx10^(25)`
D. `1.2xx10^(24)`

Answer

C

Explanation

`rho=(NxxM)/(a^(3)+10^(-30)xN_(0))`
`M=(10(200)^(3) xx 10^(-30) xx 6 xx 10^(23))/(4) =12`
Thus 12 g contain `N_(0)=6xx10^(23)` atoms
`:. 240` g will contain `(6xx10^(23))/(12) xx240=1.2xx10^(25)` atoms