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. `3 xx 10^(25)`
B. `5 xx 10^(24)`
C. `1 xx 10^(25)`
D. `5.96 xx 10^(-3)`

Answer

b

Explanation

`M = (d+ a^(3) xx N_(0))/(n)`
`= (10 xx (200 xx 10^(-3))^(3) xx (6.02 xx 10^(23)))/(4) = 12.04 g`
No of atoms in `100 gm= (6.02 xx 10^(22))/(12.04) xx 100`
`= 5 xx 10^(24)`