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. `3xx10^(25)`
B. `0.5xx10^(25)`
C. `1xx10^(25)`
D. `2xx10^(25)`

Answer

B

Explanation

`M=(d xxa^(3)xxN_(0)xx10^(-30))/(Z)`
`=(10xx(200)^(3)xx(6.02xx10^(23))xx10^(-30))/(4)`
`12.04`
No. of atomsin 100 g `=(6.02xx10^(23))/(12.04)xx100`
`=0.5xx10^(25)`.