Question

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
A. `4xx10^(25)`
B. `3xx10^(25)`
C. `2xx10^(25)`
D. `1xx10^(25)`

Answer 1

Correct Answer - A
`M=(rhoxxa^(3)xxN_(0)xx10^(-30))/(z)`
`=(10xx(100)^(3)xx(6.023xx10^(23))xx10^(-30))/(4)=15.05`
No. of atoms in 100g `=(6.023xx10^(23))/(15.05)xx100=4xx10^(25)`.