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

Options

A. `5xx10^(24)`
B. `5xx10^(25)`
C. `6xx10^(23)`
D. `2xx10^(25)`

Explanation

`rho=(Z_(eff)xxMv)/(a^(3)xx10^(-30)xxN_(A))` [For fcc,`Z_(eff)=4`/unit cell]
`therefore Mw=(rhoxxa^(3)xx10^(-30)xxN_(A))/(Z_(eff))`
`=((10.0gcm^(-3)xx(200"pm")^(3)xx10^(-30)cm^(3)xx6xx10(23)"atoms"))/(4)`
`=12g "mol"^(-1)`
Thus, 12 g `"mol"^(-1)` contains =`N_(A)` atoms
`=6xx10^(23)` atoms
`therefore 100g "contains"=(6xx10^(23))/(12)xx100=5xx10^(24)` atoms.