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. `5 xx 10^(24`
B. `5 xx 10^(25)`
C. `6 xx 10^(23)`
D. `2 xx 10^(25)`

Answer

A

Explanation

`rho=(Z_(eff)xxMw)/(a^(3)xx10^(-30)xxN_(A))`, [For fcc, `Z_(eff) = 4//"unit cell"`]
`:. Mw = (rho xx a^(3) xx 10^(-30) xx N_(A))/(Z_(eff))`
`((10.0g cm^(-3) xx (200 "pm")^(3) xx 10^(-30)cm^(3)xx6xx10^(23)"atoms"))/(4)`
`= 12 g mol^(-1)`
Thus, `12 g mol^(-1)` contains `= N_(A)` atoms `= 6 xx 10^(23)` atoms
`:. 100 g` contains `= (6 xx 10^(23))/(12) xx 100 = 5 xx 10^(24)` atoms