The number of atoms in 100 g of an FCC crystal with density `d = 10 g//cm^(3)` and cell edge equal to 100 pm, is equal to

Options

A. `1 xx 10^25`
B. `2 xx 10^25`
C. `3 xx 10^25`
D. `4 xx 10^25`

Answer

D

Explanation

`M = (rho xx a^3 xx N_A xx 10^(-30))/(Z)`
` = (10 xx (100)^3 xx 6.02 xx 10^23 xx 10^(-30))/(4) = 15.05`
Number of atoms in 100 g ` = (6.02 xx 10^23)/(15.05) xx 100`
` = 4 xx 10^25`