A metal of density `7.5 xx 10^(3) kg m^(-3)` has an fcc crystal structure with lattice parameter `a = 400 pm`. Calculater the number of unit cells present in `0.015 kg` of metal.

Options

A. `6.250 xx 10^(22)`
B. `3.125 xx 10^(23)`
C. `3.125 xx 10^(22)`
D. `1.563 xx 10^(22)`

Answer

C

Explanation

The volume availagble `= (0.015)/(7.5 xx 10^(3))`
`("Number of unit cells") xx (400 xx 10^(-12))^(3) = (0.015)/(7.5 xx 10^(3))`
`rArr` Number of unit cell `= 3.125 xx 10^(22)`