How many unit cell are present in a cubic-shaped ideal crystal of `NaCl` of mass `1.0 g`?

Options

A. `1.28 xx 10^(21)`
B. `1.71 xx 10^(21)`
C. `2.57 xx 10^(21)`
D. `5.14 xx 10^(21)`

Answer

C

Explanation

For fcc, `Z_(eff) = 4//"unit cell"`
`Mw` of `NaCl = 58.5 g mol^(-1)`
Number of atoms in `1.0 g NaCl = (6 xx 10^(23))/(58.5)`
Number of unit cells in `0.1 g NaCl`
`= (6 xx 10^(23))/(58.5 xx 4) = 2.57 xx 10^(21) "unit cells"`