A. `1.28 xx 10^(21)`
B. `1.71 xx 10^(21)`
C. `2.57 xx 10^(21)`
D. `5.14 xx 10^(21)`
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Correct Answer – C
`n=4` for `f.c.c.,`
Also no. of atoms in `1 g NaCl=(6.-23xx10^(23))/(58.5)`
No. of unit cell present in `1g NaCl`
`=(6.023xx10^(23))/(58.5xx4)`
`=2.57xx10^(21)` unit cell
Correct Answer – A
`(1)/(58.5)xx6.023xx10^(23)=1.029xx10^(22)`
A unit cell contains `4Na^(+)` ion and `4Cl^(-)` ions
`therefore` Unit cell `=(1.029xx10^(22))/(4)=2.57xx10^(21)` unit cell.
Correct Answer – C
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”`