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

Options

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

Explanation

Mass of one unit cell = `V xx d ` (V is volume, d is density ) = `a^(3) xx d` also , density `(n xx at.wt.)/(a^(3) xx" av.no.")`
`therefore` Mass of one unit cell = `(a^(3) xx n xx "at.wt.")/(a^(3) xx "av.no.") = (n xx at.wt)/("av.no.")`
`= (4 xx 58.5 )/(6.02 xx 10^(23))` `" "` (n = 4 for cubic shape)
= `38.87 xx 10^(-23)` g `" " therefore ` No of unit cell in 1 g ` = (1)/(38.87 xx10^(-23) ) = 2.57 xx 10^(21)`