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

Options

A. `2.75 xx 10^(21)`
B. `5.14 xx 10^(21)`
C. `1.28 xx 10^(21)`
D. `1.71 xx 10^(21)`

Answer

a

Explanation

Mass of the one unit cell `= Y xx d`(V is volume , d is density ) `= a^(3) xx d`
Also density `= (n + at, wt)/(a^(3) xx (av, no)`
Mass of one unit of cell `= (a^(3) xx at, wt)/(a^(3) xx av, no) = (n xx at, wt)/(av, no)`
`= (4 xx 58.3)/(6.02 xx 10^(23))`M
`= 38.87 xx 10^(-25)g`
No of unit cell is `1g = (1)/(38.87 xx 10^(-23)) = 2.57 xx 10^(21)`