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Calculate the edge length of the unit cell of sodium chloride given density of NaCI is `2.17 xx 10^(3) kgm^(-3)` and molecular weight `58.5 xx 10^(-3)kg mol^(-1)`.

Answer» NaCl is face-centred cubic lattice so that number of NaCl molecules in a unit cell (z) = 4
We know density `d = (zM)/(a^(3)N_(0))`
where a = length of the unit cell
volume `= a^(3) = (Mz)/(dN_(0)) = (4 xx 58.8 xx 10^(-3))/(2.17 xx 10^(3) xx 6.02 xx 10^(23)) = 1.79 xx 10^(-28) m^(3)`
`a = 5.64 xx 10^(-10) m = 5.64 Å = 564` pm


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