Given that density of NaCl is nearly 2×103kgm−3 and molecular weight is 58.5×10−3kgmol−1. Then calculate the value of a3 for unit cell of sodium chloride.
Here a is edge length of the fcc unit cell.
Take NA=6×1023
A
1.95×10−28m3
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B
3.5×10−27m3
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C
0.4×10−30m3
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D
4.44×10−29m3
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Solution
The correct option is A1.95×10−28m3 NaCl is face-centred cubic lattice so that number of NaCl molecules in a unit cell is (Z)=4.
We know,
Density of the unit cell, ρ=Z×MNA×a3
where, Z=No. of atoms in a unit cellM=Molar massNA=Avagadro number a=length of the unit cell
a3=MZρNA
Substituting the given values, we get a3=4×58.5×10−32×103×6×1023