Answer :

Given,


Diameter of the sodium atom, (d)= 2.5 Å


Radius, (r) = 1.25 Å


Density of sodium in crystalline phase, (ρc)= 970 kg m-3


Consider every atom of sodium to be sphere,


Volume (V) = r3 m3


V = ×3.14× (1.25×10-10)3 m3


≈ 8.18×10-30 m3


We know that,


According to Avogadro law, a mole contains 6.023×1023 atoms.


The molecular weight of sodium atom, (M) = 23 g = 23×10-3 kg


Density, (ρ) = kg/m3


ρ = ≈ 4.67×10-7 kg/m3


Thus,


The density of sodium in crystalline phase (ρc) is much higher than the density of general sodium in general configuration (ρ). Because, sodium atoms in crystalline phase have very less inter-atomic separation.


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