calculate the distance between Na+ and Cl- ions in NaCl crystal if its density is 2.165 gcm-3 (molar mass of NaCl = 58.5gmol-1 NA = 6.02 x 10 23 mol-1)

For NaCl

M = 58.44g, z=4, d=2.165 g/cm3

Substituting the values in ..

d = (Z.M)/a^{3 }N_{A}

a= 5.64 x 10^{-8} cm

Na^{+} ion is present at edge center, therefore:

2(R^{+}+R^{-})=a

Where

R^{+} is radius of sodium ion

R^{-} is radius of chloride ion

and

(R^{+}+R^{-}) is interionic distance.

(R^{+}+R^{-}) =2.82x 10^{-8 }cm

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