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The Solid State

Question
CBSEENCH12005505

Caesium chloride crystallises as a body centred cubic lattice and has a density of 4.0 g cm–3 Calculate the length of the edge of the unit cell of caesium chloride crystal.
         [Molar Mass of CsCl = 168.5 g mol–1, NA = 6.02 x 1023 mol–1]

Solution

Solution;
we have given that    

   Z =1, ρ = 4 g cm-3,
        M = 168.5 mol-1,NA = 6.02×1023mol-1   ρ = Z×Ma3×N

or          a3 = Z×Mρ×NA    = 1×168.5g mol-14×6.02×1023    = 168.5×10-2324.08
log a3 = log 168.5 + log 10-23 - log 24.083 log a = 2.2266 - 23.000-1.38163 log a = -22.1550+2-23 log a = 24+(2.1550)       a = Antilog 8¯.7183 = 5.228×10-8cm         = 5.228×10-8×1010pm = 522.8 pm.