A compound AB has a rock salt type structure .the formula weight of AB is 6.023Y amu and the closest distance is Y^1/3nm where Y is. an arbitrary constant find the density of lattice
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1. We have, AB has rock salt structure (A : B :: 1 : 1).
2. That is,
F.C.C.structure (n = 4) and formula weight of AB is 6.023 y g
closest distance A–B = y 13 nm.
Therefore, edge length of unit cell, a = 2(A+ + B–) = 2×y3×10-9m
Therefore, density of AB =n×mol.wt.NA V=4×6.023×y×10−36.023×1023×(2×y310−9)3
= 5.0 km–3
2. That is,
F.C.C.structure (n = 4) and formula weight of AB is 6.023 y g
closest distance A–B = y 13 nm.
Therefore, edge length of unit cell, a = 2(A+ + B–) = 2×y3×10-9m
Therefore, density of AB =n×mol.wt.NA V=4×6.023×y×10−36.023×1023×(2×y310−9)3
= 5.0 km–3
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