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The Solid State

Question
CBSEENCH12005502

The unit cell of an element of atomic mass 96, and density 10.3 g cm–3 is a cube with edge length of 314 pm. Find the structure of crystal lattice (simple cubic, F.C.C. or B.C.C.) Avogadro’s constant. NA = 6.023 x 1023 mol–1?

Solution

solution:
we have given

Density of element,  ρ=10.3 g cm-3
Cell edge  a= 314pm or 3.14 x 10-10 cm          

     NA =6.023×1023 mol-1

Atomic mass = 96 g mol-1

                 P = Z×Ma3×NAZ = P×a3×NAM

= 10.g cm-3×(3.14)3×10-30cm3×6.023×1023 mol-196 g mol-1=2

The structure of the crystal lattice is B.C.C.