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The Solid State

Question
CBSEENCH12010094

An element with density 11.2 g cm-3  forms a f.c.c. lattice with edge length of 4 x10-8

Calculate the atomic mass of the element. (Given:  NA = 6.022x 10-23 (mol-1

Solution

Density, d = 11.2 g cm-3

Edge length, a = 4x10-8 cm

Avogadro number, NA = 6.022x1023 mol-1

Number of atoms present per unit cell, Z (fcc) = 4


We know for a crystal system,

  straight d equals space fraction numerator straight z space straight x space straight m over denominator straight a cubed space straight x space straight N subscript straight A end fraction

straight m space equals fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A over denominator straight Z end fraction

We space get comma

straight M space equals fraction numerator 11.2 space straight x space 64 space straight x space 10 to the power of negative 24 end exponent space straight x space 6.022 straight x space 10 to the power of 23 over denominator 4 end fraction equals space 107.91 space straight g

   

Thus, the atomic mass of the element is 107.91 g.