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प्रश्न
An element with density 11.2 g cm–3 forms a f.c.c. lattice with edge length of 4 × 10–8 cm.
Calculate the atomic mass of the element.
(Given : NA = 6.022 × 1023 mol–1)
उत्तर
Given,
Density, d = 11.2 g cm-3
Edge length, a = 4 × 10-8 cm
Avogadro number, NA = 6.022 × 1023
Number of atoms present per unit cell, Z (fcc) = 4
We know for a crystal system,
\[d = \frac{z \times M}{a^3 \times N_A}\]
\[ \Rightarrow M = \frac{d \times a^3 \times N_A}{Z}\]
\[ \Rightarrow M = \frac{11 . 2 \times 64 \times {10}^{- 24} \times 6 . 022 \times {10}^{23}}{4} = 107 . 91 g\]
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