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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) - Chemistry

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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)

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Solution

Given,
Density, = 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, (fcc) = 4

We know for a crystal system,

\[d = \frac{z \times M}{a^3 \times N_A}\]

\[ \Rightarrow M = \frac{d \times a^3 \times N_A}{Z}\]

\[ \Rightarrow M = \frac{11 . 2 \times 64 \times {10}^{- 24} \times 6 . 022 \times {10}^{23}}{4} = 107 . 91 g\]

Thus, atomic mass of the element is 107.91 g.
Concept: Calculations Involving Unit Cell Dimensions
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