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 : N_{A} = 6.022 × 10^{23} mol^{–1})

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#### Solution

Given,

Density, *d *= 11.2 g cm^{-3}

Edge length,* a* = 4 × 10^{-8} cm

Avogadro number, N_{A} = 6.022 × 10^{23}

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

We know for a crystal system,

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

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

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

Thus, atomic mass of the element is 107.91 g.

Concept: Calculations Involving Unit Cell Dimensions

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