The element crystallizes in a body centred cubic lattice and the edge of unit cell is $0.351 n m$ . The density is $0.533 g / c m 3$ . What is the atomic weight?

12.0

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6.94

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9.01

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10.8

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Solution

The correct option is C $6.94$
Density $d = \frac{Z \times M}{a^{3} \times N_{A}}$
$a = 0.351 n m = 3.51 \times 10^{- 8} c m$
$d = 0.533 g / c m^{3}$
$Z = 2$
$d = \frac{Z \times M}{a^{3} \times N_{A}}$
$M = \frac{d \times a^{3} \times N_{A}}{Z}$
$M = \frac{0.533 \times (3.51 \times 10^{- 8})^{3} \times 6.023 \times 10^{23}}{2}$
$M = 6.94$

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The element crystallizes in a body centred cubic lattice and the edge of unit cell is 0.351nm. The density is 0.533g/cm3. What is the atomic weight?

The correct option is C 6.94Density d=Z×Ma3×NAa=0.351 nm=3.51×10−8 cmd=0.533 g/cm3Z=2d=Z×Ma3×NAM=d×a3×NAZM=0.533×(3.51×10−8)3×6.023×10232M=6.94

The element crystallizes in a body centred cubic lattice and the edge of unit cell is $0.351 n m$ . The density is $0.533 g / c m 3$ . What is the atomic weight?