Question

An element with density 10gcm-³ forms a cubic unit cell with edge length of 3×10^-8cm.what is the nature of the cubic unit cell if the atomic mass of the element is 81 gmol-1

Answer 1

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Answer 2

Answer:

The unit cell is body centered cubic.

Explanation:

The unit cell that has 12 coordination number, and its has 2 unit cell then this unit cell is said to be body centered cubic.

Given,

• Density(d) = $10\  g\cdot cm^{-3}$.
• Edge length(a) = $3\times 10^{-8}\ cm$.
• Molecular weight. (M) = 81 g/mol.
• Avagadro's number($N_{A})=6.02\times 10^{23}$

The expression to calculate the number of the unit cell is as follows:

$Z=\frac{d\times N_{A}\times a^{3}}{M}$

Where,

d is the density.

a is the edge length.

M is the molecular weight.

Z is the number of the unit cell.

Substitute the value in the above expression.

$Z=\frac{10\  g\cdot cm^{-3}\times 6.02\times 10^{23}\times (3\times 10^{-8} cm)^{3}}{81 g/mol}\\=2.006$

The number of the unit cell is 2. Therefore, the unit cell is body centered cubic.