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A unit cell of iron crystal has edge length 288 pm and density 7.86 g.cm-3. Find the number of atoms per unit cell and type of the crystal lattice.

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Question

A unit cell of iron crystal has edge length 288 pm and density 7.86 g.cm-3. Find the number of atoms per unit cell and type of the crystal lattice.

Given : Molar mass of iron = 56 g.mol-1; Avogadro's number NA = 6.022 x 1023.

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Solution

Edge length of unit cell (a) = 288pm

volume of unit cell = (a)3

=(288)3 = 2.389 × 10-23 cm3

density of iron = ρ = 7.86g.cm3

Mass of iron unit cell (M) = Density × Volume

= 7.86 × 2.389 × 10-23

= 18.778 × 10-23 g

molar mass of iron = 56 g.mol-1

moles of iron in a unit cell = mass of iron in unit cell / molar mass of iron

= 18.778 × 10-23 / 56

= 3.353 × 10-24 mol

number of atoms per unit cell = moles of iron in a unit cell × Avogadro's number

= 3.353 × 10-24 × 6.022 x 1023

= 2.0191

≈ 2 atoms per unit cell

In a body centered cubic structure (bcc), the total number of atoms equals 2.

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2013-2014 (October)

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