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Density

Iron crystall...

Question

Iron crystallises in a bcc system with a lattice parameter of $2.861 ˚ A$ . Calculate the density of iron in the bcc system
[atomic weight of $F e = 56, N_{A} = 6.02 \times 10^{23} {m o l}^{- 1}$ ].

7.94 g m L^{- 1}

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8.96 g m L^{- 1}

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2.78 g m L^{- 1}

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6.72 g m L^{- 1}

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Solution

The correct option is A $7.94 g m L^{- 1}$
$d = \frac{Z M}{N_{A} a^{3}}$ (for bcc, $Z = 2$ ),

$a = 2.861 ˚ A = 2.861 \times 10^{- 8} c m$

$d_{F e} = \frac{(2) \times 56.0 g {m o l}^{- 1}}{(6.02 \times 10^{23} {m o l}^{- 1}) {(2.861 \times 10^{- 8})}^{3} {c m}^{3}}$

$= 7.94 g / m L$

Hence, the correct option is $A$

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= 6.02 \times 10^{23} m o l^{- 1})

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530 {kg m}^{- 3}

6.94 {g mol}^{- 1}

(N_{A} = 6.02 \times 10^{23} {mol}^{- 1})

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