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Density

Face centered...

Question

Face centered cubic crystal lattice of copper has density of 8.966 g cm $^{- 3}$ . Calculate the volume of the unit cell.
(Given Molar mass of copper = $63.5 g {m o l}^{- 1}$ ; Avogadro number $N_{A} = 6.022 \times 10^{23} {m o l}^{- 1}$ )

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Solution

Given $d = 8.966 g / c m$

Molar mass of copper $= 63.5 g / m o l$

$N_{A} = 6.022 \times 10^{23}$

$d = \frac{Z \times M}{V \times N_{A}}$

where $Z = 4$ (For FCC)

$V = \frac{Z \times M}{d \times N_{A}}$

$V = \frac{4 \times 63.5}{8.966 \times 6.023 \times 10^{23}}$

$V = 4.704 \times 10^{- 23} c m^{3}$

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Similar questions

8.996 g . c m^{- 3}

63.5 g . m o l^{- 1}

N_{A}

6.022 \times 10^{23} m o l^{- 1}

8.930

c m^{-} 3

298

63.55

_{A}

6.02 \times 1023 m o l^{-} 1)

127.8

63.55

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

8.95

^{- 3}

= 63.5

^{- 1}

N_{A} = 6.02 \times 10^{23}

^{- 1}

40 {g mol}^{- 1}

\frac{100}{24} {g/cm}^{3}

o A

N_{A} = 6 \times 10^{23}

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