Copper crystallises into nbsp-a fcc structure and the unit cell has length of edge - 3-61 -10- -8 cm- Calculate the density of copper if the molar mass of - Cu- is - 63-5-nbsp-g mol-1- - -8-92 g cm-

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Standard XII

Chemistry

Zeff

Copper crysta...

Question

Copper crystallises into a fcc structure and the unit cell has length of edge $3.61 \times 10^{- 8} c m$ . Calculate the density of copper if the molar mass of $C u$ is $63.5 g m o l^{- 1}$ $(8.92 g c m)$

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Solution

Given,

$a = 3.61 \times 10^{- 8} c m M = 63.5 g / m o l$

Number of atoms in the unit cell of $F C C, Z = 4$

$∴$ $d e n s i t y = \frac{Z \times M}{a^{3} N A} = \frac{4 \times 63.5}{{(3.61 \times 10^{- 8})}^{3} \times 6.023 \times 10^{23}} = 8.92 g / c m^{3}$

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Copper crystallises into a fcc lattice with edge length $3.61 \times 10^{- 8} c m .$ Show that the calculated density is in agreement with its measured value of $8.92 g c m^{- 3}$

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Copper crystallises into a fcc structure and the unit cell has length of edge 3.61×10−8 cm. Calculate the density of copper if the molar mass of Cu is 63.5 g mol−1 (8.92 g cm)

Given, a=3.61×10−8cmM=63.5g/molNumber of atoms in the unit cell of FCC,Z=4∴ density=Z×Ma3NA=4×63.5(3.61×10−8)3×6.023×1023=8.92g/cm3

Copper crystallises into a fcc structure and the unit cell has length of edge $3.61 \times 10^{- 8} c m$ . Calculate the density of copper if the molar mass of $C u$ is $63.5 g m o l^{- 1}$ $(8.92 g c m)$

Given,

$a = 3.61 \times 10^{- 8} c m M = 63.5 g / m o l$

Number of atoms in the unit cell of $F C C, Z = 4$

$∴$ $d e n s i t y = \frac{Z \times M}{a^{3} N A} = \frac{4 \times 63.5}{{(3.61 \times 10^{- 8})}^{3} \times 6.023 \times 10^{23}} = 8.92 g / c m^{3}$