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The density of KBr is 2.73g/cm. The length of the unit cell is 654pm. Find the number of atoms per unit cell in the cubic structure.


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Since $Z = \large\frac{\rho \times a^3 \times N_A}{at.wt}$
$Z = \large\frac{2.73\times(654\times10^{-10})^3\times6.023\times10^{23}}{119}$
$Z \approx 4$
Hence answer is (d)
answered Feb 25, 2014 by sharmaaparna1

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