(A) The density of a unit cell is given by the formula: $\rho = \frac{Z \times M}{a^{3} \times N_{A}}$For $CsBr$ in a body-centred cubic (bcc) lattice

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(A) The density of a unit cell is given by the formula: $\rho = \frac{Z \times M}{a^{3} \times N_{A}}$For $CsBr$ in a body-centred cubic (bcc) lattice,the number of formula units per unit cell $Z = 1$.The molar mass $M = 133 + 80 = 213 \, g/mol$.The edge length $a = 436.6 \, pm = 436.6 \times 10^{-10} \, cm$.Substituting these values into the formula:$\rho = \frac{1 \times 213}{(436.6 \times 10^{-10})^{3} \times 6.02 \times 10^{23}}$$\rho = \frac{213}{83.25 \times 10^{-24} \times 6.02 \times 10^{23}}$$\rho = \frac{213}{50.11} \approx 4.25 \, g/cm^{3}$.

Class 12 Chemistry Solid State Mathematical analysis of cubic system and Bragg’s equation

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$CsBr$ crystallises in a body-centred cubic lattice. The unit cell length is $436.6 \, pm$. Given that the atomic mass of $Cs = 133$ and that of $Br = 80 \, amu$ and Avogadro number being $6.02 \times 10^{23} \, mol^{-1}$,the density of $CsBr$ is .............. $g/cm^{3}$.

AIPMT 2006 All AIPMT

A
$4.25$
B
$42.5$
C
$0.425$
D
$8.25$

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Mathematical analysis of cubic system and Bragg’s equation

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$CsBr$ crystallises in a body-centred cubic lattice. The unit cell length is $436.6 \, pm$. Given that the atomic mass of $Cs = 133$ and that of $Br = 80 \, amu$ and Avogadro number being $6.02 \times 10^{23} \, mol^{-1}$,the density of $CsBr$ is .............. $g/cm^{3}$.

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