CsBr crystallises in a body-centred cubic lat | vedclass

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Question

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

Vedclass · Accepted Answer

$4.25$

Vedclass · Answer

(A) The density of a unit cell is given by the formula: $ho = \frac{Z 	imes M}{a^{3} 	imes 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 $a = 436.6 \, pm = 436.6 	imes 10^{-10} \, cm$.Substituting these values into the formula:$ho = \frac{1 	imes 213}{(436.6 	imes 10^{-10})^{3} 	imes 6.02 	imes 10^{23}}$$ho = \frac{213}{83.25 	imes 10^{-24} 	imes 6.02 	imes 10^{23}}$$ho = \frac{213}{50.11} \approx 4.25 \, g/cm^{3}$.

$CsBr$ 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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