The crystal of CsBr has an edge length of 437 | vedclass

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Question

(A) Given: Edge length $(a) = 437 \ pm = 437 	imes 10^{-10} \ cm$. Density $(d) = 4.24 \ g \ cm^{-3}$. Molar mass $(M)$ of $CsBr = 133 + 80 = 213 \ g

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$bcc$

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(A) Given: Edge length $(a) = 437 \ pm = 437 	imes 10^{-10} \ cm$. Density $(d) = 4.24 \ g \ cm^{-3}$. Molar mass $(M)$ of $CsBr = 133 + 80 = 213 \ g \ mol^{-1}$. Avogadro's number $(N_A) = 6.022 	imes 10^{23} \ mol^{-1}$. Using the density formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$. Rearranging for $Z$: $Z = \frac{d 	imes a^3 	imes N_A}{M}$. Substituting the values: $Z = \frac{4.24 	imes (437 	imes 10^{-10})^3 	imes 6.022 	imes 10^{23}}{213}$. $Z = \frac{4.24 	imes 83.45 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{213} \approx \frac{213.1}{213} \approx 1$. Since the number of formula units per unit cell $(Z)$ is $1$, the crystal structure corresponds to a $CsCl$ type structure, which is a body-centered cubic $(bcc)$ lattice.

The crystal of $CsBr$ has an edge length of $437 \ pm$. If the density of the crystal is $4.24 \ g \ cm^{-3}$, determine the type of crystal structure of $CsBr$ (Atomic mass of $Cs = 133, Br = 80$).

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