Solid AB has NaCl type structure. If the radi | vedclass

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(B) For $NaCl$ type structure,the number of formula units per unit cell $(Z)$ is $4$. The edge length $(a)$ of the unit cell is given by $a = 2(r_{A^+

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$5$

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(B) For $NaCl$ type structure,the number of formula units per unit cell $(Z)$ is $4$. The edge length $(a)$ of the unit cell is given by $a = 2(r_{A^+} + r_{B^-}) = 2(0.5 + 1.5) \ \mathring{A} = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$. The density $(ho)$ is calculated using the formula: $ho = \frac{Z 	imes M}{N_A 	imes a^3}$. Substituting the values: $ho = \frac{4 	imes 48}{6 	imes 10^{23} 	imes (4 	imes 10^{-8})^3} \ g/cc$. $ho = \frac{192}{6 	imes 10^{23} 	imes 64 	imes 10^{-24}} \ g/cc$. $ho = \frac{192}{6 	imes 64 	imes 10^{-1}} \ g/cc = \frac{192}{38.4} \ g/cc = 5 \ g/cc$.

Solid $AB$ has $NaCl$ type structure. If the radius of $A^{+}$ and $B^{-}$ are $0.5 \ \mathring{A}$ and $1.5 \ \mathring{A}$ respectively and the formula mass of $AB$ is $48 \ g/mol$,what is the density of $AB$ solid? $(N_A = 6 \times 10^{23})$

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