The edge length of the unit cell of a metal ( | vedclass

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(B) The density formula is given by: $ho = \frac{z 	imes M}{N_A 	imes a^3}$.Given: $ho = 1.74 \ g \ cm^{-3}$, $M = 24 \ g \ mol^{-1}$, $a = 4.53

Vedclass · Accepted Answer

$160$

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(B) The density formula is given by: $ho = \frac{z 	imes M}{N_A 	imes a^3}$.Given: $ho = 1.74 \ g \ cm^{-3}$, $M = 24 \ g \ mol^{-1}$, $a = 4.53 \ \mathring{A} = 4.53 	imes 10^{-8} \ cm$, $N_A = 6 	imes 10^{23} \ mol^{-1}$.Substituting the values: $1.74 = \frac{z 	imes 24}{6 	imes 10^{23} 	imes (4.53 	imes 10^{-8})^3}$.Solving for $z$: $z = \frac{1.74 	imes 6 	imes 10^{23} 	imes 93.06 	imes 10^{-24}}{24} \approx 4$.Since $z = 4$, the structure is Face-Centered Cubic $(FCC)$.For $FCC$, the relationship between radius $(r)$ and edge length $(a)$ is: $r = \frac{a}{2\sqrt{2}}$.$r = \frac{4.53 \ \mathring{A}}{2 	imes 1.414} \approx 1.60 \ \mathring{A}$.Converting to $pm$: $1.60 \ \mathring{A} = 160 \ pm$.Thus, the correct option is $B$.

The edge length of the unit cell of a metal $(M_w = 24 \ g \ mol^{-1})$ having a cubic structure is $4.53 \ \mathring{A}$. If the density of the metal is $1.74 \ g \ cm^{-3}$, the radius of the metal atom is ............... $pm$ $(N_A = 6 \times 10^{23} \ mol^{-1})$

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