The edge length of a unit cell of a metal hav | vedclass

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(C) The formula for density is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.Given: $ho = 2 \ g/cm^3$, $M = 75 \ g/mol$, $a = 5 \ \mathring{A} = 5 	im

Vedclass · Accepted Answer

$216.5$

Vedclass · Answer

(C) The formula for density is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.Given: $ho = 2 \ g/cm^3$, $M = 75 \ g/mol$, $a = 5 \ \mathring{A} = 5 	imes 10^{-8} \ cm$, $N_A = 6 	imes 10^{23}$.Calculating $Z$:$Z = \frac{ho 	imes N_A 	imes a^3}{M} = \frac{2 	imes 6 	imes 10^{23} 	imes (5 	imes 10^{-8})^3}{75} = \frac{12 	imes 10^{23} 	imes 125 	imes 10^{-24}}{75} = \frac{1500 	imes 10^{-1}}{75} = \frac{150}{75} = 2$.Since $Z = 2$, the crystal structure is Body-Centered Cubic $(BCC)$.For a $BCC$ structure, the relation between radius $r$ and edge length $a$ is $r = \frac{\sqrt{3}}{4} a$.$r = \frac{\sqrt{3}}{4} 	imes 5 \ \mathring{A} = 1.732 	imes 1.25 \ \mathring{A} = 2.165 \ \mathring{A}$.Converting to $pm$: $2.165 \ \mathring{A} = 2.165 	imes 100 \ pm = 216.5 \ pm$.

The edge length of a unit cell of a metal having a molar mass of $75 \ g/mol$ is $5 \ \mathring{A}$. It crystallizes in a cubic lattice. If the density is $2 \ g/cm^3$, find the radius of the metal atom in $pm$. (Given: $N_A = 6 \times 10^{23}$)

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