Each edge of a cubic unit cell is 400 pm lon | vedclass

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(B) The formula for density is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given: $d = 6.25 \ g/cm^3$, $M = 120 \ g/mol$, $a = 400 \ pm = 400 	imes 10^{-

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Body-centered cubic

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(B) The formula for density is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given: $d = 6.25 \ g/cm^3$, $M = 120 \ g/mol$, $a = 400 \ pm = 400 	imes 10^{-10} \ cm = 4 	imes 10^{-8} \ cm$, $N_A = 6 	imes 10^{23} \ mol^{-1}$.Substituting the values: $6.25 = \frac{Z 	imes 120}{(4 	imes 10^{-8})^3 	imes 6 	imes 10^{23}}$.$6.25 = \frac{Z 	imes 120}{64 	imes 10^{-24} 	imes 6 	imes 10^{23}} = \frac{Z 	imes 120}{64 	imes 0.6} = \frac{Z 	imes 120}{38.4}$.$Z = \frac{6.25 	imes 38.4}{120} = \frac{240}{120} = 2$.Since $Z = 2$, the crystal lattice is $B.C.C$ (Body-centered cubic).

Each edge of a cubic unit cell is $400 \ pm$ long. If the atomic weight of the element is $120$ and its density is $6.25 \ g/cm^3$, identify the crystal lattice $(N_A = 6 \times 10^{23} \ mol^{-1})$.

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