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(D) For an $FCC$ lattice,the number of atoms per unit cell,$Z = 4$.Given edge length,$a = 4.0 	imes 10^{-8} \ cm$.Density,$d = 9.03 \ g \ cm^{-3}$.Av

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$87$

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(D) For an $FCC$ lattice,the number of atoms per unit cell,$Z = 4$.Given edge length,$a = 4.0 	imes 10^{-8} \ cm$.Density,$d = 9.03 \ g \ cm^{-3}$.Avogadro's number,$N_{A} = 6.02 	imes 10^{23} \ mol^{-1}$.The formula for density is $d = \frac{Z 	imes M}{N_{A} 	imes a^{3}}$.Rearranging for molar mass $M$: $M = \frac{d 	imes N_{A} 	imes a^{3}}{Z}$.Substituting the values: $M = \frac{9.03 	imes 6.02 	imes 10^{23} 	imes (4.0 	imes 10^{-8})^{3}}{4}$.$M = \frac{9.03 	imes 6.02 	imes 10^{23} 	imes 64 	imes 10^{-24}}{4}$.$M = \frac{9.03 	imes 6.02 	imes 6.4}{4} = 86.97 \ g/mol$.Rounding to the nearest integer,$M \approx 87 \ g/mol$.

Metal $M$ crystallizes into a $FCC$ lattice with the edge length of $4.0 \times 10^{-8} \ cm$. The atomic mass of the metal is $........ \ g/mol$. (Nearest integer). (Use: $N_{A} = 6.02 \times 10^{23} \ mol^{-1}$,density of metal,$d = 9.03 \ g \ cm^{-3}$)

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