An element with molar mass 2.7 times 10^{-2} | vedclass

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Question

(A) The density of a unit cell is given by the formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given:Density $d = 2.7 	imes 10^3 \ kg \ m^{-3} = 2.7

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face centered cubic

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(A) The density of a unit cell is given by the formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given:Density $d = 2.7 	imes 10^3 \ kg \ m^{-3} = 2.7 \ 	ext{g} \ cm^{-3}$.Molar mass $M = 2.7 	imes 10^{-2} \ kg \ mol^{-1} = 27 \ 	ext{g} \ mol^{-1}$.Edge length $a = 405 \ pm = 405 	imes 10^{-10} \ cm = 4.05 	imes 10^{-8} \ cm$.Avogadro's number $N_A = 6.02 	imes 10^{23} \ mol^{-1}$.Rearranging the formula to find $Z$:$Z = \frac{d 	imes a^3 	imes N_A}{M}$.Substituting the values:$Z = \frac{2.7 	imes (4.05 	imes 10^{-8})^3 	imes 6.02 	imes 10^{23}}{27}$.$Z = \frac{2.7 	imes 66.43 	imes 10^{-24} 	imes 6.02 	imes 10^{23}}{27}$.$Z = \frac{1079.7 	imes 10^{-1}}{27} = \frac{107.97}{27} \approx 4$.Since the number of atoms per unit cell $Z = 4$, the unit cell is face-centered cubic $(FCC)$.

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the nature of the cubic unit cell is: $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$

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