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

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(A) Given:Density, $d = 2.7 	imes 10^{3} \,kg \,m^{-3}$Molar mass, $M = 2.7 	imes 10^{-2} \,kg \,mol^{-1}$Edge length, $a = 405 \,pm = 4.05 	imes 1

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Face-centred cubic $(fcc)$

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(A) Given:Density, $d = 2.7 	imes 10^{3} \,kg \,m^{-3}$Molar mass, $M = 2.7 	imes 10^{-2} \,kg \,mol^{-1}$Edge length, $a = 405 \,pm = 4.05 	imes 10^{-10} \,m$Avogadro's number, $N_A = 6.022 	imes 10^{23} \,mol^{-1}$Using the formula for density of a unit cell:$d = \frac{z \cdot M}{a^3 \cdot N_A}$Rearranging for $z$:$z = \frac{d \cdot a^3 \cdot N_A}{M}$Substituting the values:$z = \frac{(2.7 	imes 10^{3}) \cdot (4.05 	imes 10^{-10})^3 \cdot (6.022 	imes 10^{23})}{2.7 	imes 10^{-2}}$$z = \frac{2.7 	imes 10^{3} \cdot 66.43 	imes 10^{-30} \cdot 6.022 	imes 10^{23}}{2.7 	imes 10^{-2}}$$z \approx 4$Since the number of atoms per unit cell is $4$, the unit cell is face-centred cubic $(fcc)$.

An element with molar mass $2.7 \times 10^{-2} \,kg \,mol^{-1}$ forms a cubic unit cell with edge length $405 \,pm$. If its density is $2.7 \times 10^{3} \,kg \,m^{-3}$, what is the nature of the cubic unit cell?

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