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(C) $1$. Calculate the molar mass $(M)$ of the element:Number of atoms $(N)$ = $2.03 	imes 10^{24}$Mass $(w)$ = $135 \ g$Avogadro's number $(N_A)$ =

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$78.8$

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(C) $1$. Calculate the molar mass $(M)$ of the element:Number of atoms $(N)$ = $2.03 	imes 10^{24}$Mass $(w)$ = $135 \ g$Avogadro's number $(N_A)$ = $6.022 	imes 10^{23} \ mol^{-1}$Moles $(n)$ = $N / N_A = (2.03 	imes 10^{24}) / (6.022 	imes 10^{23}) \approx 3.37 \ mol$Molar mass $(M)$ = $w / n = 135 / 3.37 \approx 40 \ g \ mol^{-1}$.$2$. Calculate the density $(d)$ using the formula:$d = (Z 	imes M) / (N_A 	imes a^3)$For $FCC$ lattice, $Z = 4$.Edge length $(a)$ = $150 \ pm = 150 	imes 10^{-10} \ cm = 1.5 	imes 10^{-8} \ cm$.$a^3 = (1.5 	imes 10^{-8})^3 = 3.375 	imes 10^{-24} \ cm^3$.$d = (4 	imes 40) / (6.022 	imes 10^{23} 	imes 3.375 	imes 10^{-24})$$d = 160 / (6.022 	imes 0.3375) = 160 / 2.0324 \approx 78.72 \ g \ cm^{-3}$.Rounding to the nearest option, the density is $78.8 \ g \ cm^{-3}$.

An element has $2.03 \times 10^{24}$ atoms in $135 \ g$. If the element crystallizes in a face-centered cubic $(FCC)$ lattice structure with an edge length of $150 \ pm$, then the density of the element is ............... $g \ cm^{-3}$.

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