Calculate number of atoms per unit cell of an element having molar mass $92.0 \mathrm{~g} \mathrm{~mol}^{-1}$ and density $8.6 \mathrm{~g} \mathrm{~cm}^{-3}$ forming cubic unit cell structure. $\left[\mathrm{a}^3 imes \mathrm{N}_{\mathrm{A}}=21.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1} ight]$

Question

Calculate number of atoms per unit cell of an element having molar mass $92.0 \mathrm{~g} \mathrm{~mol}^{-1}$ and density $8.6 \mathrm{~g} \mathrm{~cm}^{-3}$ forming cubic unit cell structure. $\left[\mathrm{a}^3 	imes \mathrm{N}_{\mathrm{A}}=21.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}ight]$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Density }(ho)=\frac{M 	imes n}{a^3 	imes N_A} \ & \mathrm{n}=\frac{ho 	imes \mathrm{a}^3 	imes \mathrm{N}_A}{\mathrm{M}}=\frac{8.6 \mathrm{~g} \mathrm{~cm}^{-3} 	imes 21.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}}{92 \mathrm{~g} \mathrm{~mol}^{-1}}=2\end{aligned}$

Calculate number of atoms per unit cell of an element having molar mass $92.0 \mathrm{~g}…

Solution