Calculate the number of atoms present in unit cell if an element having molar mass $23 \mathrm{~g}…
Calculate the number of atoms present in unit cell if an element having molar mass $23 \mathrm{~g} \mathrm{~mol}^{-1}$ and density $0.96 \mathrm{~g} \mathrm{~cm}^{-3}$.
$\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=48 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]$
(A) 1
(B) 2
(C) 4
(D) 6
1
2
4
6
Solution
$\begin{aligned}
& \text { Density }(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \\
& 0.96=\frac{\mathrm{n} \times 23}{48} \\
& \mathrm{n}=\frac{0.96 \times 48}{23}=2.003
\end{aligned}$
Number of atoms present in unit cell $=2$