Calculate the volume of unit cell of an element having molar mass $27 \mathrm{~g} \mathrm{~mol}^{-1}$ that…
Calculate the volume of unit cell of an element having molar mass $27 \mathrm{~g} \mathrm{~mol}^{-1}$ that forms fce unit cell. $\left[\rho . \mathrm{N}_{\mathrm{A}}=16.0 \times 10^{23} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}\right]$
[Note: The question has been modified to get the correct answer.]
$7.50 \times 10^{-23} \mathrm{~cm}^3$
$6.75 \times 10^{-23} \mathrm{~cm}^3$
$5.75 \times 10^{-23} \mathrm{~cm}^3$
$8.25 \times 10^{-23} \mathrm{~cm}^3$
Solution
$\begin{aligned}
& \rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}} \\
& \mathrm{a}^3=\frac{\mathrm{n} \times \mathrm{M}}{\rho \times \mathrm{N}_{\mathrm{A}}}=\frac{4 \times 27 \mathrm{~g} \mathrm{~mol}^{-1}}{16 \times 10^{23} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}} \\
& \mathrm{~V}=\mathrm{a}^3=6.75 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}$
[Note: In the question, $16.0 \times 10^{-23}$ is changed to $16.0 \times 10^{23}$ to apply appropriate textual concepts.]