An element with molar mass
Question:

An element with molar mass $2.7 \times 10^{-2} \mathrm{kgmol}^{-1}$ forms a cubic unit cell with edge length $405 \mathrm{pm}$. If its density is $2.7 \times 10^{3} \mathrm{kgm}^{-3}$, the radius of the element is approximately $\times 10^{-12} \mathrm{~m}$ (to the nearest integer).

Solution:

$\mathrm{d}=\frac{\mathrm{Z}\left(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\right)}{\mathrm{a}^{3}}$

$2.7 \times 10^{3}=\mathrm{z} \frac{\left(\frac{2.7 \times 10^{-2}}{6 \times 10^{23}}\right)}{\left(405 \times 10^{-12}\right)^{3}}$

$2.7 \times 10^{3}=\mathrm{z} \frac{\left(2.7 \times 10^{-2}\right)}{6 \times 10^{23}\left(4.05 \times 10^{-10}\right)^{3}}$

$2.7 \times 10^{3}=\mathrm{z} \frac{\left(2.7 \times 10^{-2}\right)}{6 \times 10^{23} \times 66.43 \times 10^{-30}}$

$3.98=\mathrm{z}$

$\mathrm{z} \approx 4$ structure is fcc

$\frac{a}{\sqrt{2}}=2 r$

$\mathrm{r}=\frac{\mathrm{a}}{2 \sqrt{2}}=\frac{\sqrt{2} \mathrm{a}}{4}=\frac{1.414 \times 405 \times 10^{-12}}{4}$

$r=143.16 \times 10^{-12}$

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