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A compound forms hexagonal close-packed structure. What is the total number of voids in \(0.5 \mathrm{~mol}\) of it? How many of these are tetrahedral voids?
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Verified Answer
No. of atoms in close packing \(=0.5 \mathrm{~mol}\)
\(\begin{aligned}
&=0.5 \times 6.022 \times 10^{23} \\
&=3.011 \times 10^{23}
\end{aligned}\)
No. of octahedral voids
\(=\) No. of atoms in packing
\(=3.011 \times 10^{23}\)
No. of tetrahedral voids
\(=2 \times\) No. of atoms in packing
\(=2 \times 3.011 \times 10^{23}=6.022 \times 10^{23}\)
Total no. of voids
\(\begin{aligned}
&=3.011 \times 10^{23}+6.022 \times 10^{23} \\
&=9.033 \times 10^{23}
\end{aligned}\)
\(\begin{aligned}
&=0.5 \times 6.022 \times 10^{23} \\
&=3.011 \times 10^{23}
\end{aligned}\)
No. of octahedral voids
\(=\) No. of atoms in packing
\(=3.011 \times 10^{23}\)
No. of tetrahedral voids
\(=2 \times\) No. of atoms in packing
\(=2 \times 3.011 \times 10^{23}=6.022 \times 10^{23}\)
Total no. of voids
\(\begin{aligned}
&=3.011 \times 10^{23}+6.022 \times 10^{23} \\
&=9.033 \times 10^{23}
\end{aligned}\)
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