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# A compound forms hexagonal close-packed structure.

1.15  A compound forms hexagonal close-packed structure. What is the total number of voids in $\inline 0.5 \; mol$ of it? How many of these are tetrahedral voids?

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Amount of compound given = 0.5 mol

We know that,

$Moles = \frac{No.\ of\ atoms}{6.022 \times 10^{23}}$

So, No. of atoms  $=0.5\times 6.022 \times 10^{23} = 3.011 \times10^{23} atoms$

We also know that, No. of tetrahedral voids =   2(No. of atoms in closed packing)

$=2\times 3.011 \times10^{23} = 6.011\times10^{23}$

No. of octahedral voids = No. of atoms present in closed packing.

$= 3.011 \times10^{23}$

So,   Total number of voids =   No. of tetrahedral voids + No. of octahedral voids.

$=6.022 \times10^{23} +\ 3.011 \times10^{23}$

$=9.033 \times10^{23}\ voids$

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