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The pyknometric density of NaCl crystal is $2.165\times {{10}^{-3}}$ $kg{{m}^{-3}}$ while its X-ray density is $2.178\times {{10}^{-3}}$ $kg{{m}^{-3}}$. Therefore, the fraction of unoccupied sites in the NaCl crystal is:
$\begin{align}
 & a)5.96\times {{10}^{-4}} \\
 & b)5.96\times {{10}^{-2}} \\
 & c)5.96\times {{10}^{-1}} \\
 & d)5.96\times {{10}^{-3}} \\
\end{align}$

Answer Verified Verified
Hint: $Fraction\text{ of unoccupied sites = }\dfrac{\text{X-ray density(X) - Pyknometric Density(p)}}{X-ray\text{ density (X)}}$
Use this formula to try and calculate the required fraction. Be very careful of the units while you’re at it.

Complete step-by-step answer:
Before we solve this question, let us try and understand the important terminologies involved in this question.
Pyknometric density is the closest approximation to true density calculated from the molecular weight and crystalline lattice of a substance.
The density of an object measured with the help of diffraction of X-rays by the crystal in question is referred to as its X-ray density.
Since the real density is smaller than the one established from diffraction (using the volume of the unit cell) apparently not all atoms are present where you would expect them. The ratio of the two densities can help us find how many are empty.
Let us now put all this information to good use and plug the given data into the formula used to calculate the fraction of unoccupied sites in a crystal which is,
$Fraction\text{ of unoccupied sites = }\dfrac{\text{X-ray density(X) - Pyknometric Density(p)}}{X-ray\text{ density (X)}}$
$Fraction\text{ of unoccupied sites = }\dfrac{2.178\times {{10}^{-3}}\text{ - }2.165\times {{10}^{-3}}}{2.178\times {{10}^{-3}}}\approx 5.96\times {{10}^{-3}}$
Therefore, we can conclude that the fraction of empty spaces in a Sodium Chloride (NaCl) crystal is $5.96\times {{10}^{-3}}$.
Thus, we can conclude that the answer to this question is d) $5.96\times {{10}^{-3}}$

Note: It is extremely important to know the exact definitions of pyknometric and x-ray densities to be able to solve this question. Therefore, you are advised to be extremely careful when putting these two densities into the required calculation.
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