Question

The pyknometric density of the sodium chloride crystal is $2.165\times {{10}^{3}}kg{{m}^{-3}}$ while its X-ray density is $2.178\times {{10}^{3}}kg{{m}^{-3}}$. The fraction of unoccupied sites in the sodium chloride crystal is:
A. $5.96\times {{10}^{-3}}$
B. $5.96$
C. $5.96\times {{10}^{-2}}$
D. $5.96\times {{10}^{-1}}$

Answer 1

Think about what the pyknometric and the X-ray densities indicate about the lattices of the crystal. Calculate the volume that is unoccupied by subtracting one type of density from the other.

Complete step by step answer:
We know that the pyknometric density is the actual density of the crystal and takes into account the unoccupied sites. This density is calculated by taking into account only the sites that are occupied and thus gives us information about the actual sites that are occupied.
X-ray density on the other hand gives us information about the crystal in an ideal case where all sites are considered to be occupied. It is also known as the theoretical density of the molecules.
So, to find the volume of unoccupied sites in the sodium chloride crystals we will subtract the actual molar volume from the theoretical molar volume. We can find the molar volume using the formula that relates density volume and mass. Let us consider that the total mass of $NaCl$ crystals present is $M$.

So, we get the following formula:
$\text{density = }\dfrac{\text{mass}}{\text{molar volume}}$
So, we will modify the equations to suit our needs. First let us consider the theoretical molar volume.

& \text{X-ray density = }\dfrac{M}{\text{theoretical molar volume}} \\\ & \text{theoretical molar volume = }\dfrac{M}{2.178\times {{10}^{3}}} \\\ \end{aligned}$$ Now we will calculate the actual molar volume using the pyknometric density. $$\begin{aligned} & \text{pyknometric density = }\dfrac{M}{\text{actual molar volume}} \\\ & \text{actual molar volume = }\dfrac{M}{2.165\times {{10}^{3}}} \\\ \end{aligned}$$ So, now that we have the theoretical and actual molar volumes of the crystal, we can find the volume of the unoccupied sites in the crystal by subtracting the theoretical molar volume from the actual molar volume. Volume of unoccupied sites = actual molar volume - theoretical molar volume $$\begin{aligned} & \text{volume of unoccupied sites = }\dfrac{M}{2.165\times {{10}^{3}}}-\dfrac{M}{2.178\times {{10}^{3}}} \\\ & \text{volume of unoccupied sites = }M\times {{10}^{-3}}\left( \dfrac{1}{2.165}-\dfrac{1}{2.178} \right) \\\ & \text{volume of unoccupied sites = }M\times \dfrac{0.013}{2.165\times 2.178}\times {{10}^{-3}}{{m}^{3}} \\\ \end{aligned}$$ We can find the fraction of the unoccupied sites by dividing the volume of the unoccupied sites by the actual volume. So, the fraction of unoccupied sites will be: $$\begin{aligned} & \text{fraction of unoccupied sites =}\dfrac{\text{volume of unoccupied sites}}{\text{actual volume}} \\\ & \text{fraction of unoccupied sites = }\dfrac{M\times \dfrac{0.013}{2.165\times 2.178}\times {{10}^{-3}}}{\dfrac{M}{2.165\times {{10}^{3}}}} \\\ & \text{fraction of unoccupied sites =}\dfrac{0.013}{2.178} \\\ & \text{fraction of unoccupied sites =}5.96\times {{10}^{-3}} \\\ \end{aligned}$$ Hence, the correct answer is. $5.96\times {{10}^{-3}}$ **So, the correct answer is “Option A”.** **Additional Information:** We can find the fraction of unoccupied sites by an easier method using the following formula: $$\text{fraction of unoccupied sites = }\dfrac{\text{X-ray density }-\text{ Pyknometric density}}{\text{X-ray density}}$$ **Note:** Note that the actual volume of the crystal will be greater than the theoretical calculated value of volume. This happens due to the presence of unoccupied sites. They reduce the interaction between two unit cells and make the crystal slightly loosely packed, so that the volume will increase.