Question

The molar volume of KCl and NaCl are \[37.46\] mL and \[27.94\] ml respectively. Find the ratio of the unit cube edges of the two crystals (assume both have the same packing fraction)

Answer 1

Hint:A cube has twelve edges; all the faces are square hence all the edges are of the same length. We shall substitute the values in the formula given to calculate the radius and then take their ratios.
Formula used: Molar volume = N $ \times \dfrac{{4\pi {r^3}}}{3}$
Where N is Avogadro number.

Complete step by step answer:
Molar volume of KCl is \[37.46\] so by using the formula given above we can find the radii.
Molar volume of NaCl is \[27.94\] so by using the formula given above we can find radii.
Then we find the ratio of both the radii that is the ratio of radii of KCl and ratio of radii of NaCl.
The ratio of radii will be equal to the ratio of edge length as both belong to the same crystal structure.
Molar volume = N $ \times \dfrac{{4\pi {r^3}}}{3}$
Substituting:
\[37.46 = 6.022 \times {10^{^{23}}}\dfrac{{4{r^{^3}}}}{3}\]
r of KCl = 2.458 $ \times {10^{ - 8}}$
Similarly,
Molar volume = N$ \times \dfrac{{4\pi {r^3}}}{3}$
Substituting:
\[27.94 = 6.022 \times {10^{^{23}}}\dfrac{{4{r^{^3}}}}{3}\]
r of NaCl = 2.22$ \times {10^{ - 8}}$
Taking the ratio, we get:
$\dfrac{{2.458}}{{2.22}} = 1.108$
By taking the ratio of radii of KCl and NaCl we get an answer as \[1.108\].

Note:
The smallest repeating unit of the crystal lattice unit cell, which is the building block of a crystal is a unit cell.
Unit cells fill the space without overlapping as they are identical. They type of unit cell are:
Primitive cubic cells in this type of unit cell atom are present only at the corner.
Body Centred cubic unit cells in this type of atom are present in all corners but also has one atom at the centre.
Face centred cubic unit cell: in this type all the atoms are present in corners plus one atom at the centre of each face.