Question

KF has NaCl structure. If the distance between ${{K}^{+}}$ and ${{F}^{-}}$ is 269 pm, find the density of KF. (Atomic mass of K=39 and F=19)

Answer 1

Hint: As we know that unit cells are basically the smallest repeating atoms in a crystal. There are different types of unit cell like body centred unit cell represented as BCC, face centred unit cell (FCC). We will solve the question by using the formula of density which is:
\[\rho =\dfrac{M\times Z}{V\times {{N}_{A}}}\]

complete Step by step solution:
- We are being provided with the values of distance between ${{K}^{+}}$ and ${{F}^{-}}$ that is 269 pm.
- KF has NaCl structure. As we know that the structure of NaCl is FCC that is Face Centred Cubic unit cell structure. So, we can say that the value of its atomic number will be 4 that is Z=4.
- And we know that the formula of Density is:
\[\rho =\dfrac{M\times Z}{V\times {{N}_{A}}}\]
Where,
M= Molar mass , Z= Atomic number , V= Volume, $\rho$= Density,${{N}_{A}}$= Avogadro number
As, we know that the formula of volume is $V=2\sqrt{2}\times {{(r)}^{3}}$
Now, we will put all the values given in the equation of density as:
\[\begin{align}
& \rho =\dfrac{M\times Z}{V\times {{N}_{A}}} \\
& =\dfrac{58\times 4}{2\times \sqrt{2}{{(r)}^{3}}\times {{N}_{A}}} \\
& =\dfrac{58\times 4}{2\times \sqrt{2}\times {{\left( 269 \right)}^{3}}\times 6\times {{10}^{23}}} \\
 & =8.9\text{ }g\text{ }c{{m}^{-3}} \\
\end{align}\]

Hence, we can conclude that the density of KF is 8.9$g\text{ }c{{m}^{3}}$.

Note: - We must not get confused in terms of FCC and BCC. FCC is face centred cubic unit cell, it
contains four atoms per unit cell. Whereas, BCC is a body centred unit cell, it contains 2
atoms per unit cell.
- We should write the unit after solving any question.