Question

How many unit cells are present in a) $2.5g$ cubic crystal of $KCl$ and b) with each edge of the crystal? [Given $KCl = 74.5gmo{l^{ - 1}}$ , Cell constant is $4$ ]

Answer 1

Hint: A unit cell is the smallest and most basic repeating unit of any solid. It is used to simplify the visual crystalline pattern in which solids arrange themselves. When the unit cell is repeated a network is formed which is called a lattice. It is also used to define the properties of crystalline lattices.

Complete answer:
We are given the cell constant whose value is $4$ , which indicates that in the cubic structure of $KCl$ a unit cell contains $4$ molecules of $KCl$ .
Now, molar mass of potassium chloride ($KCl$) is also given which is:
$KCl = 74.5gmo{l^{ - 1}}$
So, we will now calculate the mass of one unit cell $KCl$ as follows:
$74.5gmo{l^{ - 1}}$ is the mass of $6.022 \times {10^{23}}$ molecules.
Hence the mass of one unit cell of potassium chloride is m.
$m = \dfrac{{4 \times 74.5}}{{6.022 \times {{10}^{23}}}}$
Now, the number of unit cells in $2.5g$ of cubic crystal of potassium chloride is divided by dividing the given mass by the mass of one unit cell.
Hence, it is calculated as:
$$n = \dfrac{{2.5 \times 6.022 \times {{10}^{23}}}}{{298}}$$
Therefore, the number of unit cells, n in the $2.5g$ of cubic crystal of potassium chloride is:
$n = 5.052 \times {10^{21}}$
Hence, $5.052 \times {10^{21}}$ molecules are present in $2.5g$ of cubic crystals of potassium chloride.

Note:
Both $NaCl$ and $KCl$ crystallize in the face-centred cubic lattice. In this type of lattice eight atoms are present at all the corners of the unit cell which contributes one-eighth in each unit cell and each face-centre has an atom which contributes half of it in an individual unit cell. There are two FCC lattices one for cations and other for anions.