Question

The total number of diamond unit cells in $\text{ 1}\text{.2 g }$ of diamond sample is:
A) $\text{ 6}\text{.0 }\times \text{1}{{\text{0}}^{\text{21}}}\text{ }$
B) $\text{ 6}\text{.0 }\times \text{1}{{\text{0}}^{\text{22}}}\text{ }$
C) $\text{ 7}\text{.5 }\times \text{1}{{\text{0}}^{\text{21}}}\text{ }$
D) $\text{ 5}\text{.0 }\times \text{1}{{\text{0}}^{\text{22}}}\text{ }$

Answer 1

Diamond exists in the face-centred cubic structure. The unit cell contains the 4 atoms per unit cell and it has four tetrahedral voids that are occupied by the carbon atoms. We know that the one mole of a substance contains the $\text{ 6}\text{.023 }\times \text{1}{{\text{0}}^{\text{22}}}\text{ }$ particles in it. This is also known as the Avogadro's number. Calculate the total number of particles in the sample and find out the total number of unit cells.

Complete step by step answer:
We are given the following data: The weight of the diamond given is $\text{ 1}\text{.2 g }$
We have given the weight of the diamond. The number of moles of a substance is equal to the ratio of the weight of the substance to the molar mass of the substance. The carbon has a molar mass equal to 12. Thus, the number of moles of carbon in the diamond is given as,
$\,\text{No}\text{. of moles of C = }\dfrac{1.2}{12}\text{ = 0}\text{.1 }$
Thus, the $\text{ 0}\text{.1 }$ moles of carbons are present in $\text{ 1}\text{.2 g }$ of diamond samples. We know that the one mole of a substance contains the Avogadro's number of particles .the relation is stated as follows,
$\text{ 1 mole = 6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{ particles }$
Where, $\text{6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{ }$ particles are called the Avogadro's number. Therefore, the number of particles present in the $\text{ 1}\text{.2 g }$ diamond sample is,
$\begin{aligned} & \text{ No}\text{.of C atoms in 1}\text{.2 g of diamond sample = moles }\times \text{ }{{\text{N}}_{\text{A}}} \\\ & \Rightarrow \text{ 0}\text{.1 }\times \text{ 6}\text{.023 }\times \text{1}{{\text{0}}^{\text{23}}} \\\ & \Rightarrow \text{ 6}\text{.023 }\times \text{1}{{\text{0}}^{\text{22}}}\text{ particles } \\\ \end{aligned}$
Thus, the given weight of the compound contains the $\text{ 6}\text{.023 }\times \text{1}{{\text{0}}^{\text{22}}}\text{ }$ particles. We have asked the number of atoms in the diamond unit cell. Since each unit cell of diamond exists in the FCC structure. Thus diamond has 4 carbon atoms in the unit cell and 4 atoms in the unit cell.
Therefore the total number of carbon atoms per unit cell of the diamond is given as:
$\text{ 4 + 4 = 8 carbon atoms }$
One unit cell contains 8 carbon atoms, thus the number of atoms in the unit cell are,
$\text{ Number of unit cells = }\dfrac{6.022\text{ }\times \text{ 1}{{\text{0}}^{\text{22}}}\text{ atoms}}{8}\text{ = 7}\text{.5 }\times \text{1}{{\text{0}}^{\text{21}}}\text{ }$
Therefore, the total number of diamond unit cells in $\text{ 1}\text{.2 g }$ of diamond sample is $\text{7}\text{.5 }\times \text{1}{{\text{0}}^{\text{21}}}$.

Hence, (C) is the correct option.

Note: Note that this is an advanced case of the FCC structure where 4 atoms are present in the unit cell and additionally the unit cell contains the four atoms which are inside the unit cell. This atom occupies the tetrahedral voids. The unit cell contains a total of 8 carbon atoms in the diamond unit cell. Each carbon atom in the diamond has 4 bonds to other carbon atoms and this results into the tetrahedral array of atoms.