Question

The number of atoms in $100g$ of an fcc crystal with density $= 10.0gc{m^{ - 3}}$and cell edge equal to $200pm$ is equal to:
A.$5 \times {10^{24}}$
B.$5 \times {10^{25}}$
C.$6 \times {10^{23}}$
D.$2 \times {10^{25}}$

Answer 1

Here in the question we are talking about the fcc crystal that in this type of crystal is considered to have a shape of cube and the face centre is further shared between the two opposite sides of the cube .

Complete step-by-step answer: As we know that all the substances are made up of so many particles but one question that arises in our mind is how many particles are there in each atom . for that we must calculate the number of atoms in the structure.
Here in the question we are talking about the fcc crystal . so here we will discuss it in detail in order to find the number of atoms in the given question using the given figures.
The full form of fcc is face centered cubic unit cell . When we are talking about the fcc crystal then we must know that this type of crystal is considered to have a shape of cube and the face centre is further shared between the two opposite sides of the cube . it further acts like a shield in between the centre of each atom. So when we talk about how much each atom contributed is half in number.
As we know there are six faces and eight corners in the cube so when each side gets half then it becomes a total four in number. This is further called the ${Z_{eff}}$.
For the calculation of the number of atoms this given formula is used.

= \frac{{{Z_{eff}} \times {M_v}}}{{{a^3} \times {{10}^{ - 30}} \times {N_A}}} \\\ {M_v} = \frac{{\rho \times {a^3} \times {{10}^{ - 30}} \times {N_A}}}{\begin{gathered} {Z_{eff}} \\\ = \frac{{10.0gc{m^{ - 3}} \times 200pm \times {{10}^{ - 30}} \times 6}}{\begin{gathered} 4 \\\ = 12gmo{l^{ - 1}} \\\ \end{gathered} } \\\ \end{gathered} } \\\ \end{gathered} $$ as we know that $${Z_{eff}}$$ is calculated to be four in number as explained above. Further $$\rho $$ the coefficient used. For converting the answer in Avogadro number we found that $$\begin{gathered} 6 \times {10^{23}} \\\ \frac{{6 \times {{10}^{23}}}}{{12}} \times 100 = 5 \times {10^{24}} \\\ \end{gathered} $$ **Note:** If we further calculate the percentage of total space filled by the particles then it is called packing efficiency. Hence option A is correct.