Question

Calculate the value of Avogadro’s number from the following data:
Density of \[NaCl\] is $2.165 g cm^{-3}$ and distance between $N{a^ + }$ and $C{l^ - }$ in \[NaCl\] is 281 pm
A.$6.01 \times {10^{21}}$
B.$6.01 \times {10^{ - 23}}$
C.$6.01 \times {10^{23}}$
D.None of the above

Answer 1

Hint: To solve this question we have to use the concept of lattice of \[NaCl\] and to find the Avogadro number we need a relationship between Avogadro number, bond length and density, which is stated below:
$\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_A} \times {{10}^{ - 30}}}}$where,
\[\rho = {\text{density}}\]
$Z = {\text{number of molecules per unit cell}}$
$M = {\text{molar mass}}$
$a = {\text{side length of unit cell}}$
${N_A} = {\text{Avogadro number}}$

Complete step by step answer:
According to the given question we need to calculate the following terms according to our relation:
 $\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_A} \times {{10}^{ - 30}}}}$ where,
$\rho = {\text{density}}$
$Z = {\text{number of molecules per unit cell}}$
$M = {\text{molar mass}}$
$a = {\text{side length of unit cell}}$
${N_A} = {\text{Avogadro number}}$

Let us 1st calculate all the parameters required in the above formula:
Bond length in the given question is = $281×10^{-10}$ cm
The number of molecules per unit cell Z = 4 (for \[NaCl\])
The molar mass for \[NaCl\] M = 58.5
Given the density of\[NaCl\] is $\rho $ = $2.165 g cm^{-3}$
Now “a”, can be calculated as shown below:
$a = 2 × bond length$
$a = 2 × 281 × 10^{-10} cm$
Therefore, a = $562 × 10^{-10}$ cm


Now substituting the above value in the following formula we get:
 $\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_A} \times {{10}^{ - 30}}}}$
$
   \Rightarrow 2.165 = \dfrac{{4 \times 58.5}}{{{{562}^3} \times {N_A} \times {{10}^{ - 30}}}} \\
   \Rightarrow {N_A} = 6.01 \times {10^{23}} \\
$
Therefore, the value of Avogadro's number is $6.01 \times {10^{23}}$.
Thus, C is the correct option.

Additional information:
On the basis of geometrical consideration, there are 32 different combinations of elements of symmetry of a crystal. These combinations are called 32 systems. Among all, 7 types of basic or primitive unit cells have been recognized. They are cubic, orthorhombic, tetragonal, monoclinic, triclinic, hexagonal and rhombohedral. There are 14 different ways in which similar points can be arranged in a 3-D space. So, the total number of space lattices belonging to the 7 crystal system are 14.


Note:
Crystals belonging to cubic system have 3 kinds of Bravais lattices, they are:
Simple cubic
Face-centered cubic
Body-centered cubic
These need to be taken into account while solving the question because side length, number of molecules per unit cell depends upon the Bravais lattice and can only be determined when we know that given molecule belongs to which system.