Question

The density of crystalline sodium chloride is $5.85{\text{ gc}}{{\text{m}}^{ - 3}}$. What is the edge length of the unit cell?
A.$4.04 \times {10^{ - 9}}$ cm
B.$1.32 \times {10^{ - 14}}$ cm
C.$7.8 \times {10^{ - 23}}$ cm9
D.$9.6 \times {10^{ - 24}}$ cm

Answer 1

Hint:The relationship between density in a unit cell and the edge length is given as-
$ \Rightarrow d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$ Where d is density, Z is No. of atoms in a unit cell, ‘a’ is edge length, M is Atomic mass and ${{\text{N}}_{\text{A}}}$ is Avogadro number. The value of Z for simple cubic is $1$, for bcc is $2$and for fcc is $4$. Sodium chloride has an fcc type unit cell.

Complete step-by-step answer:Given, density of crystalline sodium chloride d=$5.85{\text{ gc}}{{\text{m}}^{ - 3}}$
We have to find the edge length of the unit cell.
We will use the relationship between density in a unit cell and the edge length which is given as-
$ \Rightarrow d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$ Where d is density, Z is No. of atoms in a unit cell, ‘a’ is edge length, M is Atomic mass and ${{\text{N}}_{\text{A}}}$ is Avogadro number.
We know that sodium chloride is made up of sodium ions and chloride ions.
The atomic mass of sodium Na=$23$ and the Atomic mass of chlorine Cl=$35.5$
So gram formula mass (M) of NaCl=$23 + 35.5 = 58.5gmo{l^{ - 1}}$
Since NaCl has fcc type unit cell so Z=$4$
And Avogadro number=$6.023 \times {10^{23}}$
On putting the given values in the above formula, we get-
$ \Rightarrow 5.85 = \dfrac{{4 \times 58.5}}{{6.023 \times {{10}^{23}} \times {a^3}}}$
On rearranging, we get-
$ \Rightarrow {a^3} = \dfrac{{4 \times 58.5}}{{6.023 \times {{10}^{23}} \times 5.85}}$
On simplifying, we get-
$ \Rightarrow {a^3} = \dfrac{{3.886 \times {{10}^{ - 22}}}}{{5.85}}$
On simplifying further, we get-
$ \Rightarrow {a^3} = 0.66427350427 \times {10^{ - 22}}$
$ \Rightarrow {a^3} = 66.427350427 \times {10^{ - 24}}$
On further solving, we get-
$ \Rightarrow a = 4.0499 \times {10^{ - 8}}$cm

Hence the correct answer is $A= 4.0499 \times {10^{ - 8}}$.

Note:In this formula, five parameters are given so value of any one can be calculated if values of other parameters are known. Also, remember that generally edge length is calculated in pm (picometre) and in questions if it is expressed in any other unit then convert it into pm $\left( {1{\text{ pm = 1}}{{\text{0}}^{ - 10}}{\text{cm}}} \right)$.