Solveeit Logo
Login

Question

Question: The density of crystalline sodium chloride is \(5.85{\text{ gc}}{{\text{m}}^{ - 3}}\). What is the e...

The density of crystalline sodium chloride is 5.85 gcm−35.85{\text{ gc}}{{\text{m}}^{ - 3}}. What is the edge length of the unit cell?
A.4.04×10−94.04 \times {10^{ - 9}} cm
B.1.32×10−141.32 \times {10^{ - 14}} cm
C.7.8×10−237.8 \times {10^{ - 23}} cm9
D.9.6×10−249.6 \times {10^{ - 24}} cm

Explanation

Solution

The relationship between density in a unit cell and the edge length is given as-
⇒d=Z×MNA×a3\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 NA{{\text{N}}_{\text{A}}} is Avogadro number. The value of Z for simple cubic is 11, for bcc is 22and for fcc is 44. Sodium chloride has an fcc type unit cell.

Complete step-by-step answer: Given, density of crystalline sodium chloride d=5.85 gcm−35.85{\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-
⇒d=Z×MNA×a3\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 NA{{\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=2323 and the Atomic mass of chlorine Cl=35.535.5
So gram formula mass (M) of NaCl=23+35.5=58.5gmol−123 + 35.5 = 58.5gmo{l^{ - 1}}
Since NaCl has fcc type unit cell so Z=44
And Avogadro number=6.023×10236.023 \times {10^{23}}
On putting the given values in the above formula, we get-
⇒5.85=4×58.56.023×1023×a3\Rightarrow 5.85 = \dfrac{{4 \times 58.5}}{{6.023 \times {{10}^{23}} \times {a^3}}}
On rearranging, we get-
⇒a3=4×58.56.023×1023×5.85\Rightarrow {a^3} = \dfrac{{4 \times 58.5}}{{6.023 \times {{10}^{23}} \times 5.85}}
On simplifying, we get-
⇒a3=3.886×10−225.85\Rightarrow {a^3} = \dfrac{{3.886 \times {{10}^{ - 22}}}}{{5.85}}
On simplifying further, we get-
⇒a3=0.66427350427×10−22\Rightarrow {a^3} = 0.66427350427 \times {10^{ - 22}}
⇒a3=66.427350427×10−24\Rightarrow {a^3} = 66.427350427 \times {10^{ - 24}}
On further solving, we get-
⇒a=4.0499×10−8\Rightarrow a = 4.0499 \times {10^{ - 8}}cm

Hence the correct answer is A=4.0499×10−8A= 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 (1 pm = 10−10cm)\left( {1{\text{ pm = 1}}{{\text{0}}^{ - 10}}{\text{cm}}} \right).