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Question: Formula mass of NaCl is \(58.45{\text{g mo}}{{\text{l}}^{ - 1}}\) and density of its pure form is \(...

Formula mass of NaCl is 58.45g mol−158.45{\text{g mo}}{{\text{l}}^{ - 1}} and density of its pure form is 2.167g cm−32.167{\text{g c}}{{\text{m}}^{ - 3}}. The average distance between adjacent sodium and chloride ions in the crystal is 2.814×10−82.814 \times {10^{ - 8}} cm. Calculate Avogadro's constant.
(A) 0.6×1023mol−10.6 \times {10^{23}}{\text{mo}}{{\text{l}}^{ - 1}}
(B) 6.05×1023mol−16.05 \times {10^{23}}{\text{mo}}{{\text{l}}^{ - 1}}
(C) 60.5×1023mol−160.5 \times {10^{23}}{\text{mo}}{{\text{l}}^{ - 1}}
(D) 0.06×1023mol−10.06 \times {10^{23}}{\text{mo}}{{\text{l}}^{ - 1}}

Explanation

Solution

Density of a unit is given as the ratio of mass of a unit cell and volume of unit cell. The mass of the unit cell is equal to the product of the number of atoms in the unit cell and the mass of each atom in the unit cell.

Complete step by step answer:
Mass of unit cell = number of atoms in unit cell ×\timesmass of each atom = z×mz \times m
Where, z is the number of atoms in the unit cell and m = mass of each atom.
Mass of an atom can be written in the terms of Avogadro number and molar mass as: MNA\dfrac{M}{{{N_A}}}
Where, M = molar mass and NA{N_A}= Avogadro’s number
Volume of unit cell = V =a3{a^3}, a = side length of cell
Density of unit cell = mass of unit cell volume of unit cell\dfrac{{{\text{mass of unit cell }}}}{{{\text{volume of unit cell}}}}= mV=z×ma3=z×Ma3×NA\dfrac{m}{V} = \dfrac{{z \times m}}{{{a^3}}} = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}
Thus, we can determine any one’s value when the values of all of them are given except the required value.
According to the question, NaCl is given and we know that NaCl has F.C.C. structure as z=4z = 4 for F.C.C.
So,z=4z = 4,d=2.167d = 2.167, M = 58.45g/mol58.45g/mol
Edge length = 2(r++r−)=2(2.814×10−8)2\left( {{r^ + } + {r^ - }} \right) = 2\left( {2.814 \times {{10}^{ - 8}}} \right)
a=5.628×10−8a = 5.628 \times {10^{ - 8}}
Therefore, a3=(5.628×10−8)3=1.78×10−22cm3{a^3} = {\left( {5.628 \times {{10}^{ - 8}}} \right)^3} = 1.78 \times {10^{ - 22}}c{m^3}
We know that, d=z×Ma3×NAd = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}
⇒NA=z×Ma3×d=4×58.451.78×10−22×2.167=6.06×1023\Rightarrow {N_A} = \dfrac{{z \times M}}{{{a^3} \times d}} = \dfrac{{4 \times 58.45}}{{1.78 \times {{10}^{ - 22}} \times 2.167}} = 6.06 \times {10^{23}}
Hence, option B is the correct answer.

Note:
Z has value of 22 for B.C.C and 44 value for F.C.C. and 66 value for H.C.P structure.The hexagonal closest packed (hcp) has a coordination number of 12 and contains 6 atoms per unit cell. The face-centered cubic (fcc) has a coordination number of 12 and contains 4 atoms per unit cell. The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell.