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Question: An element ‘X’ (At. Mass = 40g \(mo{{l}^{-1}}\)) having f.c.c structure, has unit cell edge length o...

An element ‘X’ (At. Mass = 40g mol−1mo{{l}^{-1}}) having f.c.c structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4g of ‘X’. (NA{{N}_{A}}= 6.022×1023 mol−1\times {{10}^{23}}\,mo{{l}^{-1}})

Explanation

Solution

f.c.c stands for face centered cubic. An f.c.c unit cell has atoms at the corners as well as on at the centre of each face of the cube. Number of atoms present per unit cell in a face centered cubic unit cell (Z) is 4.

Complete answer:
An element ‘X’ has an f.c.c unit cell. Therefore, the number of atoms present per unit, Z = 4.
Given, atomic mass of the element, M = 40g mol−1mo{{l}^{-1}}
If ‘a’ is the edge length of a unit cell. Then, the volume of the unit cell will be equal to ‘a3{{\text{a}}^{3}}’
Given, edge length of unit cell of element ‘X’ having f.c.c structure, a = 400 pm = 400 ×10−12\times {{10}^{-12}} m.
Density of a unit cell can be given as, d = mass of unit cellvolume of unit cell\dfrac{\text{mass of unit cell}}{\text{volume of unit cell}}
Mass of the unit cell will be equal to the total number of atoms present in the unit cell multiplied by the mass of each atom, i.e. m×\times Z. For f.c.c unit cell, Z = 4.
Therefore, the density of the unit cell can be calculated asd=m×ZV =m×4(400×10−12m)3 ∵m×NA=M⇒m=MNA ⇒d=M×4(400×10−12m)3×NA d=40g×4(400×10−12m)3×6.022×1023mol−1=4.2×10−6g m−3 \begin{aligned} & d=\dfrac{m\times Z}{V} \\\ & =\dfrac{m\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}} \\\ & \because m\times {{N}_{A}}=M\Rightarrow m=\dfrac{M}{{{N}_{A}}} \\\ & \Rightarrow d=\dfrac{M\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times {{N}_{A}}} \\\ & d=\dfrac{40g\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times 6.022\times {{10}^{23}}mo{{l}^{-1}}}=4.2\times {{10}^{-6}}g\,{{m}^{-3}} \\\ \end{aligned}
Therefore, the density of the unit cell is 4.2×10−6\times {{10}^{-6}} gm−3{{m}^{-3}}.
One mole of the element ‘X’ has 6.022×1023\times {{10}^{23}} atoms.
40 g of ‘X’ = 6.022×1023\times {{10}^{23}} atoms.
1 g of ‘X’ = 6.022×102340\dfrac{6.022\times {{10}^{23}}}{40} atoms
4 g of ‘X’ contains = 4×6.022×102340=6.022×1022\times \dfrac{6.022\times {{10}^{23}}}{40}=6.022\times {{10}^{22}} atoms
We know that in an f.c.c unit cell one unit cell has 4 atoms.
1 unit cell = 4 atoms
1 atom = 14\dfrac{1}{4} unit cell
Then, 6.022×1022\times {{10}^{22}} atoms = 14×6.022×1022\dfrac{1}{4}\times 6.022\times {{10}^{22}} = 1.505×10221.505\times {{10}^{22}} unit cell.

Note: An atom at the centre of a face is shared between two unit cells and that at the corners is shared between 8 unit cells. There are 8 corners and 6 faces. So to find the total number of atoms present in an f.c.c unit cell = 8×18+6×128\times \dfrac{1}{8}+6\times \dfrac{1}{2}= 4.