Answer
Verified
333.2k+ views
Hint: 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 $mo{{l}^{-1}}$
If ‘a’ is the edge length of a unit cell. Then, the volume of the unit cell will be equal to ‘${{\text{a}}^{3}}$’
Given, edge length of unit cell of element ‘X’ having f.c.c structure, a = 400 pm = 400 $\times {{10}^{-12}}$ m.
Density of a unit cell can be given as, d = $\dfrac{\text{mass of unit cell}}{\text{volume of unit cell}}$
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 as\[\begin{align}
& d=\dfrac{m\times Z}{V} \\
& =\dfrac{m\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}} \\
& \because m\times {{N}_{A}}=M\Rightarrow m=\dfrac{M}{{{N}_{A}}} \\
& \Rightarrow d=\dfrac{M\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times {{N}_{A}}} \\
& d=\dfrac{40g\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times 6.022\times {{10}^{23}}mo{{l}^{-1}}}=4.2\times {{10}^{-6}}g\,{{m}^{-3}} \\
\end{align}\]
Therefore, the density of the unit cell is 4.2$\times {{10}^{-6}}$ g${{m}^{-3}}$.
One mole of the element ‘X’ has 6.022$\times {{10}^{23}}$ atoms.
40 g of ‘X’ = 6.022$\times {{10}^{23}}$ atoms.
1 g of ‘X’ = $\dfrac{6.022\times {{10}^{23}}}{40}$ atoms
4 g of ‘X’ contains = 4$\times \dfrac{6.022\times {{10}^{23}}}{40}=6.022\times {{10}^{22}}$ atoms
We know that in an f.c.c unit cell one unit cell has 4 atoms.
1 unit cell = 4 atoms
1 atom = $\dfrac{1}{4}$ unit cell
Then, 6.022$\times {{10}^{22}}$ atoms = $\dfrac{1}{4}\times 6.022\times {{10}^{22}}$ = $1.505\times {{10}^{22}}$ 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 = $8\times \dfrac{1}{8}+6\times \dfrac{1}{2}$= 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 $mo{{l}^{-1}}$
If ‘a’ is the edge length of a unit cell. Then, the volume of the unit cell will be equal to ‘${{\text{a}}^{3}}$’
Given, edge length of unit cell of element ‘X’ having f.c.c structure, a = 400 pm = 400 $\times {{10}^{-12}}$ m.
Density of a unit cell can be given as, d = $\dfrac{\text{mass of unit cell}}{\text{volume of unit cell}}$
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 as\[\begin{align}
& d=\dfrac{m\times Z}{V} \\
& =\dfrac{m\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}} \\
& \because m\times {{N}_{A}}=M\Rightarrow m=\dfrac{M}{{{N}_{A}}} \\
& \Rightarrow d=\dfrac{M\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times {{N}_{A}}} \\
& d=\dfrac{40g\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times 6.022\times {{10}^{23}}mo{{l}^{-1}}}=4.2\times {{10}^{-6}}g\,{{m}^{-3}} \\
\end{align}\]
Therefore, the density of the unit cell is 4.2$\times {{10}^{-6}}$ g${{m}^{-3}}$.
One mole of the element ‘X’ has 6.022$\times {{10}^{23}}$ atoms.
40 g of ‘X’ = 6.022$\times {{10}^{23}}$ atoms.
1 g of ‘X’ = $\dfrac{6.022\times {{10}^{23}}}{40}$ atoms
4 g of ‘X’ contains = 4$\times \dfrac{6.022\times {{10}^{23}}}{40}=6.022\times {{10}^{22}}$ atoms
We know that in an f.c.c unit cell one unit cell has 4 atoms.
1 unit cell = 4 atoms
1 atom = $\dfrac{1}{4}$ unit cell
Then, 6.022$\times {{10}^{22}}$ atoms = $\dfrac{1}{4}\times 6.022\times {{10}^{22}}$ = $1.505\times {{10}^{22}}$ 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 = $8\times \dfrac{1}{8}+6\times \dfrac{1}{2}$= 4.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE