Question

An element has FCC structure with edge length $200$pm. Calculate density if $200$g of this element contains $24 \times {10^{23}}$ atoms.
A. $4.16{\text{ }}gc{m^{ - 3}}$
B. $41.6{\text{ }}gc{m^{ - 3}}$
C. $4.16{\text{ }}kgc{m^{ - 3}}$
D. $41.6{\text{ }}kgc{m^{ - 3}}$

Answer 1

Hint: First find the mass of the unit cell using formula-
Mass of a unit cell = $\dfrac{{ZM}}{N}$ where Z is the effective number of atoms in the unit cell, M is molar mass; N is the Avogadro number or number of atoms. Then calculate the density by putting the given values in the following formula-
Density= $\dfrac{{{\text{Mass of unit cell}}}}{{{\text{Volume of unit cell}}}}$

Step-by-Step Solution-
Given, Molar Mass M= $200$g
Number of atoms N= $24 \times {10^{23}}$
Edge length a= $200$pm
Element having FCC structure has total $4$ atoms in a unit cell. The effective number of atoms in unit cell Z= $4$
 So the mass of the unit cell is equal to the mass of the $4$atoms.
Then we know that Mass of a unit cell is given by= $\dfrac{{ZM}}{N}$ where Z is the effective number of atoms in unit cell, M is molar mass, N is the Avogadro number or number of atoms.
On putting the values of the formula we get,
Mass of unit cell= \[4 \times \dfrac{{200}}{{24 \times {{10}^{23}}}}\]
On solving we get,
Mass of unit cell= $3.33 \times {10^{ - 22}}$ g--- (i)
Now the volume of the unit cell= \[{\left( a \right)^3}\]
On putting the values we get,
Volume of unit cell= ${\left( {200 \times {{10}^{ - 10}}} \right)^3}$ $c{m^3}$
Then on solving we get,
Volume of unit cell= $8 \times {10^{ - 24}}$ $c{m^3}$--- (ii)
Now we have to calculate density
And we know that Density= $\dfrac{{{\text{Mass of unit cell}}}}{{{\text{Volume of unit cell}}}}$
On putting the values from eq. (i) and (ii) in the formula we get,
Density= $\dfrac{{3.33 \times {{10}^{ - 22}}}}{{8 \times {{10}^{ - 24}}}}$
On solving we get,
Density = $0.416 \times {10^2}$
Density= $41.6gc{m^{ - 3}}$

Answer-Hence the correct answer is B.

Note: In FCC structure, following points are to be noted-
- The atoms in a unit cell are all present in the corners of the crystal lattice.
- One atom is present at the centre of every face of the cube
- This face centered atom is shared between two adjacent units’ cells in the crystal lattice.
- Only half of each atom belongs to the unit cell.