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. $4.16gc{m}^{-3}$
B. $41.6gc{m}^{-3}$
C. $4.16kgc{m}^{-3}$
D. $41.6kgc{m}^{-3}$

Answer 1

Hint: First find the mass of the unit cell using formula-
Mass of a unit cell = ${\displaystyle \frac{ZM}{N}}$ 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= ${\displaystyle \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}}$

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= ${\displaystyle \frac{ZM}{N}}$ 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= $$4\times {\displaystyle \frac{200}{24\times {10}^{23}}}$$
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= ${\displaystyle \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}}$
On putting the values from eq. (i) and (ii) in the formula we get,
Density= ${\displaystyle \frac{3.33\times {10}^{-22}}{8\times {10}^{-24}}}$
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.