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Question: An element has FCC structure with edge length \(200\)pm. Calculate density if \(200\)g of this eleme...

An element has FCC structure with edge length 200200pm. Calculate density if 200200g of this element contains 24×102324 \times {10^{23}} atoms.
A. 4.16 gcm−34.16{\text{ }}gc{m^{ - 3}}
B. 41.6 gcm−341.6{\text{ }}gc{m^{ - 3}}
C. 4.16 kgcm−34.16{\text{ }}kgc{m^{ - 3}}
D. 41.6 kgcm−341.6{\text{ }}kgc{m^{ - 3}}

Explanation

Solution

First find the mass of the unit cell using formula-
Mass of a unit cell = ZMN\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= Mass of unit cellVolume of unit cell\dfrac{{{\text{Mass of unit cell}}}}{{{\text{Volume of unit cell}}}}

Step-by-Step Solution-
Given, Molar Mass M= 200200g
Number of atoms N= 24×102324 \times {10^{23}}
Edge length a= 200200pm
Element having FCC structure has total 44 atoms in a unit cell. The effective number of atoms in unit cell Z= 44
So the mass of the unit cell is equal to the mass of the 44atoms.
Then we know that Mass of a unit cell is given by= ZMN\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= 4×20024×10234 \times \dfrac{{200}}{{24 \times {{10}^{23}}}}
On solving we get,
Mass of unit cell= 3.33×10−223.33 \times {10^{ - 22}} g--- (i)
Now the volume of the unit cell= (a)3{\left( a \right)^3}
On putting the values we get,
Volume of unit cell= (200×10−10)3{\left( {200 \times {{10}^{ - 10}}} \right)^3} cm3c{m^3}
Then on solving we get,
Volume of unit cell= 8×10−248 \times {10^{ - 24}} cm3c{m^3}--- (ii)
Now we have to calculate density
And we know that Density= Mass of unit cellVolume of unit cell\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= 3.33×10−228×10−24\dfrac{{3.33 \times {{10}^{ - 22}}}}{{8 \times {{10}^{ - 24}}}}
On solving we get,
Density = 0.416×1020.416 \times {10^2}
Density= 41.6gcm−341.6gc{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.