Question

Metal has an fcc lattice. The edge length of the unit cell is 404pm. The density of the metal is $2.72gc{{m}^{-3}}$. The molar mass ($gmo{{l}^{-1}}$) of the metal is:
[Avogadro's constant=${{N}_{A}}$=$6.02\times {{10}^{23}}mo{{l}^{-1}}$
A. 40
B. 30
C. 27
D. 20

Answer 1

To solve the given problem we need to know a simple basic formula i.e
$d=\text{ }(z\times m)\div ({{a}^{3}}\times {{N}_{a}})$
Where ,
$d$= density of the unit cell
$z$= no of atoms in an unit cell
$a$= edge length of the unit cell
${{N}_{A}}$= Avogadro’s number
$M$= molecular weight

Complete step by step solution:
We first need to know about the FCC structure .
Fcc structure is also known as face centered cubic structure . In fcc structures there are 4 atoms per unit cell and 12 coordination numbers.
In fcc structures there are 8 atoms present in all the corners of the lattice or unit cell which is shared by 8 atoms and 6 atoms present in the center of the faces of the unit cell which is shared by 2 atoms .
That is $(1\div 8)\times 8=1$ and $(1\div 2)\times 6=3$
Hence, we can see that there are total 4 (3+1)atoms per unit cell.
Now let's come to the given problem ,
Using the formula ,
$d=\text{ }(z\times M)/({{a}^{3}}\times {{N}_{a}})$
$M=(d\times {{a}^{3}}\times {{N}_{a}})/z$
$d=2.72gc{{m}^{-3}}$
{{N}_{A}}=$$$6.02\times {{10}^{23}}mo{{l}^{-1}}$$ M=? Z=4 a=$$404pm= $$404\times {{10}^{-10}}cm$$ Now, $$M=(2.72\times 404\times 404\times 404\times {{10}^{-30}}\times 6.02\times {{10}^{-23}})\div 4$$ M=26.992gram$$=27 gram$

So the answer is 27 gram per mole which matches with the answer option C.

Hint: Unit cell is the smallest portion of a crystal lattice which is repeated in different directions generates the entire lattice for example primitive unit cell, face centered unit cell .
A regular three dimensional arrangement of constituent particles in a crystal in which each particle is depicted as a point is called a crystal lattice. There are 14 possible three dimensional lattices which are called bravais lattices.