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
609.3k+ views
Hint: 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.
\[d=\text{ }(z\times m)\div ({{a}^{3}}\times {{N}_{a}})\]
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.
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