Question

Calculate the theoretical density of iron, and then determine the number of vacancies per cm3 needed for a BCC iron crystal to have a density of 7.874 g/cm3 . The lattice parameter of iron is 2.866 * 10-8 cm.

Answer 1

Answer:

Therefore the theoretical density of iron is 7.877 g/cm³ .

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$

Explanation:

BCC structure contains 2 atoms per cell.

BCC : Body centered cubic.

Atomic mass of iron = 55.845 gram/mole.

Atomic mass of a chemical element is the mass of 1 mole of the chemical element.

Density: Density of a matter is the ratio of mass of the matter and volume of the matter.

Avogrdro Number is number of atoms per 1 mole.

Avogrdro Number= 6.023×10²³

Lattice parameter of iron = 2.866×10⁻⁸ cm

The theoretical density:

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

 $=\frac{2\times 55.845 }{(6.023\times 10^{23})\times (2.866\times 10^{-8})^3}$   g/cm³      [x= 2,Since it is BCC structure]

=7.877 g/cm³

Therefore the theoretical density of iron is 7.877 g/cm³ .

Now we have to find out the number of unit cell of iron crystal having density  7.874 g/cm³.

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

$\Rightarrow 7.874 =\frac{x\times 55.845}{6.023\times 10^{23}\times (2.866\times 10^{-8})^3}$

$\Rightarrow x= \frac{7.874\times 6.023\times 10^{23}\times (2.866\times 10^{-8})^3}{55.845}$

⇒ x =1.9923 atom/cell

Therefore the vacancy of atom per cell = (2- 1.9923)=0.0077

$Vacancy\ per\ cm^3=\frac{\textrm{Vacancy per cm} ^3 }{\textrm{( lattice parameter)} ^3}$

                          $=\frac{0.0077}{(2.866\times 10^-8)^3}$

                           $=3.27 \times 10^{20}$

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$

   

   

Answer 2

In this exercise we have to use the density knowledge to calculate the number of vacancies, in this way we can find that:

Density is $7.877 g/cm^2$

Number of vacancy is $3.27*10^{20$

So from the information given in the text we find that:

• BCC structure contains 2 atoms per cell and is a body centered cubic.

• Atomic mass of iron = 55.845 gram/mole.

• Avogrdro Number= 6.023×10²³

• Lattice parameter of iron = 2.866×10⁻⁸ cm

Knowing that the density formula is given by:

$\rho= (atom*atomic\ mass)/(avogadro's \ number* lattice \ parameter)^3$

Now substituting the values ​​in the given formula we find that:

$= (2*55.845)/(6*10^{23})*(2.866*10^{-8}) \\=7.874 g/cm^3$

Now we have to find out the number of unit cell of iron crystal having density, we have:

$7.874=(X*55.845)/(6.023*10^{23})*2.866*10^{-8})^3\\X=1.9923 atom/cell$

Therefore the vacancy of atom per cell :

$(2- 1.9923)=0.0077$

That will be:

$V= (0.0077)/(2.866*10^{-8})^3\\=3.27*10^{20$

See more about density at brainly.com/question/952755