Question

An element crytallises into structures which may be described by a cubr type of unit cell having one atom on each corner of the cube and two atoms on one of its diagonal.

If the volume of this unit cell is 24 * 10ki power -24 cm cube and density = 7.2 g/cm3.

Calculate the number of atoms present in 200g of the element​

Answer 1

Answer: $3.4725*10^{24}$

Given,

An element crystallizes in simple cubic centre , thus the number of atoms = $\frac{1}{8}$ * 8 = 1

Now,

Given, in additional two atoms are on one of the diagonal .

Total number of atoms = 2+1 = 3

As per the question,

Volume = $24 *10^{-24} cm^{3}$

Density = 7.2 $g cm^{-3}$

Volume of element =  $\frac{mass_of_element}{Density_of_element}$

                            = $\frac{200}{7.2}$

                            = $27.78$

Volume of unit cells = volume of element / volume of unit cell

                                  =$\frac{27.78}{24*10^{-24}}$

                                                = $1.575 * 10^{24}$

Thus number of atoms present in $1.575 * 10^{24}$

atoms =

=  3 × $1.575 * 10^{24}$

=   $3.4725*10^{24}$

=  $3.471*10^{24}$