An element crystallizes into a structure which may be describes by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of this unit cell is `24xx10^(-24)cm^(3)` and density of element is `7.2g cm^(-3)` . Calculate the number of atoms present in `200g` of element.

Answer

`3.14xx10^(24)` atoms

Explanation

Step I. Calculation of atomic mass of the element.
Atomic mass of element `(M) = (rhoxxa^(3)xxN_(0)xx10^(-30))/(Z)`
No. of atoms per unit cell (Z) = 1 (corner ) + 2(diagonals) = 3
Volume of the unit cell `(a^(3))=24xx10^(-24)cm^(3)`
`=24xx10^(-24)xx10^(30)=24xx10^(6)("pm")^(3)`
Avogadro's no. `(N_(0))=6.022xx10^(23)mol^(-1)`
Density of unit cell or element `(rho)=7.2"g cm"^(-3)`
`M=((7.29"g cm"^(-3))xx(24xx10^(6))xx(6.022xx10^(23)mol^(-1))xx(10^(-30)cm^(3)))/(3)`
`=35.12"g mol"^(-1)`
Step II. No. of atoms in 200 g of the element.
No. of atoms in 35.12 g of the element`(N_(0))=6.022xx10^(23)`
No. of atoms in 200 g of the element `=((6.022xx10^(23))xx(200g))/((35.12g))=3.43xx10^(24)"atoms"`