Question

An element has a bcc structure with a cell edge of 288 pm. The density of the element is 7.2gcm-3. How many atoms are present in 208 g of the element

Answer 1

Formula units per unit cell Z = 2 for BCC

cubic unit cell lattice parameter a = 288 pm = 288 x10-8 cm

Volume V =  a3 =2.39X10-23cm3

Density d = 7.2g/cm3  

 

N­A = Avogadro constant = 6.022x10²³

Molecular mass M =?

We know that

Density d = ZM/NA X a3

M = dxNA x a3/Z

On Substituting values  

M= 7.2g/cm3 x(6.022x10²³)X (6.022x10²³)/2

= 51.8gmol-1

51.8 g of element contains 6.022X1023

208g of this element contains=?

= 6.022X1023X208/51.8

=2.42X1024 atoms.

Answer 2

${\huge{\underline{\underline{\mathcal{\red{♡ANSWER♡}}}}}}$.

• ❤.. ${Volume of the unit cell}$=${(288pm)^3}$
• ❤..=${(288×10^-12 m)^-3 = (288×10^-10 cm)^3}$
• ❤..=${2.39×10^-23 cm^3}$
• ❤..${Volume of208 g of element}$
• ❤..$\frac{mass}{density}$=$\frac{208g}{7.2gcm^-3}$=${28.88cm^3}$
• ❤..${Number of unit cells in this volume}$

=$\frac{28.88cm^3}{2.39×10^-23Cm^3/unit cell}$=${12.08×10^23 unit cells}$

• ⭐.. Since each${bcc}$cubic unit cell contains ${2}$ atoms,therefore,the total number of atoms in ${208g = 2}$${(atoms/unit cell)×12.08×10^23}$ unit cells=${24.16×10^23}$ atoms.