Question

An element has a face centered cubic unit cell with a length of 352.4 pm along an edge.The density of the element is 8.9 gcm^-3.How many atoms are present in 100g of an element?

 

Answer 1

Unit Cell: A basic repeating structural unit of a crystalline solid.

Number of atoms in a face centered cubic unit cell (fcc) n = 4.

Density of the unit cell = Mass of the unit cell / Volume of the unit cell

\( Density of the unit cell \) = \( \frac{nM}{a^3 Avogadro Number} \)

n -> Number of atoms

M -> Molar Mass

a -> edge length

 

Number of atoms = (Mass of the element /Molar mass) x Avogadro's number.

Given:

1. Density = 8.9 gcm^-3
2. Edge length a= 352.4 pm = 352.4 x 10^-10cm (1 pm = 10^-12m = 10^-10 cm)
3. Mass of the element = 100g
4. Avogadro constant = 6.023 x 10²³mol^-1

Solution:

           

 

 

 

The number of atoms present in 100g of the element  is 10.23 x 10²³.