#### Question

Answer in Brief

An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm^{-3}. calculate the number of atoms in 108 g of the element.

#### Solution

The volume of a unit cell

=(300 pm)^{3}

=(3.00 × 10^{-8} cm)^{3}

= 2.7 × 10^{-23} cm^{3}

Volume of 108 g of element

= `"mass"/"density"= (108 "g")/(10.8 "cm"^-3) = 10 "cm"^3`

Number of unit cells in this volume =`(10"cm"^3)/(2.7xx10^-23 "cm"^3//"unit cell")= ( 10^24)/(2.7) "Unit cells"`

Since each FCC unit cell contains 4 atoms, therefore the total number of atoms in 108 g 4 atoms/unit cell × `10^24/2.7`unit cell

= 1.48 × 10^{24} atoms

Concept: Calculations Involving Unit Cell Dimensions

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