Question

A domain in ferromagnetic iron is in the form of a cube of side length $2$ $\mu m$ then the number of iron atoms in the domain are (Molecular mass of iron $=55$g $mo{l}^{-1}$ and density $=7.92$g $c{m}^{-3}$)

• A. $6.92\times {10}^{12}$ atoms
• B. $6.92\times {10}^{11}$ atoms
• C. $6.92\times {10}^{10}$ atoms
• D. $6.92\times {10}^{13}$ atoms

Answer 1

The correct option is B $6.92\times {10}^{11}$ atoms

The volume of the cubic domain
$V=(2\mu m{)}^3=(2\times {10}^{-6}m{)}^3$
$=8\times {10}^{-18}{m}^3=8\times {10}^{-12}c{m}^3$
and mass $=$volume $\times$ density
$=8\times {10}^{-12}c{m}^3\times 7.9gc{m}^{-3}$
$=63.2\times {10}^{-12}$g
Now the Avagadro number $(6.023\times {10}^{23})$ of iron atoms have a mass of $55$g. Hence the number of atoms in the domain are.
$N=\frac{63.2\times {10}^{-12}\times 6.023\times {10}^{23}}{55}=6.92\times {10}^{11}$ atoms.