Ask Questions, Get Answers


An element has a body centred cubic structure with a cell edge of 288 pm.The density of the element is $7.2gcm^{-3}$.calculate the number of atoms present in 280g of the element.

$\begin{array}{1 1}2.418\times 10^{24}\\4.418\times 10^{24}\\6.418\times 10^{24}\\1.418\times 10^{24}\end{array} $

1 Answer

Volume of unit cell =$(288 pm)^3$
$\Rightarrow (288\times 10^{-10}cm)^3$
$\Rightarrow 2.389\times 100^{-23}cm^3$
Volume of 208g of the element =$\large\frac{\text{Mass}}{\text{Density}}$
$\Rightarrow \large\frac{208g}{7.2gcm^{-3}}$
$\Rightarrow 28.89cm^3$
Number of unit cells =$\large\frac{\text{Total volume}}{\text{Volume of a unit cell}}$
$\Rightarrow \large\frac{28.89cm^3}{2.389\times 10^{-23}cm^3}$
$\Rightarrow 12.09\times 10^{23}$
For a bcc structure,number of atom per unit cell = 2
Number of atoms present in 208g=No. of atoms per unit cell $\times$ No. of unit cells
$\Rightarrow 2\times 12.09\times 10^{23}$
$\Rightarrow 24.18\times 10^{23}$
$\Rightarrow 2.418\times 10^{24}$
answered Aug 4, 2014 by sreemathi.v