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An element E crystallizes in body-centred cubic structure.If the edge centre of the cell is $1.469\times 10^{-10}$m and the density is $19.3gcm^{-3}$.Calculate the atomic mass of this element.Also calculate the radius of an atom of this element.

$\begin{array}{1 1}183.5g,0.634\times 10^{-8}cm\\173.5g,0.534\times 10^{-8}cm\\163.5g,0.434\times 10^{-8}cm\\143.5g,0.334\times 10^{-8}cm\end{array} $

1 Answer

The radius of an atom of this element $d=\large\frac{xM}{a^3N_A}$
Where
$d$=Density of the element
$z$=no. of atoms
$a=1.469\times 10^{-10}$ atom= $146.9\times 10^{-10}cm$
$N_A=6.023\times 10^{23}$
$M = \large\frac{da^3N_A}{z}$
$\;\;\;\;=\large\frac{19.3\times 146.9\times 10^{-10}\times 6.023 \times 10^{23}}{2}$
$\;\;\;\;=183.5g$
$r=\large\frac{\sqrt 3}{4}$$a$
$\;\;=\large\frac{\sqrt 3\times 1.469\times 10^{-8}}{4}$
$r=0.634\times 10^{-8}$cm
answered Aug 4, 2014 by sreemathi.v
 

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