Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Solutions

The atomic radius of stronitum is $215\;pm$ and it crystallizes with a cubical closest - packed structure . Calculate the density of strontium. The molar mass of strontium is $ 87.6\;g\;mol^{-1}$

$\begin{array}{1 1} 2.586\;g\;cm^{-3} \\ 7 .34\;g\;cm^{-3} \\ 8\;g\;cm^{-3} \\ 9\;g\;cm^{-3} \end{array} $

1 Answer

Answer : $ 2.586\;g\;cm^{-3}$
The cubical closest - packed structure has a face- centered cubic unit cell.
In the latter atoms touch along the face diagonal of the cube. hence,
$\sqrt 2 a= 4r= 4(215\;pm)=860\;pm$
or $a= \large\frac{860\;pm}{\sqrt 2}$
$\qquad= 608.2 \;pm$
Now using the expression $\rho =\large\frac{N}{a^3} \bigg( \large\frac{M}{N_A}\bigg)$
We get, $\rho = \large\frac{4}{(6.023 \times 10^{-12} m)^3} \bigg( \large\frac{87.6\;g\;mol^{-1}}{6.023 \times 10^{23} mol^{-1}} \bigg)$
$\qquad= 2.586 \times 10^{6}g \;m^{-3}=2.586\;g\;cm^{-3}$
answered Aug 8, 2014 by meena.p

Related questions