Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

$CsBr$ crystallises in a body centered cubic lattice. The unit cell length is $436.6\; pm$ . Given that the atomic mass of $Cs = 133$ and of $Br = 80\; amu$. Avogadro's number being $6.02\times10^{23}mol^{-1}$ The density of $CsBr$ is


Can you answer this question?

1 Answer

0 votes
Density = $\large\frac{Z\times mol.wt}{a^3\times N_A}$
$\therefore$ Z = 2 for b.c.c and mol.wt of CsBr = 213
$\rho = \large\frac{2\times 213}{(4.366\times10^{-8})^3\times6.032\times10^{23}}$
$= 8.50 g/cm^3$
Hence answer is (a)
answered Feb 27, 2014 by sharmaaparna1
edited Jun 22 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App