$CsBr$ crystallises in a body-centred cubic lattice. The unit cell length is $436.6 \, pm$. Given that the atomic mass of $Cs = 133$ and that of $Br = 80 \, amu$ and Avogadro number being $6.02 \times 10^{23} \, mol^{-1}$,the density of $CsBr$ is .............. $g/cm^{3}$.

  • A
    $4.25$
  • B
    $42.5$
  • C
    $0.425$
  • D
    $8.25$

Explore More

Similar Questions

Calculate the density of a metal having a unit cell volume of $64 \times 10^{-24} \ cm^3$ and a molar mass of $192 \ g \ mol^{-1}$,containing $4$ particles per unit cell. (in $g \ cm^{-3}$)

If an element having atomic number $96$ crystallises in a cubic lattice with a density of $10.3 \ g \ cm^{-3}$ and an edge length of $314 \ pm$, then the structure of the solid is:

An element has a density of $6.8 \ g \ cm^{-3}$ and crystallizes in a $bcc$ structure with a unit cell edge length of $290 \ pm$. The number of atoms in $200 \ g$ of the element is:

Difficult
View Solution

Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0 \ \mathring{A}$. Assuming the density of the oxide is $4.0 \ g \ cm^{-3}$,the number of $Fe^{2+}$ and $O^{2-}$ ions present in each unit cell will be:

In a body-centred cubic $(bcc)$ lattice of potassium, the correct relation between the atomic radius $(r)$ of potassium and the edge-length $(a)$ of the cube is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo