$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

  • A
    $6.022 \times 10^{23} \,mol^{-1}$
  • B
    $6.045 \times 10^{23} \,mol^{-1}$
  • C
    $5.980 \times 10^{23} \,mol^{-1}$
  • D
    $6.120 \times 10^{23} \,mol^{-1}$

Explore More

Similar Questions

Find the molar mass of an element that crystallizes forming a unit cell structure having an edge length of $4 \times 10^{-8} \ cm$ and containing $4$ particles. (Density $\rho = 19.7 \ g \ cm^{-3}$)

$A$ substance has a density of $2 \ g \ cm^{-3}$. It crystallizes in the $fcc$ crystal with an edge length of $600 \ pm$. The molar mass of the substance (in $g \ mol^{-1}$) is
$(N_{A} = 6 \times 10^{23} \ mol^{-1})$

Copper crystallises in $fcc$ structure with a unit cell length of $361 \ pm$. What is the radius of copper atom? ............ $pm$

Calculate the number of atoms in $0.3 \ g$ of a metal if it forms a $bcc$ structure,given that $[\rho \times a^3 = 3 \times 10^{-22} \ g]$.

The distance between $Na^{+}$ and $Cl^{-}$ ions in solid $NaCl$ of density $2.165 \ g \ cm^{-3}$ is $........ \times 10^{-10} \ m$. (Nearest Integer)
(Given : $N_{A} = 6.02 \times 10^{23} \ mol^{-1}$,Molar mass of $NaCl = 58.5 \ g \ mol^{-1}$)

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