$A$ metal $(X)$ of atomic weight $M \ g \ mol^{-1}$ crystallizes in a $bcc$ lattice. Its density is $d \ g \ cm^{-3}$. What is the equation for the unit cell edge length $(a)$? ($N=$ Avogadro number)
- A$a = \left( \frac{2M}{Nd} \right)^{\frac{1}{3}}$
- B$a = \left( \frac{2M}{Nd} \right)^{\frac{1}{2}}$
- C$a = \left( \frac{4M}{Nd} \right)^{\frac{1}{3}}$
- D$a = \left( \frac{M}{Nd} \right)^{\frac{1}{3}}$
Explore More
Similar Questions
$KF$ has a structure similar to $NaCl$. If the density of the crystal is $2.48 \, g/cm^3$, the distance between $K^+$ and $F^-$ ions will be ............. $pm$. (Atomic masses: $K = 39 \, amu, F = 19 \, amu$)
Medium
View SolutionAn element (atomic mass $100 \ g/mol$) having $bcc$ structure has unit cell edge $400 \ pm$. Then density of the element is (in $g/cm^3$)
MediumAIIMS 2002
View Solution$A$ solid having density of $9 \times 10^3 \ kg \ m^{-3}$ forms face-centred cubic crystals of edge length $200 \sqrt{2} \ pm$. What is the molar mass of the solid? [Avogadro constant $\cong 6 \times 10^{23} \ mol^{-1}$,$\pi \cong 3$]
DifficultJEE MAIN 2019
View SolutionWhat is the number of unit cells present in a cubic crystal lattice having $4$ atoms per unit cell and weighing $0.60 \ g$ (molar mass $60 \ g \ mol^{-1}$)?
$KF$ has a $NaCl$ type structure. What is the distance between $K^+$ and $F^-$ in $KF$ if the density is $2.48 \ g \ cm^{-3}$?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo