Calculate the number of atoms per unit cell of an element having a molar mass of $92.0 \ g \ mol^{-1}$ and a density of $8.6 \ g \ cm^{-3}$ forming a cubic unit cell structure. Given $[a^3 \times N_{A} = 21.5 \ cm^3 \ mol^{-1}]$.
- A$1$
- B$2$
- C$3$
- D$4$
Explore More
Similar Questions
The density of $Li$ metal is $0.53 \ g/cm^3$ and the edge length of its unit cell is $3.5 \ \mathop A\limits^o$. Calculate the number of $Li$ atoms in the unit cell. $(N_A = 6.02 \times 10^{23} \ mol^{-1}, M = 6.94 \ g \ mol^{-1})$
Medium
View Solution$NaCl$ has a face-centered cubic structure. The edge length of the unit cell is $0.564 \ nm$. What is the density of sodium chloride? $.............. \ g/cm^3$ [$1 \ nm = 10^{-7} \ cm$]
Difficult
View SolutionAt $T \ K$,copper (atomic mass $= 63.5 \ u$) has $fcc$ unit cell structure with edge length of $x \ \mathring{A}$. What is the approximate density of $Cu$ in $g \ cm^{-3}$ at that temperature? $(N_A = 6.0 \times 10^{23} \ mol^{-1})$
Niobium crystallizes in a $bcc$ structure. If its density is $8.55 \, g/cm^3$, calculate the atomic radius of niobium. [Atomic mass of $Nb = 93 \, u$] (in $pm$)
Easy
View SolutionCalculate the density of an element having molar mass $27 \ g \ mol^{-1}$ that forms $fcc$ unit cell. $[a^3 \cdot N_A = 38.5 \ cm^3 \ mol^{-1}]$ (in $g \ cm^{-3}$)
Vedclass 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