An element (atomic mass = 100 g/mol) having | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a $bcc$ structure,the number of atoms per unit cell,$z = 2$. Given: Atomic mass $M = 100 \ g/mol$,edge length $a = 400 \ pm = 400 	imes 10^{-

Vedclass · Accepted Answer

$5.188$

Vedclass · Answer

(B) For a $bcc$ structure,the number of atoms per unit cell,$z = 2$. Given: Atomic mass $M = 100 \ g/mol$,edge length $a = 400 \ pm = 400 	imes 10^{-10} \ cm$,and Avogadro's number $N_A = 6.023 	imes 10^{23} \ mol^{-1}$. The formula for density is $ho = \frac{z 	imes M}{N_A 	imes a^3}$. Substituting the values: $ho = \frac{2 	imes 100}{6.023 	imes 10^{23} 	imes (400 	imes 10^{-10})^3}$. $ho = \frac{200}{6.023 	imes 10^{23} 	imes 64 	imes 10^{-24}} = \frac{200}{38.5472} \approx 5.188 \ g/cm^3$.

An element (atomic mass $= 100 \ g/mol$) having $bcc$ structure has a unit cell edge of $400 \ pm$. The density of the element is ................. $g/cm^3$.

Similar Questions

How many unit cells are present in a cube-shaped ideal crystal of $NaCl$ of mass $1.00 \ g$? [Atomic masses: $Na = 23, Cl = 35.5$]

How can the crystal form of ice be examined?

Copper (atomic mass $= 63.5$) crystallises in a $fcc$ lattice and has density $8.93 \, g \, cm^{-3}$. The radius of copper atom is closest to $.... \, pm$.

Calculate the edge length of a $bcc$ unit cell if the radius of the metal atom is $227 \ pm$.

Calculate the radius of a metal atom if it forms a $bcc$ unit cell having an edge length of $530 \ pm$. (in $pm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module