An element crystallises in a bcc type of unit | vedclass

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

Question

(D) For a $bcc$ unit cell, the number of atoms per unit cell $n = 2$. Given density $ho = 4 \ g \ cm^{-3}$ and edge length $a = 500 \ pm = 500 	ime

Vedclass · Accepted Answer

$150.5$

Vedclass · Answer

(D) For a $bcc$ unit cell, the number of atoms per unit cell $n = 2$. Given density $ho = 4 \ g \ cm^{-3}$ and edge length $a = 500 \ pm = 500 	imes 10^{-10} \ cm = 5 	imes 10^{-8} \ cm$. The volume of the unit cell $a^{3} = (5 	imes 10^{-8} \ cm)^{3} = 125 	imes 10^{-24} \ cm^{3}$. The formula for atomic mass $M$ is $M = \frac{ho 	imes a^{3} 	imes N_{A}}{n}$. Substituting the values: $M = \frac{4 	imes 125 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{2} = 2 	imes 125 	imes 0.6022 = 150.55 \ g \ mol^{-1}$. Thus, the atomic mass is approximately $150.5 \ g \ mol^{-1}$.

An element crystallises in a $bcc$ type of unit cell. The density and edge length of the unit cell are $4 \ g \ cm^{-3}$ and $500 \ pm$ respectively. What is the atomic mass of the element?

Similar Questions

Sodium metal crystallizes as a body-centered cubic $(bcc)$ lattice with the cell edge $4.29 \ \mathring{A}$. What is the radius of a sodium atom?

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 $pm$)

Calculate the volume of the unit cell when a metal having a density of $1 \ g \ cm^{-3}$ and a molar mass of $23 \ g \ mol^{-1}$ crystallizes to form a $bcc$ structure.

Which is the volume of unit cell of a metal (atomic mass $25 \ g \ mol^{-1}$) having $BCC$ structure and density $3 \ g \ cm^{-3}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module