An element has a bcc structure. If its edge l | vedclass

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

Question

(A) Given: $bcc$ structure,$Z = 2$,$a = 250 \, pm = 250 	imes 10^{-10} \, cm$,$d = 8 \, g/cm^3$,$N_A = 6.022 	imes 10^{23} \, mol^{-1}$.Using the fo

Vedclass · Accepted Answer

$M = 37.64 \, g/mol, r = 108.25 \, pm$

Vedclass · Answer

(A) Given: $bcc$ structure,$Z = 2$,$a = 250 \, pm = 250 	imes 10^{-10} \, cm$,$d = 8 \, g/cm^3$,$N_A = 6.022 	imes 10^{23} \, mol^{-1}$.Using the formula $d = \frac{Z 	imes M}{a^3 	imes N_A}$:$M = \frac{d 	imes a^3 	imes N_A}{Z} = \frac{8 	imes (250 	imes 10^{-10})^3 	imes 6.022 	imes 10^{23}}{2}$.$M = 4 	imes 15.625 	imes 10^{-24} 	imes 6.022 	imes 10^{23} = 37.6375 \approx 37.64 \, g/mol$.For $bcc$ structure,the relation between radius $r$ and edge length $a$ is $r = \frac{\sqrt{3}}{4} 	imes a$.$r = \frac{1.732}{4} 	imes 250 \, pm = 0.433 	imes 250 = 108.25 \, pm$.

An element has a $bcc$ structure. If its edge length is $250 \, pm$ and its density is $8 \, g/cm^3$,calculate its molar mass and the radius of the atom.

Similar Questions

The mass of an atom present in a unit cell is $4.4 \times 10^{-23} \ g$ and the product of density and volume of the unit cell is $1.792 \times 10^{-22} \ g$. What is the type of cubic unit cell?

Calculate the number of unit cells in $0.9 \ g$ of a metal if it forms a $bcc$ structure. Given: $\rho \times a^3 = 3 \times 10^{-22} \ g$.

If the radii of $Na^{+}$ and $Cl^{-}$ are $95 \, pm$ and $181 \, pm$ respectively, then the edge length of the $NaCl$ unit cell is equal to ...... $pm$.

$CsBr$ has $bcc$ structure with edge length $4.3 \ \mathring{A}$. The shortest interionic distance in between $Cs^{+}$ and $Br^{-}$ (in $\mathring{A}$) is :

The number of atoms in $100 \ g$ of an $fcc$ crystal with density $d = 10 \ g/cm^3$ and cell edge equal to $100 \ pm$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module