An element has a bcc structure with an atomic | vedclass

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

Question

(A) For a $bcc$ unit cell, the number of atoms per unit cell $(n)$ is $2$.Atomic mass $(M)$ = $50 \ 	ext{g/mol}$.Edge length $(a)$ = $290 \ pm = 290

Vedclass · Accepted Answer

$6.81$

Vedclass · Answer

(A) For a $bcc$ unit cell, the number of atoms per unit cell $(n)$ is $2$.Atomic mass $(M)$ = $50 \ 	ext{g/mol}$.Edge length $(a)$ = $290 \ pm = 290 	imes 10^{-10} \ cm = 2.9 	imes 10^{-8} \ cm$.Avogadro's number $(N_A)$ = $6.022 	imes 10^{23} \ 	ext{mol}^{-1}$.The formula for density $(d)$ is: $d = \frac{n 	imes M}{a^3 	imes N_A}$.Substituting the values: $d = \frac{2 	imes 50}{(2.9 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}$.$d = \frac{100}{24.389 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}$.$d = \frac{100}{14.687} \approx 6.81 \ 	ext{g/cm}^{-3}$.

An element has a $bcc$ structure with an atomic mass of $50$. If the edge length of the unit cell is $290 \ pm$, calculate the density of the unit cell in $\text{g/cm}^{-3}$.

Similar Questions

The number of atoms in $2.4 \ g$ of body-centred cubic $(BCC)$ crystal with edge length $200 \ pm$ is (density = $10 \ g \ cm^{-3}$,$N_A = 6 \times 10^{23} \ atoms \ mol^{-1}$)

Calculate the volume of a simple cubic unit cell if the radius of the particle in it is $400 \ pm$.

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 volume of a unit cell having four particles in it with a density of $19.0 \ g \ cm^{-3}$ [molar mass of element $= 190 \ g \ mol^{-1}$].

Calculate the volume of the unit cell for an element having a molar mass of $56 \ g \ mol^{-1}$ that forms $bcc$ unit cells. $\left[\rho \cdot N_{A} = 4.8 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module