A metal has a fcc lattice. The edge length of | vedclass

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

Question

(D) The formula for density is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.For an $fcc$ lattice, the number of atoms per unit cell $Z = 4$.Given edge l

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(D) The formula for density is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.For an $fcc$ lattice, the number of atoms per unit cell $Z = 4$.Given edge length $a = 404 \, pm = 404 	imes 10^{-10} \, cm$.Given density $ho = 2.72 \, g \, cm^{-3}$ and $N_A = 6.02 	imes 10^{23} \, mol^{-1}$.Substituting the values: $2.72 = \frac{4 	imes M}{6.02 	imes 10^{23} 	imes (404 	imes 10^{-10})^3}$.Solving for $M$: $M = \frac{2.72 	imes 6.02 	imes 10^{23} 	imes 66.0 	imes 10^{-24}}{4} \approx 27 \, g \, mol^{-1}$.

$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is :- ............ $g \, mol^{-1}$ ( $N_A$ Avogadro constant $= 6.02 \times 10^{23} \, mol^{-1}$ )

Similar Questions

Calculate the edge length of a simple cubic unit cell if the radius of an atom is $167.3 \ pm$. (in $pm$)

If the edge length of a cubic unit cell in a $bcc$ structure is $400 \, pm$, then the atomic radius of the metal will be ........... $pm$.

Calculate the number of unit cells in $10 \, g$ of $CsCl$ which crystallizes in a $bcc$ structure. $(CsCl = 168.5 \, amu)$

The number of atoms in $4.5 \ g$ of a face-centred cubic crystal with edge length $300 \ pm$ is (Given: Density $= 10 \ g \ cm^{-3}$ and $N_A = 6.022 \times 10^{23}$)

$A$ metal crystallises in two cubic phases, $fcc$ and $bcc$ with edge lengths $3.5 \ \mathring{A}$ and $3 \ \mathring{A}$ respectively. The ratio of densities of $fcc$ and $bcc$ is approximately

Vedclass Test Series

Exam Paper Generator

Online Exam Module