A metal M crystallizes into two lattices: fac | vedclass

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

Question

(D) The density $d$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N_{A} 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The density $d$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N_{A} 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_{A}$ is Avogadro's number,and $a$ is the edge length.For $fcc$,$Z = 4$ and $a = 2.0 \ \mathring{A}$.For $bcc$,$Z = 2$ and $a = 2.5 \ \mathring{A}$.The ratio of densities is: $\frac{d_{fcc}}{d_{bcc}} = \frac{4 	imes M / (N_{A} 	imes 2.0^3)}{2 	imes M / (N_{A} 	imes 2.5^3)}$.$\frac{d_{fcc}}{d_{bcc}} = \frac{4}{8} 	imes \frac{15.625}{1} = 0.5 	imes 15.625 = 7.8125$.The nearest integer is $8$.

Similar Questions

What is the number of unit cells present in a cubic crystal lattice having $4$ atoms per unit cell and weighing $0.60 \ g$ (molar mass $60 \ g \ mol^{-1}$)?

Niobium crystallises in a body-centred cubic $(bcc)$ structure. If the density is $8.55 \, g \, cm^{-3}$,calculate the atomic radius of niobium using its atomic mass $93 \, u$.

Calculate the number of atoms in $1 \ g$ of a metal if it forms an $fcc$ crystal structure,given that $\varrho \times a^3 = 1.728 \times 10^{-22} \ g$.

An element crystallizes in an $f.c.c.$ structure with a unit cell edge length of $200 \, pm$. If $200 \, g$ of this element contains $24 \times 10^{23}$ atoms,calculate its density in $\text{g cm}^{-3}$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module