An element crystallizes in a body-centred cub | vedclass

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

Question

(D) For a body-centred cubic $(BCC)$ lattice, the number of atoms per unit cell is $Z = 2$. The density formula is given by $ho = \frac{Z 	imes M}{

Vedclass · Accepted Answer

$5.0 	imes 10^{24}$

Vedclass · Answer

(D) For a body-centred cubic $(BCC)$ lattice, the number of atoms per unit cell is $Z = 2$. The density formula is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$. Given: $ho = 5.0 \ g \ cm^{-3}$, $a = 200 \ pm = 200 	imes 10^{-10} \ cm = 2 	imes 10^{-8} \ cm$. $5.0 = \frac{2 	imes M}{6.022 	imes 10^{23} 	imes (2 	imes 10^{-8})^3}$. $M = \frac{5.0 	imes 6.022 	imes 10^{23} 	imes 8 	imes 10^{-24}}{2} = 12.044 \ g \ mol^{-1}$. Number of moles in $100 \ g = \frac{100}{M} = \frac{100}{12.044} \approx 8.303 \ mol$. Number of atoms $= 	ext{moles} 	imes N_A = \frac{100}{M} 	imes N_A$. Substituting $M = \frac{5.0 	imes N_A 	imes 8 	imes 10^{-24}}{2}$: Number of atoms $= \frac{100 	imes 2}{5.0 	imes 8 	imes 10^{-24}} = \frac{200}{40 	imes 10^{-24}} = 5 	imes 10^{24}$.

An element crystallizes in a body-centred cubic lattice. The edge length of the unit cell is $200 \ pm$ and the density of the element is $5.0 \ g \ cm^{-3}$. Calculate the number of atoms in $100 \ g$ of this element.

Similar Questions

Calculate the volume of the unit cell for an element having a molar mass of $92 \ g \ mol^{-1}$ that forms a $bcc$ structure,given $\left[\varrho \times N_{A} = 5.0 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$.

$X$-rays of wavelength $1 \ \mathring{A}$ are diffracted by the planes of particles in a metal. The second-order diffraction occurs at an angle of $60^{\circ}$. Calculate the distance between two parallel planes in $\mathring{A}$.

Aluminium crystallises in a cubic close-packed structure. Its metallic radius is $125 \ pm$.
$(i)$ What is the length of the side of the unit cell?
$(ii)$ How many unit cells are there in $1.00 \ cm^3$ of aluminium?

$A$ metal (atomic mass $= 50$) has a body-centered cubic $(bcc)$ crystal structure. If the density of the metal is $5.96 \, g \, cm^{-3}$,then the volume of the unit cell is ............ $\times 10^{-24} \, cm^3$.

An element has an $fcc$ structure. If its edge length is $200 \, pm$, calculate the density of this element having a mass of $200 \, g$. $[200 \, g$ of the element contains $24 \times 10^{23}$ atoms.$]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module