How many atoms of niobium are present in 2.43 | 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

$2.0 	imes 10^{22}$

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 $a^3$ is the volume of the unit cell $(V)$.Alternatively,$d = \frac{Z 	imes 	ext{mass of one atom}}{V}$.For a $bcc$ structure,the number of atoms per unit cell is $Z = 2$.Given: $d = 9 \ g \ cm^{-3}$,$V = 2.7 	imes 10^{-23} \ cm^3$,and total mass $= 2.43 \ g$.Let $N$ be the total number of atoms in $2.43 \ g$.The mass of one atom is $\frac{2.43}{N}$.Substituting the values into the density formula: $9 = \frac{2 	imes (2.43 / N)}{2.7 	imes 10^{-23}}$.Rearranging for $N$: $N = \frac{2 	imes 2.43}{9 	imes 2.7 	imes 10^{-23}}$.$N = \frac{4.86}{24.3 	imes 10^{-23}} = 0.2 	imes 10^{23} = 2.0 	imes 10^{22}$ atoms.

How many atoms of niobium are present in $2.43 \ g$ if it forms $bcc$ structure with density $9 \ g \ cm^{-3}$ and volume of unit cell $2.7 \times 10^{-23} \ cm^3$?

Similar Questions

An element (atomic weight $= 250 \ u$) crystallises in a simple cubic lattice. If the density of the unit cell is $7.2 \ g \ cm^{-3}$,what is the radius (in $\mathring{A}$) of the atom of the element? $(N_A = 6.02 \times 10^{23} \ mol^{-1})$

An element crystallising in $fcc$ lattice has a density of $8.92 \ g \ cm^{-3}$ and edge length of $3.61 \times 10^{-8} \ cm$. What is the atomic weight of the element (in $u$)? $(N_A = 6.022 \times 10^{23} \ mol^{-1})$

The substance $CuCl$ has a $ZnS$ $(ccp)$ structure. If its density is $3.4 \ g \ cm^{-3}$, what is the edge length of the unit cell?

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 $93 \, u$.

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 $\text{g cm}^{-3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module