Ferrous oxide has a cubic structure and each | vedclass

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

Question

(A) The density formula for a unit cell is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Here,$d = 4.0 \, g \, cm^{-3}$,$M = 72 \, g \, mol^{-1}$,$a = 5.0 \

Vedclass · Accepted Answer

Four $Fe^{2+}$ and four $O^{2-}$

Vedclass · Answer

(A) The density formula for a unit cell is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Here,$d = 4.0 \, g \, cm^{-3}$,$M = 72 \, g \, mol^{-1}$,$a = 5.0 \, \mathring{A} = 5.0 	imes 10^{-8} \, cm$,and $N_A = 6.022 	imes 10^{23} \, mol^{-1}$.Rearranging for $Z$:$Z = \frac{d 	imes a^3 	imes N_A}{M}$$Z = \frac{4.0 	imes (5.0 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{72}$$Z = \frac{4.0 	imes 125 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{72}$$Z = \frac{500 	imes 0.6022}{72} = \frac{301.1}{72} \approx 4.18 \approx 4$.Since the formula is $FeO$,the ratio of $Fe^{2+}$ to $O^{2-}$ is $1:1$. Thus,there are $4$ $Fe^{2+}$ ions and $4$ $O^{2-}$ ions per unit cell.

Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0 \, \mathring{A}$. Assuming the density of the oxide is $4.0 \, g \, cm^{-3}$,the number of $Fe^{2+}$ and $O^{2-}$ ions present in each unit cell will be ($M_w$ of $FeO = 72$)

Similar Questions

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the number of atoms present in one unit cell of it is (Given : $N_{A}=6.023 \times 10^{23} \ mol^{-1}$)

$Au$ crystallizes in an $fcc$ lattice. If the edge length of the unit cell is $4.07 \ \mathring{A}$,then the distance between two nearest $Au$ atoms is ............... $\mathring{A}$.

If an element having atomic number $96$ crystallises in a cubic lattice with a density of $10.3 \ g \ cm^{-3}$ and an edge length of $314 \ pm$, then the structure of the solid is:

Sodium crystallizes in $bcc$ structure with radius $1.86 \times 10^{-8} \text{ cm}$. What is the edge length of unit cell of sodium?

$A$ metal crystallizes in a simple cubic lattice. The volume of one unit cell is $6.4 \times 10^7 \ pm^3$. What is the radius of the metal atom in $pm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module