(A) For a $bcc$ unit cell, the number of atoms per unit cell $Z = 2$. The density formula is given by $d = \frac{Z \times M}{a^3 \times N_A}$. Here, $

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$6.022 \times 10^{23} \,mol^{-1}$

(A) For a $bcc$ unit cell, the number of atoms per unit cell $Z = 2$. The density formula is given by $d = \frac{Z \times M}{a^3 \times N_A}$. Here, $d = 7.874 \,g/cm^3$, $Z = 2$, $M = 55.845 \,g/mol$, and $a = 286.65 \,pm = 286.65 \times 10^{-10} \,cm$. Rearranging for $N_A$: $N_A = \frac{Z \times M}{d \times a^3}$. $N_A = \frac{2 \times 55.845}{7.874 \times (286.65 \times 10^{-10})^3}$. $N_A = \frac{111.69}{7.874 \times 23.556 \times 10^{-24}}$. $N_A \approx 6.022 \times 10^{23} \,mol^{-1}$.

Class 12 Chemistry Solid State Mathematical analysis of cubic system and Bragg’s equation

Easy

$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

A
$6.022 \times 10^{23} \,mol^{-1}$
B
$6.045 \times 10^{23} \,mol^{-1}$
C
$5.980 \times 10^{23} \,mol^{-1}$
D
$6.120 \times 10^{23} \,mol^{-1}$

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Mathematical analysis of cubic system and Bragg’s equation

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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 ........... $pm$.

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$A$ substance forms a face-centered cubic $(FCC)$ crystal. Its density is $1.984 \ g \ cm^{-3}$ and the edge length of the unit cell is $630 \ pm$. Calculate the molar mass of the substance in $g \ mol^{-1}$.

Difficult

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An element crystallizes in an $fcc$ unit cell structure with an edge length of $200 \, pm$. If $200 \, g$ of the element contains $24 \times 10^{23}$ atoms, what is its density in $g/cm^3$?

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Calculate the number of atoms in $1 \text{ g}$ of a metal that forms a $bcc$ crystal structure. Given that the product of density and unit cell volume is $\rho \times a^3 = 6.6 \times 10^{-22} \text{ g}$.

DifficultMHT CET 2026

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$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

Similar Questions

The mass of an atom present in a unit cell is $4.4 \times 10^{-23} \ g$ and the product of density and volume of the unit cell is $1.792 \times 10^{-22} \ g$. What is the type of cubic unit cell?

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 ........... $pm$.

$A$ substance forms a face-centered cubic $(FCC)$ crystal. Its density is $1.984 \ g \ cm^{-3}$ and the edge length of the unit cell is $630 \ pm$. Calculate the molar mass of the substance in $g \ mol^{-1}$.

An element crystallizes in an $fcc$ unit cell structure with an edge length of $200 \, pm$. If $200 \, g$ of the element contains $24 \times 10^{23}$ atoms, what is its density in $g/cm^3$?

Calculate the number of atoms in $1 \text{ g}$ of a metal that forms a $bcc$ crystal structure. Given that the product of density and unit cell volume is $\rho \times a^3 = 6.6 \times 10^{-22} \text{ g}$.

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