(B) Number of moles $= \frac{0.6 \ g}{60 \ g \ mol^{-1}} = 0.01 \ mol$ Total number of atoms $= 0.01 \times 6.022 \times 10^{23} = 6.022 \times 10^{21

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(B) Number of moles $= \frac{0.6 \ g}{60 \ g \ mol^{-1}} = 0.01 \ mol$ Total number of atoms $= 0.01 \times 6.022 \times 10^{23} = 6.022 \times 10^{21} \ \text{atoms}$ Number of atoms per unit cell $= 4$ Number of unit cells $= \frac{\text{Total number of atoms}}{\text{Atoms per unit cell}} = \frac{6.022 \times 10^{21}}{4} \approx 1.5 \times 10^{21} \ \text{unit cells}$

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

Medium

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 $60 \ g \ mol^{-1}$)?

MHT CET 2023 All MHT CET

A
$1 \times 10^{21}$
B
$1.5 \times 10^{21}$
C
$3.0 \times 10^{21}$
D
$6.0 \times 10^{21}$

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Mathematical analysis of cubic system and Bragg’s equation

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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 $60 \ g \ mol^{-1}$)?

Similar Questions

$A$ metal has a $fcc$ structure. If the edge length of the unit cell is $200 \text{ pm}$, calculate the volume occupied by the particles in a unit cell.

In $CsCl$ type structure, if the radius of the cation and anion are $80 \ pm$ and $100 \ pm$ respectively, then the closest distance between two cations is:

$A$ metal crystallizes in both $fcc$ and $bcc$ lattice structures with unit cell edge lengths of $380 \ pm$ and $300 \ pm$ respectively. The ratio of their densities $(bcc/fcc)$ is:

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