(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

Sodium metal crystallizes as a body-centered cubic $(bcc)$ lattice with the cell edge $4.29 \ \mathring{A}$. What is the radius of a sodium atom?

$CsBr$ has $bcc$ structure with edge length $4.3 \ \mathring{A}$. The shortest interionic distance in between $Cs^{+}$ and $Br^{-}$ (in $\mathring{A}$) is :

$A$ metal $(X)$ of atomic weight $M \ g \ mol^{-1}$ crystallizes in a $bcc$ lattice. Its density is $d \ g \ cm^{-3}$. What is the equation for the unit cell edge length $(a)$? ($N=$ Avogadro number)

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