(B) The formula for density is $d = \frac{Z \times M}{a^3 \times N_A}$.For a $bcc$ structure, the number of atoms per unit cell $Z = 2$.The edge lengt

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(B) The formula for density is $d = \frac{Z \times M}{a^3 \times N_A}$.For a $bcc$ structure, the number of atoms per unit cell $Z = 2$.The edge length $a = 420 \ pm = 420 \times 10^{-10} \ cm = 4.20 \times 10^{-8} \ cm$.The Avogadro number $N_A = 6.022 \times 10^{23} \ mol^{-1}$.Substituting the values: $1 = \frac{2 \times M}{(4.20 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.$M = \frac{1 \times (74.088 \times 10^{-24}) \times 6.022 \times 10^{23}}{2}$.$M = \frac{44.61}{2} \approx 22.3 \ g \ mol^{-1}$.

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

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Calculate the molar mass of a metal with density $1 \ g \ cm^{-3}$ forming a $bcc$ structure with an edge length of $420 \ pm$.

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A
$32.2 \ g \ mol^{-1}$
B
$22.3 \ g \ mol^{-1}$
C
$25.5 \ g \ mol^{-1}$
D
$43.3 \ g \ mol^{-1}$

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Mathematical analysis of cubic system and Bragg’s equation

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Calculate the molar mass of a metal with density $1 \ g \ cm^{-3}$ forming a $bcc$ structure with an edge length of $420 \ pm$.

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$A$ face-centered cubic $(FCC)$ solid of an element (atomic mass $60$) has a cubic edge length of $4 \times 10^{-8} \, cm$. If Avogadro's number is $6 \times 10^{23} \, mol^{-1}$,then the density of the solid is:

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