(A) For a $CCP$ lattice,the number of atoms per unit cell is $z = 4$.The formula for density is $d = \frac{z \times M}{N_{A} \times a^{3}}$.Given: $d

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(A) For a $CCP$ lattice,the number of atoms per unit cell is $z = 4$.The formula for density is $d = \frac{z \times M}{N_{A} \times a^{3}}$.Given: $d = 7.62 \ g \ cm^{-3}$,$a = 0.4518 \ nm = 0.4518 \times 10^{-7} \ cm$,$N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.Rearranging for molar mass $M$: $M = \frac{d \times N_{A} \times a^{3}}{z}$.$M = \frac{7.62 \times 6.022 \times 10^{23} \times (0.4518 \times 10^{-7})^{3}}{4}$.$M = \frac{7.62 \times 6.022 \times 10^{23} \times 9.223 \times 10^{-23}}{4} \approx 105.8 \ g \ mol^{-1}$.Rounding to the nearest integer,we get $106 \ g \ mol^{-1}$.

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

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$A$ copper complex crystallising in a $CCP$ lattice with a cell edge of $0.4518 \ nm$ has been revealed by employing $X$-ray diffraction studies. The density of the copper complex is found to be $7.62 \ g \ cm^{-3}$. The molar mass of the copper complex is $..... \ g \ mol^{-1}$. (Nearest integer)
[Given : $N_{A} = 6.022 \times 10^{23} \ mol^{-1}$]

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A
$106$
B
$560$
C
$780$
D
$230$

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Mathematical analysis of cubic system and Bragg’s equation

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Element $X$ crystallizes in a $12$ coordination face-centered cubic $(fcc)$ lattice. On applying high temperature,it changes to an $8$ coordination body-centered cubic $(bcc)$ lattice. Find the ratio of the density of the crystal lattice before and after applying high temperature. The atomic radius of $X$ is the same in both crystals.

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