Mathematical analysis of cubic system and Bragg’s equation Questions in English

(C) For $BCC$ unit cell,the relation between radius $r$ and edge length $a$ is $r = \frac{\sqrt{3}}{4} a$.
The volume of one spherical particle is given by $V = \frac{4}{3} \pi r^3$.
Substituting the value of $r$:
$V = \frac{4}{3} \pi \left( \frac{\sqrt{3}}{4} a \right)^3 = \frac{4}{3} \pi \left( \frac{3 \sqrt{3} a^3}{64} \right) = \frac{\sqrt{3} \pi a^3}{16}$.

(B) The density formula is given by $\rho = \frac{n \times M}{a^3 \times N_A}$.
Rearranging for the number of atoms per unit cell $(n)$: $n = \frac{\rho \times N_A \times a^3}{M}$.
Given: $\rho = 4 \ g \ cm^{-3}$,$M = 149 \ g \ mol^{-1}$,$a = 5 \mathring{A} = 5 \times 10^{-8} \ cm$,and $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $n = \frac{4 \times 6.022 \times 10^{23} \times (5 \times 10^{-8})^3}{149}$.
$n = \frac{4 \times 6.022 \times 10^{23} \times 125 \times 10^{-24}}{149} = \frac{301.1}{149} \approx 2.02$.
Since $n \approx 2$,the crystal structure is Body-centred cubic $(BCC)$.

(C) The mass of the unit cell is given by the product of density $(\rho)$ and volume $(a^3)$,which is $\rho \times a^3 = 3 \times 10^{-22} \ g$.
Number of unit cells = $\frac{\text{Total mass of metal}}{\text{Mass of one unit cell}}$
Number of unit cells = $\frac{0.9 \ g}{3 \times 10^{-22} \ g}$
Number of unit cells = $0.3 \times 10^{22} = 3.0 \times 10^{21}$.

(C) The mass of the metal is given as $m = 0.4 \ g$.
The product of density $(\rho)$ and volume of the unit cell $(V = a^3)$ is given as $\rho \times a^3 = 1.2 \times 10^{-22} \ g$.
The number of unit cells is calculated by dividing the total mass of the metal by the mass of one unit cell.
$\text{Number of unit cells} = \frac{\text{Total mass}}{\text{Mass of one unit cell}} = \frac{m}{\rho \times a^3}$.
Substituting the values: $\frac{0.4 \ g}{1.2 \times 10^{-22} \ g} = 3.33 \times 10^{21}$.
Therefore,the number of unit cells is $3.3 \times 10^{21}$.

(A) For a simple cubic unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $a = 2r$.
Therefore, $r = \frac{a}{2}$.
Given $a = 334.7 \ pm$.
$r = \frac{334.7 \ pm}{2} = 167.35 \ pm$.

(B) The density formula for a unit cell is given by $\rho = \frac{M \cdot n}{a^3 \cdot N_A}$.
For an $fcc$ unit cell,the number of atoms per unit cell $(n)$ is $4$.
Given: $\rho = 21 \ g \ cm^{-3}$ and $a^3 \cdot N_A = 36 \ cm^3 \ mol^{-1}$.
Substituting the values: $21 = \frac{M \times 4}{36}$.
Solving for $M$: $M = \frac{21 \times 36}{4} = 21 \times 9 = 189.00 \ g \ mol^{-1}$.

(C) For an $fcc$ unit cell,the number of atoms per unit cell,$n = 4$.
The formula for density is $\rho = \frac{M \times n}{a^3 \cdot N_A}$.
Given $\rho = 2.7 \ g \ cm^{-3}$ and $a^3 \cdot N_A = 40 \ cm^3 \ mol^{-1}$.
Substituting the values: $2.7 = \frac{M \times 4}{40}$.
Solving for $M$: $M = \frac{2.7 \times 40}{4} = 27 \ g \ mol^{-1}$.

(B) For a $bcc$ unit cell, the relationship between radius $r$ and edge length $a$ is given by $r = \frac{\sqrt{3}}{4} a$.
Therefore, $a = \frac{4r}{\sqrt{3}}$.
Substituting the given value $r = 173 \ pm$:
$a = \frac{4 \times 173}{1.732} \approx 400 \ pm$.
Converting to centimeters:
$a = 400 \times 10^{-10} \ cm = 4.00 \times 10^{-8} \ cm$.

(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 $= \frac{\text{Total number of atoms}}{\text{Atoms per unit cell}} = \frac{6.022 \times 10^{21}}{4} \approx 1.5 \times 10^{21} \ \text{unit cells}$

(C) For a $BCC$ unit cell, the number of atoms per unit cell is $n = 2$.
Density formula: $\rho = \frac{M \times n}{a^3 \times N_A}$
Rearranging for molar mass $M$: $M = \frac{\rho \times a^3 \times N_A}{n}$
Given: $\rho = 10 \ g \ cm^{-3}$, $a = 300 \ pm = 3 \times 10^{-8} \ cm$, $N_A = 6.022 \times 10^{23} \ mol^{-1}$, $n = 2$.
$M = \frac{10 \times (3 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2}$
$M = \frac{10 \times 27 \times 10^{-24} \times 6.022 \times 10^{23}}{2}$
$M = \frac{162.594}{2} \approx 81.3 \ g \ mol^{-1}$.

(D) In a $bcc$ (body-centered cubic) unit cell,the atoms touch each other along the body diagonal.
The length of the body diagonal is given by $\sqrt{3} a$.
Since the body diagonal consists of two radii of the corner atoms and one full diameter of the central atom,we have $\sqrt{3} a = 4r$.
Rearranging this for the edge length $a$,we get $a = \frac{4r}{\sqrt{3}}$.

(C) For $fcc$ crystal structure, the relationship between radius $r$ and edge length $a$ is given by $r = \frac{a}{2\sqrt{2}} = \frac{\sqrt{2}}{4} a$.
Given $a = 405 \ pm$.
Substituting the value: $r = \frac{1.414 \times 405}{4} = \frac{572.67}{4} = 143.1675 \ pm \approx 143.2 \ pm$.

(A) For a simple cubic unit cell,the number of atoms per unit cell is $n = 1$.
The formula for density $(\rho)$ is given by $\rho = \frac{M \cdot n}{a^3 \cdot N_A}$.
Substituting the given values: $9.3 = \frac{M \cdot 1}{22.6}$.
Solving for molar mass $M$: $M = 9.3 \times 22.6 = 210.2 \ g \ mol^{-1}$.

(A) For a $bcc$ unit cell,the number of atoms per unit cell is $n = 2$.
The formula for density is $\rho = \frac{M \cdot n}{a^3 \cdot N_{A}}$.
Given $\rho = 7.8 \ g \ cm^{-3}$ and $a^3 \cdot N_{A} = 16.2 \ cm^3 \ mol^{-1}$.
Substituting the values: $7.8 = \frac{M \times 2}{16.2}$.
Solving for $M$: $M = \frac{7.8 \times 16.2}{2} = 63.18 \ g \ mol^{-1}$.

(B) In a simple cubic unit cell, the atoms touch each other along the edge of the cube.
Therefore, the relationship between the edge length $a$ and the atomic radius $r$ is given by $a = 2r$.
Given $r = 167.3 \ pm$.
Thus, $a = 2 \times 167.3 \ pm = 334.6 \ pm$.

(A) The density $\rho$ is given by the formula: $\rho = \frac{M \times Z}{a^3 \cdot N_A}$
For an $fcc$ unit cell,the number of atoms per unit cell,$Z = 4$.
Given: $M = 27 \ g \ mol^{-1}$ and $a^3 \cdot N_A = 38.5 \ cm^3 \ mol^{-1}$.
Substituting the values: $\rho = \frac{27 \ g \ mol^{-1} \times 4}{38.5 \ cm^3 \ mol^{-1}}$
$\rho = \frac{108}{38.5} \ g \ cm^{-3} \approx 2.8 \ g \ cm^{-3}$.

(A) For a $bcc$ unit cell, the relationship between the radius $r$ and the edge length $a$ is given by $4r = \sqrt{3}a$.
Substituting the given value of $a = 287 \ pm$:
$r = \frac{\sqrt{3}}{4} \times 287$
$r = \frac{1.732 \times 287}{4} = 124.27 \ pm$.

(B) For a $bcc$ unit cell, the relationship between the radius of the atom $(r)$ and the edge length $(a)$ is given by $4r = \sqrt{3}a$.
Substituting the given edge length $a = 450 \ pm$:
$r = \frac{\sqrt{3} \times 450}{4} = \frac{1.732 \times 450}{4} = 194.85 \ pm$.

(A) For an $fcc$ unit cell,the number of atoms per unit cell $(n)$ is $4$.
The density formula is given by $\rho = \frac{M \times n}{a^3 \times N_A}$.
Rearranging for the volume of the unit cell $(a^3)$,we get $a^3 = \frac{M \times n}{\rho \times N_A}$.
Substituting the given values: $M = 180 \ g \ mol^{-1}$,$n = 4$,and $\rho \cdot N_A = 120 \times 10^{21} \ g \ cm^{-3} \ mol^{-1}$.
$a^3 = \frac{180 \times 4}{120 \times 10^{21}} = \frac{720}{120 \times 10^{21}} = 6.00 \times 10^{-21} \ cm^3$.

(D) The density of a unit cell is given by $\rho = \frac{M \cdot n}{a^3 \cdot N_A}$.
Here,$M = 56 \ g \ mol^{-1}$,$n = 2$ (for $bcc$ structure),and $\rho \cdot N_A = 4.8 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}$.
The volume of the unit cell is $V = a^3 = \frac{M \cdot n}{\rho \cdot N_A}$.
Substituting the values: $V = \frac{56 \times 2}{4.8 \times 10^{24}} \ cm^3$.
$V = \frac{112}{4.8} \times 10^{-24} \ cm^3 = 23.33 \times 10^{-24} \ cm^3 = 2.33 \times 10^{-23} \ cm^3$.

(A) For a simple cubic unit cell,the number of atoms per unit cell $(n)$ is $1$.
Density $(\rho)$ is given by the formula: $\rho = \frac{n \cdot M}{a^3 \cdot N_{A}}$.
Substituting the given values: $\rho = \frac{1 \times 210}{21.5} \ g \ cm^{-3}$.
$\rho = 9.77 \ g \ cm^{-3}$.

(B) For a $bcc$ unit cell, the relationship between the radius $r$ and the edge length $a$ is given by $r = \frac{\sqrt{3}}{4} a$.
Therefore, $a = \frac{4r}{\sqrt{3}}$.
Substituting the value $r = 227 \ pm$:
$a = \frac{4 \times 227}{1.732} \approx 524.83 \ pm$.
Converting $pm$ to $cm$: $1 \ pm = 10^{-10} \ cm$.
$a = 524.83 \times 10^{-10} \ cm = 5.2483 \times 10^{-8} \ cm$.
Rounding to two decimal places, we get $5.24 \times 10^{-8} \ cm$.

(D) The density formula for a unit cell is given by $d = \frac{Z \times M}{a^3 \times N_A}$.
Here,$d = 22.24 \ g \ cm^{-3}$,$Z = 4$,$a^3 = 5.6 \times 10^{-23} \ cm^3$,and $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $22.24 = \frac{4 \times M}{5.6 \times 10^{-23} \times 6.022 \times 10^{23}}$.
$22.24 = \frac{4 \times M}{5.6 \times 6.022}$.
$M = \frac{22.24 \times 5.6 \times 6.022}{4}$.
$M = 187.43 \ g \ mol^{-1}$.

(B) For a simple cubic structure, the number of atoms per unit cell, $Z = 1$.
The density formula is $d = \frac{Z \times M}{a^3 \times N_A}$.
Given: $d = 9.8 \ g \ cm^{-3}$, $a = 340 \ pm = 3.40 \times 10^{-8} \ cm$, $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting values: $9.8 = \frac{1 \times M}{(3.40 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = 9.8 \times (3.9304 \times 10^{-23}) \times 6.022 \times 10^{23} \approx 231.97 \ g \ mol^{-1}$.
Number of moles in $20 \ g = \frac{20}{231.97} \approx 0.08622 \ mol$.
Number of atoms $= \text{moles} \times N_A = 0.08622 \times 6.022 \times 10^{23} \approx 5.19 \times 10^{22}$ atoms.

(D) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{a^3 \times N_{A}}$
For a $bcc$ structure, the number of atoms per unit cell $(Z)$ is $2$.
The molar mass $(M)$ is $56 \ g \ mol^{-1}$.
The edge length $(a)$ is $288 \ pm = 288 \times 10^{-10} \ cm$.
Avogadro's number $(N_{A})$ is $6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values:
$d = \frac{2 \times 56}{(288 \times 10^{-10} \ cm)^3 \times 6.022 \times 10^{23} \ mol^{-1}}$
$d = \frac{112}{2.3887 \times 10^{-23} \times 6.022 \times 10^{23}}$
$d = \frac{112}{14.385} \approx 7.78 \ g \ cm^{-3}$

(B) The formula for density of a unit cell is given by: $d = \frac{Z \times M}{a^3 \times N_{A}}$
Given: $d = 19.7 \ g \ cm^{-3}$,$Z = 4$,$a = 4 \times 10^{-8} \ cm$,$N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values:
$19.7 = \frac{4 \times M}{(4 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$
$19.7 = \frac{4 \times M}{64 \times 10^{-24} \times 6.022 \times 10^{23}}$
$19.7 = \frac{4 \times M}{38.5408}$
$M = \frac{19.7 \times 38.5408}{4} \approx 189.8 \ g \ mol^{-1}$.

(A) For a simple cubic structure, the number of atoms per unit cell, $Z = 1$.
The edge length $a = 336 \ pm = 3.36 \times 10^{-8} \ cm$.
The molar mass $M$ can be calculated from the given data: $2.64 \times 10^{23}$ atoms have a mass of $90 \ g$. Therefore, $6.022 \times 10^{23}$ atoms (Avogadro's number, $N_A$) have a mass $M = \frac{90 \times 6.022 \times 10^{23}}{2.64 \times 10^{23}} \approx 205.3 \ g \ mol^{-1}$.
Using the density formula $d = \frac{Z \times M}{a^3 \times N_A}$:
$d = \frac{1 \times 205.3}{(3.36 \times 10^{-8})^3 \times 6.022 \times 10^{23}} \approx 8.98 \ g \ cm^{-3}$.

(A) For a simple cubic unit cell, the number of atoms per unit cell, $Z = 1$.
The volume of one unit cell, $V = a^3 = (336 \times 10^{-10} \ cm)^3 = 3.793 \times 10^{-23} \ cm^3$.
The mass of one unit cell is given by $m_{uc} = \text{Density} \times \text{Volume} = 9.4 \ g \ cm^{-3} \times 3.793 \times 10^{-23} \ cm^3 = 3.565 \times 10^{-22} \ g$.
The number of unit cells in $3 \ g$ of metal is calculated as $\frac{\text{Total mass}}{\text{Mass of one unit cell}} = \frac{3 \ g}{3.565 \times 10^{-22} \ g} \approx 8.41 \times 10^{21}$.

(B) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{a^3 \times N_{A}}$
Given: $Z = 4$,$M = 27 \ g \ mol^{-1}$,$a = 405 \ pm = 4.05 \times 10^{-8} \ cm$,$N_{A} = 6.022 \times 10^{23} \ mol^{-1}$
Substituting the values: $d = \frac{4 \times 27}{(4.05 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$
$d = \frac{108}{66.43 \times 10^{-24} \times 6.022 \times 10^{23}}$
$d = \frac{108}{40.01} \approx 2.699 \ g \ cm^{-3}$
Thus,the density is approximately $2.69 \ g \ cm^{-3}$.

(C) For a $BCC$ unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $\sqrt{3} a = 4 r$.
Given: $r = 2.17 \times 10^{-8} \ cm$ and $\sqrt{3} = 1.732$.
Substituting the values: $a = \frac{4 \times 2.17 \times 10^{-8}}{1.732} \ cm$.
$a = \frac{8.68 \times 10^{-8}}{1.732} \ cm = 5.011 \times 10^{-8} \ cm$.
Rounding to two significant figures,we get $a = 5.0 \times 10^{-8} \ cm$.

(D) The formula for density is $d = \frac{Z \times M}{a^3 \times N_A}$.
Given: $d = 7.2 \ g \ cm^{-3}$, $a = 288 \ pm = 2.88 \times 10^{-8} \ cm$, $M = 52 \ g \ mol^{-1}$, $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values:
$7.2 = \frac{Z \times 52}{(2.88 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$
$7.2 = \frac{Z \times 52}{23.887 \times 10^{-24} \times 6.022 \times 10^{23}}$
$7.2 = \frac{Z \times 52}{14.385}$
$Z = \frac{7.2 \times 14.385}{52} \approx 1.991 \approx 2$.
Since the number of atoms per unit cell $Z = 2$, the crystal has a body-centered cubic $(BCC)$ structure.

(A) The density of a unit cell is given by the ratio of the mass of the unit cell to the volume of the unit cell.
The mass of the unit cell is $\frac{n \times M}{N_A}$,where $n$ is the number of atoms per unit cell,$M$ is the molar mass,and $N_A$ is Avogadro's number.
The volume of the cubic unit cell is $a^3$.
Therefore,the density $\rho = \frac{n M}{a^3 N_A}$.

(D) 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 $a = 288 \ pm = 2.88 \times 10^{-8} \ cm$.
The Avogadro constant $N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $7.8 = \frac{2 \times M}{(2.88 \times 10^{-8} \ cm)^3 \times 6.022 \times 10^{23} \ mol^{-1}}$.
$M = \frac{7.8 \times (2.88 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2}$.
$M \approx 56.1 \ g \ mol^{-1}$.

(C) For a simple cubic unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $a = 2r$.
Given that $r = 174 \ pm$.
Therefore, $a = 2 \times 174 \ pm = 348 \ pm$.

(C) The density formula for a unit cell is given by $d = \frac{Z \times M}{V \times N_A}$.
Given: $Z = 4$,$d = 19.0 \ g \ cm^{-3}$,$M = 190 \ g \ mol^{-1}$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $19.0 = \frac{4 \times 190}{V \times 6.022 \times 10^{23}}$.
Rearranging for $V$: $V = \frac{4 \times 190}{19.0 \times 6.022 \times 10^{23}}$.
$V = \frac{40}{6.022} \times 10^{-23} \ cm^3$.
$V \approx 6.64 \times 10^{-23} \ cm^3$.

(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: $1 = \frac{2 \times M}{(4.20 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = \frac{1 \times (74.088 \times 10^{-24}) \times 6.022 \times 10^{23}}{2}$.
$M = \frac{44.61}{2} \approx 22.3 \ g \ mol^{-1}$.

(B) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{a^3 \times N_A}$
Where:
$Z = 4$ (number of particles per unit cell)
$M = 192 \ g \ mol^{-1}$ (molar mass)
$a^3 = 64 \times 10^{-24} \ cm^3$ (volume of unit cell)
$N_A = 6.022 \times 10^{23} \ mol^{-1}$ (Avogadro constant)
Substituting the values:
$d = \frac{4 \times 192}{64 \times 10^{-24} \times 6.022 \times 10^{23}}$
$d = \frac{768}{38.5408} \approx 19.93 \ g \ cm^{-3}$

(A) For a simple cubic unit cell, the number of atoms per unit cell $(z)$ $= 1$.
Edge length $(a)$ $= 336 \ pm = 3.36 \times 10^{-8} \ cm$.
Density $(d)$ $= \frac{z \times M}{a^3 \times N_A}$, where $M$ is the molar mass and $N_A$ is Avogadro's number $(6.022 \times 10^{23} \ mol^{-1})$.
$9.4 = \frac{1 \times M}{(3.36 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = 9.4 \times (3.793 \times 10^{-23}) \times 6.022 \times 10^{23} \approx 214.65 \ g \ mol^{-1}$.
Number of atoms in $5 \ g = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{5}{214.65} \times 6.022 \times 10^{23} \approx 1.4 \times 10^{22}$ atoms.

(A) For a $bcc$ unit cell,the number of atoms per unit cell $(Z)$ is $2$.
The mass of one atom is given by $\frac{\text{Molar mass}}{N_A} = \frac{56}{6.022 \times 10^{23}} \ g$.
The mass of the unit cell is $Z \times \text{mass of one atom} = 2 \times \frac{56}{6.022 \times 10^{23}} \ g$.
$= \frac{112}{6.022 \times 10^{23}} \ g \approx 1.86 \times 10^{-22} \ g$.

(D) For a $BCC$ structure, the number of atoms per unit cell is $Z = 2$.
The formula for density is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.
Given: $d = 10 \ g \ cm^{-3}$, $a = 200 \ pm = 200 \times 10^{-10} \ cm = 2 \times 10^{-8} \ cm$, and $N_A \approx 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $10 = \frac{2 \times M}{6.022 \times 10^{23} \times (2 \times 10^{-8})^3}$.
$10 = \frac{2 \times M}{6.022 \times 10^{23} \times 8 \times 10^{-24}}$.
$10 = \frac{2 \times M}{4.8176}$.
$M = \frac{10 \times 4.8176}{2} = 24.088 \ g \ mol^{-1}$.
Thus, the molar mass is approximately $24.0 \ g \ mol^{-1}$.

(D) The atomic mass of potassium $(K)$ is $39 \ g/mol$.
Number of moles of potassium $= \frac{3.9 \ g}{39 \ g/mol} = 0.1 \ mol$.
Number of atoms $= \text{moles} \times N_{A} = 0.1 \ N_{A}$.
In a $BCC$ unit cell,the number of atoms per unit cell $(n)$ is $2$.
Therefore,the number of unit cells $= \frac{\text{Total number of atoms}}{n} = \frac{0.1 \ N_{A}}{2} = \frac{N_{A}}{20}$.

(C) Given: Molar mass $M = 180 \ g \ mol^{-1}$,Density $\rho = 18 \ g \ cm^{-3}$,Avogadro constant $N_A \approx 6.022 \times 10^{23} \ mol^{-1}$.
For a $BCC$ crystal,the number of atoms per unit cell $z = 2$.
The formula for density is $\rho = \frac{M \times z}{a^3 \times N_A}$.
Rearranging for the edge length $a$: $a^3 = \frac{M \times z}{\rho \times N_A}$.
Substituting the values: $a^3 = \frac{180 \times 2}{18 \times 6.022 \times 10^{23}} \approx \frac{360}{108.396 \times 10^{23}} \approx 3.32 \times 10^{-23} \ cm^3$.
To match the format,we write $a^3 = 33.2 \times 10^{-24} \ cm^3$.
Therefore,$a = \sqrt[3]{33.2} \times 10^{-8} \ cm$.

(B) In a simple cubic structure,the atoms touch along the edge of the unit cell.
Therefore,the relationship between the edge length $a$ and the atomic radius $r$ is $a = 2r$.
Given $a = 3.86 \ \mathring{A}$.
$r = \frac{a}{2} = \frac{3.86 \ \mathring{A}}{2} = 1.93 \ \mathring{A}$.
Converting to centimeters: $1 \ \mathring{A} = 10^{-8} \ cm$.
So,$r = 1.93 \times 10^{-8} \ cm$.

(C) The density formula for a unit cell is given by: $\rho = \frac{Z \times M}{N_A \times a^3}$,where $a^3$ is the volume of the unit cell $(V)$.
For a $BCC$ structure,the number of atoms per unit cell $(Z)$ is $2$.
Given: Atomic mass $(M)$ = $25 \ g \ mol^{-1}$,Density $(\rho)$ = $3 \ g \ cm^{-3}$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging the formula for volume $(V)$: $V = \frac{Z \times M}{\rho \times N_A}$.
Substituting the values: $V = \frac{2 \times 25}{3 \times 6.022 \times 10^{23}}$.
$V = \frac{50}{18.066 \times 10^{23}} \approx 2.767 \times 10^{-23} \ cm^3$.

(C) For a $BCC$ unit cell,the number of atoms per unit cell,$Z = 2$.
The edge length $a = 400 \ pm = 400 \times 10^{-10} \ cm = 4 \times 10^{-8} \ cm$.
The molar mass $M = 100 \ g \ mol^{-1}$.
The density $d$ is given by the formula: $d = \frac{Z \times M}{N_A \times a^3}$.
Substituting the values: $d = \frac{2 \times 100}{6.022 \times 10^{23} \times (4 \times 10^{-8})^3}$.
$d = \frac{200}{6.022 \times 10^{23} \times 64 \times 10^{-24}}$.
$d = \frac{200}{6.022 \times 64 \times 10^{-1}} = \frac{200}{38.54} \approx 5.18 \ g \ cm^{-3}$.

(B) The packing efficiency of $68 \%$ corresponds to a Body-Centered Cubic $(BCC)$ unit cell.
For a $BCC$ unit cell,the number of atoms per unit cell is $Z = 2$.
The edge length $a = 3.3 \ \mathring{A} = 3.3 \times 10^{-8} \ cm$.
The density formula is given by $d = \frac{Z \cdot M}{N_A \cdot a^3}$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging for molar mass $M$: $M = \frac{d \cdot N_A \cdot a^3}{Z}$.
Substituting the values: $M = \frac{8.57 \times 6.022 \times 10^{23} \times (3.3 \times 10^{-8})^3}{2}$.
$M = \frac{8.57 \times 6.022 \times 10^{23} \times 35.937 \times 10^{-24}}{2}$.
$M = \frac{185.48}{2} \approx 92.74 \ g \ mol^{-1}$.
Rounding to the nearest integer,$M = 93 \ g \ mol^{-1}$.

(D) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{N_A \times a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_A$ is Avogadro's number,and $a^3$ is the volume of the unit cell $(V)$.
Alternatively,$d = \frac{Z \times \text{mass of one atom}}{V}$.
For a $bcc$ structure,the number of atoms per unit cell is $Z = 2$.
Given: $d = 9 \ g \ cm^{-3}$,$V = 2.7 \times 10^{-23} \ cm^3$,and total mass $= 2.43 \ g$.
Let $N$ be the total number of atoms in $2.43 \ g$.
The mass of one atom is $\frac{2.43}{N}$.
Substituting the values into the density formula: $9 = \frac{2 \times (2.43 / N)}{2.7 \times 10^{-23}}$.
Rearranging for $N$: $N = \frac{2 \times 2.43}{9 \times 2.7 \times 10^{-23}}$.
$N = \frac{4.86}{24.3 \times 10^{-23}} = 0.2 \times 10^{23} = 2.0 \times 10^{22}$ atoms.

(D) The formula for density is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.
Given:
$Z = 2$ (for $BCC$ structure)
$a = 400 \ pm = 400 \times 10^{-10} \ cm = 4 \times 10^{-8} \ cm$
$d = 4 \ g \ cm^{-3}$
$N_A = 6.022 \times 10^{23} \ mol^{-1}$
Rearranging for molar mass $M$:
$M = \frac{d \cdot N_A \cdot a^3}{Z}$
Substituting the values:
$M = \frac{4 \cdot 6.022 \times 10^{23} \cdot (4 \times 10^{-8})^3}{2}$
$M = \frac{4 \cdot 6.022 \times 10^{23} \cdot 64 \times 10^{-24}}{2}$
$M = 2 \cdot 6.022 \cdot 64 \cdot 10^{-1}$
$M = 77.08 \ g \ mol^{-1} \approx 77 \ g \ mol^{-1}$.

(A) Given: Edge length $a = 500 \ pm = 500 \times 10^{-10} \ cm = 5 \times 10^{-8} \ cm$.
Density $d = 4 \ g \ cm^{-3}$.
For $BCC$ structure, the number of atoms per unit cell $Z = 2$.
The formula for density is $d = \frac{Z \times M}{N_A \times a^3}$.
Rearranging for molar mass $M$: $M = \frac{d \times N_A \times a^3}{Z}$.
Substituting the values: $M = \frac{4 \times 6.022 \times 10^{23} \times (5 \times 10^{-8})^3}{2}$.
$M = \frac{4 \times 6.022 \times 10^{23} \times 125 \times 10^{-24}}{2}$.
$M = 2 \times 6.022 \times 125 \times 10^{-1} = 150.55 \ g \ mol^{-1} \approx 150 \ g \ mol^{-1}$.

(B) For a simple cubic structure, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $a = 2r$.
Given that the edge length $a = 336 \ pm$.
Substituting the value into the formula: $336 \ pm = 2r$.
Therefore, $r = \frac{336 \ pm}{2} = 168 \ pm$.