MHT CET 2025 Chemistry Question Paper with Answer and Solution

(A) Surgical sutures are typically made from biodegradable polymers or specific synthetic fibers that are biocompatible. $Nylon-6$ is widely used in the medical field for the production of surgical sutures due to its high tensile strength and durability. Therefore,the correct option is $A$.

(C) $Dacron$ (also known as $Terylene$) is a polyester fiber. It is prepared by the condensation polymerization of ethylene glycol $(HO-CH_2-CH_2-OH)$ and terephthalic acid ($benzene-1,4-dicarboxylic \ acid$,$HOOC-C_6H_4-COOH$). The reaction involves the elimination of water molecules to form an ester linkage.

(C) Surgical sutures are typically made from biocompatible and biodegradable or non-biodegradable synthetic polymers. Nylon $6,6$ is a polyamide that is widely used in the medical field for surgical sutures due to its high tensile strength and durability. Therefore,the correct option is $C$.

(A) The given polymer structure is $[-CH_2-C(CH_3)=CH-CH_2-]_n$,which is natural rubber (cis-$1,4$-polyisoprene).
The monomer unit of natural rubber is $2-methyl-1,3-butadiene$,commonly known as Isoprene.
Therefore,the correct option is $A$.

(A) Amorphous solids are isotropic in nature,meaning their physical properties (such as refractive index,electrical conductivity,etc.) are the same in all directions.
Glass is an amorphous solid,whereas ceramics,graphite,and ice are crystalline solids,which are anisotropic.

(B) The tetragonal crystal system consists of two types of unit cells: $1$. Primitive (simple) and $2$. Body-centered. Therefore,the total number of different types of unit cells is $2$.

(B) For a $bcc$ unit cell,the number of atoms per unit cell $(Z)$ is $2$.
The mass of one unit cell is given by the product of density $(\varrho)$ and volume $(a^3)$,which is $\varrho \times a^3 = 1.58 \times 10^{-22} \ g$.
The number of unit cells in $1.58 \ g$ of the metal is calculated as:
$\text{Number of unit cells} = \frac{\text{Total mass}}{\text{Mass of one unit cell}} = \frac{1.58 \ g}{1.58 \times 10^{-22} \ g} = 10^{22}$.
Since each $bcc$ unit cell contains $2$ atoms,the total number of atoms is:
$\text{Total atoms} = 2 \times 10^{22}$.

(C) For an $fcc$ unit cell, the relation between edge length $(a)$ and atomic radius $(r)$ is $a = 2\sqrt{2}r$.
Given $r = 141.4 \ pm = 1.414 \times 10^{-8} \ cm$.
Substituting the value of $r$: $a = 2 \times 1.414 \times 1.414 \times 10^{-8} \ cm = 4.0 \times 10^{-8} \ cm$.
The volume of the unit cell is $V = a^3 = (4.0 \times 10^{-8} \ cm)^3 = 64 \times 10^{-24} \ cm^3 = 6.4 \times 10^{-23} \ cm^3$.

(D) In a close-packed structure with $N$ atoms:
$1$. The number of tetrahedral voids is $2N$. Thus,there are two tetrahedral voids per atom.
$2$. The number of octahedral voids is $N$. Thus,there is one octahedral void per atom.
$3$. $A$ tetrahedral void is formed by $4$ spheres,and an octahedral void is formed by $6$ spheres.
Comparing these with the options:
- Option $A$ is correct ($4$ spheres).
- Option $B$ is correct ($6$ spheres).
- Option $C$ is correct ($2$ voids per atom).
- Option $D$ is incorrect because there is one octahedral void per atom,not one for every two atoms.

(C) Let the number of anions $(B)$ in the $ccp$ array be $N$.
Since the number of tetrahedral voids is equal to $2N$,and cations $(A)$ occupy $1/3$ of these voids,the number of cations is $A = \frac{1}{3} \times 2N = \frac{2N}{3}$.
The ratio of cations to anions is $A:B = \frac{2N}{3} : N = \frac{2}{3} : 1 = 2 : 3$.
Therefore,the formula of the ionic compound is $A_2B_3$.

(C) In a simple cubic unit cell,the number of particles per unit cell $(Z)$ is $1$.
The volume of the unit cell $(V_{cell})$ is related to the volume occupied by a particle $(V_{particle})$ by the relation: $V_{cell} = Z \times V_{particle}$.
Given $Z = 1$ and $V_{particle} = 2.1 \times 10^{-23} \ cm^3$.
Therefore,$V_{cell} = 1 \times 2.1 \times 10^{-23} \ cm^3 = 2.1 \times 10^{-23} \ cm^3$.
Comparing this with $x \times 10^{-23} \ cm^3$,we get $x = 2.1$.
Note: The provided options do not contain the correct value $2.1$. Based on the standard definition of a simple cubic unit cell where the edge length $a = 2r$,the volume is $a^3 = (2r)^3 = 8r^3$. If the volume occupied by a particle (sphere) is $\frac{4}{3}\pi r^3 = 2.1 \times 10^{-23} \ cm^3$,then $r^3 = \frac{2.1 \times 10^{-23} \times 3}{4 \times 3.14} \approx 0.5016 \times 10^{-23} \ cm^3$.
Thus,$V_{cell} = 8 \times 0.5016 \times 10^{-23} \approx 4.01 \times 10^{-23} \ cm^3$.
Therefore,$x \approx 4.0$.

(A) For an $fcc$ unit cell,the number of atoms per unit cell $(Z)$ is $4$.
The mass of a unit cell is given by the formula: $\text{Mass of unit cell} = \frac{Z \times M}{N_A}$,where $M$ is the molar mass.
Rearranging the formula to solve for $M$: $M = \frac{\text{Mass of unit cell} \times N_A}{Z}$.
Substituting the given values: $M = \frac{1.8 \times 10^{-22} \ g \times 6.022 \times 10^{23} \ mol^{-1}}{4}$.
$M = \frac{108.396}{4} \ g \ mol^{-1} = 27.099 \ g \ mol^{-1} \approx 27.0 \ g \ mol^{-1}$.

(C) In an $hcp$ structure,the number of atoms is $N = 0.6 \times 6.022 \times 10^{23} = 3.6132 \times 10^{23}$.
Number of octahedral voids = $N = 3.6132 \times 10^{23}$.
Number of tetrahedral voids = $2N = 2 \times 3.6132 \times 10^{23} = 7.2264 \times 10^{23}$.
Total number of voids = $N + 2N = 3N = 3 \times 3.6132 \times 10^{23} = 1.08396 \times 10^{24} \approx 1.084 \times 10^{24}$.

(A) The volume of one unit cell is given by $V_{cell} = a^3$,where $a$ is the edge length of the unit cell.
Given $a = 4.0 \times 10^{-8} \ cm$,so $V_{cell} = (4.0 \times 10^{-8} \ cm)^3 = 64 \times 10^{-24} \ cm^3 = 6.4 \times 10^{-23} \ cm^3$.
The number of unit cells in $1 \ cm^3$ volume is calculated as:
$\text{Number of unit cells} = \frac{\text{Total Volume}}{\text{Volume of one unit cell}} = \frac{1 \ cm^3}{6.4 \times 10^{-23} \ cm^3} = 0.15625 \times 10^{23} = 1.5625 \times 10^{22}$.
Thus,the correct option is $A$.

(C) In a $bcc$ (body-centered cubic) unit cell,the total number of particles $(Z)$ is $2$.
The packing efficiency of a $bcc$ unit cell is $68\%$,which means $68\%$ of the total volume of the unit cell is occupied by the particles.
Total volume of unit cell = $8.2 \times 10^{-23} \ cm^3$.
Volume occupied by particles = $0.68 \times 8.2 \times 10^{-23} \ cm^3 = 5.576 \times 10^{-23} \ cm^3$.
Since there are $2$ particles in the $bcc$ unit cell,the volume occupied by a single particle is $\frac{5.576 \times 10^{-23}}{2} \ cm^3 = 2.788 \times 10^{-23} \ cm^3$.

(B) Let the number of atoms in the close-packed structure be $N = 0.4 \ mol = 0.4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23}$ atoms.
Number of octahedral voids = $N = 2.4088 \times 10^{23}$.
Number of tetrahedral voids = $2N = 2 \times 2.4088 \times 10^{23} = 4.8176 \times 10^{23}$.
Total number of voids = $N + 2N = 3N = 3 \times 2.4088 \times 10^{23} = 7.2264 \times 10^{23}$.

(B) The density of a unit cell is given by $\rho = \frac{Z \times M}{N_A \times a^3}$,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 $(V)$.
Rearranging the formula,we get $\rho \times V = \frac{Z \times M}{N_A}$.
We know that the mass of one atom is $m = \frac{M}{N_A} = 4.4 \times 10^{-23} \ g$.
Given $\rho \times V = 1.792 \times 10^{-22} \ g$.
Substituting the values: $1.792 \times 10^{-22} = Z \times (4.4 \times 10^{-23})$.
$Z = \frac{1.792 \times 10^{-22}}{4.4 \times 10^{-23}} = \frac{17.92}{4.4} \approx 4.07$.
Since $Z \approx 4$,the unit cell is a Face centred unit cell $(FCC)$.

(C) The density of a unit cell is given by the formula: $\rho = \frac{Z \times M}{N_A \times V}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_A$ is Avogadro's number,and $V$ is the volume of the unit cell.
Given that the product of density $(\rho)$ and volume $(V)$ is $1.8 \times 10^{-22} \ g$,we have $\rho \times V = 1.8 \times 10^{-22} \ g$.
We know that the mass of a unit cell is equal to $\rho \times V$.
Also,the mass of a unit cell is equal to the number of atoms $(Z)$ multiplied by the mass of a single atom $(m_{atom})$.
Therefore,$Z \times m_{atom} = \rho \times V$.
Substituting the given values: $Z \times (4.5 \times 10^{-23} \ g) = 1.8 \times 10^{-22} \ g$.
$Z = \frac{1.8 \times 10^{-22}}{4.5 \times 10^{-23}} = \frac{18 \times 10^{-23}}{4.5 \times 10^{-23}} = 4$.
Thus,the number of atoms per unit cell is $4$.

(B) For an $fcc$ unit cell,the packing efficiency is $74\%$.
This means the volume occupied by atoms is $0.74 \times V_{cell}$.
The void volume is the remaining volume,which is $100\% - 74\% = 26\%$.
Therefore,the void volume is $0.26 \times V_{cell}$.
Given that the void volume is $4.16 \times 10^{-24} \ cm^3$,we have:
$0.26 \times V_{cell} = 4.16 \times 10^{-24} \ cm^3$.
$V_{cell} = \frac{4.16 \times 10^{-24}}{0.26} \ cm^3$.
$V_{cell} = 16 \times 10^{-24} \ cm^3 = 1.6 \times 10^{-23} \ cm^3$.
Thus,the correct option is $B$.

(C) The number of unit cells is calculated by dividing the total volume by the volume of a single unit cell.
Number of unit cells = $\frac{\text{Total Volume}}{\text{Volume of one unit cell}}$
Number of unit cells = $\frac{1 \ cm^3}{3.448 \times 10^{-23} \ cm^3}$
Number of unit cells = $0.29002 \times 10^{23} = 2.9 \times 10^{22}$
Therefore,the correct option is $C$.

(A) The packing efficiency of an $fcc$ unit cell is $74\%$.
This means that $74\%$ of the total volume of the unit cell is occupied by the particles (atoms).
Volume occupied by particles = $0.74 \times \text{Volume of unit cell}$.
Given,Volume of unit cell = $1.6 \times 10^{-23} \ cm^3$.
Volume occupied = $0.74 \times 1.6 \times 10^{-23} \ cm^3 = 1.184 \times 10^{-23} \ cm^3$.

(D) For an $fcc$ unit cell,the number of atoms per unit cell $(Z)$ is $4$.
The mass of one unit cell is given by the product of density $(\varrho)$ and volume $(a^3)$,which is $\varrho \times a^3 = \frac{Z \times M}{N_A}$.
Given $\varrho \times a^3 = 6.8 \times 10^{-22} \ g$,this represents the mass of one unit cell.
The number of unit cells in $1 \ g$ of the element is $\frac{1 \ g}{6.8 \times 10^{-22} \ g/\text{unit cell}} \approx 1.4706 \times 10^{21} \ \text{unit cells}$.
Since each $fcc$ unit cell contains $4$ atoms,the total number of atoms in $1 \ g$ is $4 \times 1.4706 \times 10^{21} = 5.8824 \times 10^{21}$ atoms.
Thus,the correct option is $D$.

(B) In an $fcc$ (face-centered cubic) unit cell,the number of particles per unit cell $(Z)$ is $4$.
The packing efficiency of an $fcc$ unit cell is $74\%$.
The volume occupied by all particles in the unit cell is $0.74 \times \text{Volume of unit cell} = 0.74 \times 1.6 \times 10^{-23} \ cm^3 = 1.184 \times 10^{-23} \ cm^3$.
The volume occupied by a single particle is $\frac{1.184 \times 10^{-23} \ cm^3}{4} = 2.96 \times 10^{-24} \ cm^3$.

(C) In a $bcc$ (body-centered cubic) unit cell,the number of particles $(Z)$ is $2$.
The packing efficiency of a $bcc$ unit cell is $68\%$.
The volume occupied by the particles is given by: $\text{Volume} = \text{Packing Efficiency} \times \text{Volume of unit cell}$.
$\text{Volume} = 0.68 \times 8.0 \times 10^{-23} \ cm^3$.
$\text{Volume} = 5.44 \times 10^{-23} \ cm^3$.

(A) For an $fcc$ crystal structure,the number of atoms per unit cell $(Z)$ is $4$.
The density formula is $\varrho = \frac{Z \times M}{N_A \times a^3}$,where $M$ is the molar mass and $N_A$ is Avogadro's number $(6.022 \times 10^{23} \ mol^{-1})$.
Rearranging the formula,we get $M = \frac{\varrho \times a^3 \times N_A}{Z}$.
Substituting the given values: $M = \frac{1.728 \times 10^{-22} \times 6.022 \times 10^{23}}{4} = \frac{104.06}{4} = 26.015 \ g/mol$.
The number of atoms in $1 \ g$ of the metal is calculated as: $\text{Number of atoms} = \frac{\text{mass}}{M} \times N_A = \frac{1}{26.015} \times 6.022 \times 10^{23} \approx 2.315 \times 10^{22}$ atoms.

(D) The mass of a single unit cell is given by the product of its density $(\rho)$ and its volume $(V)$.
Given: $\rho \times V = 1.58 \times 10^{-22} \ g$.
Total mass of the metal = $0.79 \ g$.
The number of unit cells = $\frac{\text{Total mass}}{\text{Mass of one unit cell}}$.
Number of unit cells = $\frac{0.79 \ g}{1.58 \times 10^{-22} \ g} = 0.5 \times 10^{22} = 5.0 \times 10^{21}$.

(C) For a $bcc$ (body-centered cubic) unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$.
Therefore,$a = \frac{4r}{\sqrt{3}}$.
Given $r = 1.86 \times 10^{-8} \ cm$.
$a = \frac{4 \times 1.86 \times 10^{-8}}{\sqrt{3}} \approx \frac{7.44 \times 10^{-8}}{1.732} \approx 4.2956 \times 10^{-8} \ cm$.
The volume of the unit cell $(V)$ is $a^3$.
$V = (4.2956 \times 10^{-8} \ cm)^3 \approx 79.28 \times 10^{-24} \ cm^3 = 7.928 \times 10^{-23} \ cm^3$.
Rounding to the nearest option,the correct answer is $7.951 \times 10^{-23} \ cm^3$.

$(A)$ For an $fcc$ (face-centered cubic) unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2\sqrt{2}r$.
Given $r = 128 \ pm$.
$a = 2 \times 1.414 \times 128 \ pm = 361.98 \ pm \approx 362 \ pm$.
To convert $pm$ to $cm$: $362 \ pm = 362 \times 10^{-10} \ cm = 3.62 \times 10^{-8} \ cm$.
Therefore, the correct option is $A$.

(C) The density of a unit cell is given by the formula: $\varrho = \frac{Z \times m}{a^3}$,where $Z$ is the number of particles per unit cell,$m$ is the mass of one particle,and $a^3$ is the volume of the unit cell.
Rearranging the formula to solve for $Z$: $Z = \frac{\varrho \times a^3}{m}$.
Given values: $\varrho \times a^3 = 3.2 \times 10^{-22} \ g$ and $m = 8.0 \times 10^{-23} \ g$.
Substituting the values: $Z = \frac{3.2 \times 10^{-22}}{8.0 \times 10^{-23}} = \frac{32 \times 10^{-23}}{8.0 \times 10^{-23}} = 4$.
Therefore,the number of particles per unit cell is $4$.

(C) In a crystal lattice,a $bcc$ (body-centered cubic) unit cell is a type of cubic system.
In any cubic unit cell,each corner particle is shared by $8$ adjacent unit cells.
This is because each corner is a point where $8$ cubes meet in a three-dimensional space.

(D) In an $fcc$ (face-centered cubic) unit cell,the number of atoms per unit cell $(Z)$ is $4$.
The packing efficiency of an $fcc$ unit cell is $74\%$.
Total volume occupied by particles = $\text{Packing Efficiency} \times \text{Volume of unit cell}$.
Total volume = $0.74 \times 6.4 \times 10^{-23} \ cm^3$.
Total volume = $4.736 \times 10^{-23} \ cm^3$.

(C) The volume of one unit cell is given by $V_{cell} = a^3$,where $a$ is the edge length of the unit cell.
Given $a = 1.25 \times 10^{-8} \ cm$.
$V_{cell} = (1.25 \times 10^{-8} \ cm)^3 = 1.953125 \times 10^{-24} \ cm^3$.
The number of unit cells in a total volume $V_{total} = 1 \ cm^3$ is calculated as:
$\text{Number of unit cells} = \frac{V_{total}}{V_{cell}} = \frac{1 \ cm^3}{1.953125 \times 10^{-24} \ cm^3} \approx 5.12 \times 10^{23}$.

(A) For a $bcc$ (body-centered cubic) unit cell, the relationship between the edge length $(a)$ and the radius of the particle $(r)$ is given by: $4r = \sqrt{3}a$ or $a = \frac{4r}{\sqrt{3}}$.
Given $r = 186 \ pm$.
$a = \frac{4 \times 186 \ pm}{\sqrt{3}} = \frac{744}{1.732} \ pm \approx 429.56 \ pm$.
To convert $pm$ to $cm$: $1 \ pm = 10^{-10} \ cm$.
$a = 429.56 \times 10^{-10} \ cm = 4.2956 \times 10^{-8} \ cm \approx 4.296 \times 10^{-8} \ cm$.

(C) In a $ccp$ (cubic close-packed) structure,the number of tetrahedral voids is equal to twice the number of atoms present in the lattice.
Let $N$ be the number of atoms in the $ccp$ structure.
Given,$N = 9.6 \times 10^{23}$.
The number of tetrahedral voids $= 2 \times N$.
Number of tetrahedral voids $= 2 \times (9.6 \times 10^{23}) = 19.2 \times 10^{23} = 1.92 \times 10^{24}$.

(C) In a simple cubic $(SC)$ unit cell,each particle is at the corner of the cube.
Each particle is in contact with $6$ nearest neighbors (one along each axis: $x, -x, y, -y, z, -z$).
Therefore,the coordination number of a particle in a simple cubic structure is $6$.

(A) For a $bcc$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $r = \frac{\sqrt{3}}{4} a$.
Given edge length $a = 4.3 \times 10^{-8} \ cm$.
Convert $a$ to $pm$: $a = 4.3 \times 10^{-8} \ cm = 4.3 \times 10^{-8} \times 10^{10} \ pm = 430 \ pm$.
Now, substitute the value of $a$ in the formula: $r = \frac{\sqrt{3}}{4} \times 430 \ pm$.
$r = 0.433 \times 430 \ pm = 186.19 \ pm \approx 186.2 \ pm$.

(C) The volume of one unit cell is given by $V_{cell} = a^3$,where $a$ is the edge length.
Given $a = 2.0 \times 10^{-8} \ cm$.
$V_{cell} = (2.0 \times 10^{-8} \ cm)^3 = 8.0 \times 10^{-24} \ cm^3$.
The number of unit cells in $1 \ cm^3$ is calculated as:
$\text{Number of unit cells} = \frac{\text{Total Volume}}{\text{Volume of one unit cell}} = \frac{1 \ cm^3}{8.0 \times 10^{-24} \ cm^3} = 0.125 \times 10^{24} = 1.25 \times 10^{23}$.

(A) In a face-centered cubic $(fcc)$ unit cell,the number of atoms per unit cell $(Z)$ is $4$.
The packing efficiency of an $fcc$ unit cell is $74\%$.
The total volume occupied by the particles is equal to the packing efficiency multiplied by the volume of the unit cell.
$\text{Volume occupied} = 0.74 \times \text{Volume of unit cell}$.
$\text{Volume occupied} = 0.74 \times 5.2 \times 10^{-23} \ cm^3 = 3.848 \times 10^{-23} \ cm^3$.

(A) For an $fcc$ structure,the number of atoms per unit cell $(Z)$ is $4$.
The density formula is $\varrho = \frac{Z \times M}{N_A \times a^3}$,which can be rearranged as $M = \frac{\varrho \times a^3 \times N_A}{Z}$.
Given $\varrho \times a^3 = 7.2 \times 10^{-22} \ g$,$Z = 4$,and $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
The molar mass $M = \frac{7.2 \times 10^{-22} \times 6.022 \times 10^{23}}{4} \approx 108.4 \ g/mol$.
The number of moles $n = \frac{\text{mass}}{M} = \frac{5.4}{108.4} \approx 0.0498 \ mol$.
The number of atoms $= n \times N_A = 0.0498 \times 6.022 \times 10^{23} \approx 3.0 \times 10^{22}$ atoms.

(A) Let the number of anions $(B)$ in the $hcp$ array be $n$.
Since the number of octahedral voids is equal to the number of atoms in the $hcp$ arrangement,the number of octahedral voids is $n$.
Given that cations $(A)$ occupy $\frac{2}{3}$ of the octahedral voids,the number of cations $(A)$ is $\frac{2}{3}n$.
The ratio of cations $(A)$ to anions $(B)$ is $\frac{2}{3}n : n$,which simplifies to $2:3$.
Therefore,the formula of the ionic compound is $A_2B_3$.

$(A)$ For a simple cubic unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $a = 2r$.
Given, $r = 170 \ pm$.
Therefore, $a = 2 \times 170 \ pm = 340 \ pm$.
To convert the edge length from $pm$ to $cm$:
$1 \ pm = 10^{-10} \ cm$.
$a = 340 \times 10^{-10} \ cm = 3.40 \times 10^{-8} \ cm$.

(A) Stainless steel is an alloy of iron $(Fe)$ with other elements like chromium $(Cr)$,nickel $(Ni)$,and carbon $(C)$.
In the crystal lattice of iron,some iron atoms are replaced by atoms of other metals like chromium or nickel.
This type of defect,where foreign atoms replace the host atoms in the lattice,is known as a $Substitutional \ impurity \ defect$.

(B) In a $Frenkel$ defect,an ion (usually the smaller cation) is displaced from its normal lattice site and occupies an interstitial site.
This defect does not change the stoichiometry of the crystal because the number of ions remains the same.
It is commonly observed in ionic compounds where there is a large difference in the size of ions,such as $AgCl$ or $ZnS$.

(A) To form an $n-$type semiconductor,a pentavalent impurity (Group $15$ element) is added to a tetravalent semiconductor like germanium $(Ge)$.
Among the given options,$As$ (Arsenic) is a Group $15$ element,while $B$ (Boron),$In$ (Indium),and $Ga$ (Gallium) are Group $13$ elements.
Adding a Group $15$ element provides an extra electron,which is responsible for $n-$type conductivity.
Therefore,$As$ is the correct dopant.

(A) To produce a $p-$type semiconductor,a group $13$ element (trivalent impurity) is added to a group $14$ element like germanium.
$B$ (Boron) belongs to group $13$,while $P$ (Phosphorus),$As$ (Arsenic),and $Sb$ (Antimony) belong to group $15$ (pentavalent impurities).
Adding a group $13$ element creates an electron hole,which is responsible for $p-$type conductivity.
Therefore,$B$ is the correct dopant.

(A) To produce a $p$-type semiconductor, silicon (a group $14$ element) is doped with a group $13$ element.
Group $13$ elements have $3$ valence electrons, which create an electron deficiency or 'hole' in the crystal lattice.
Among the given options, $Ga$ (Gallium) belongs to group $13$, while $Sb$ (Antimony), $As$ (Arsenic), and $P$ (Phosphorus) belong to group $15$.
Group $15$ elements are used to produce $n$-type semiconductors.
Therefore, $Ga$ is the correct dopant.

(A) To obtain an $n$-type semiconductor, a group-$15$ element (pentavalent) is added as a dopant to a group-$14$ element like silicon $(Si)$.
Among the given options, $As$ (Arsenic) belongs to group-$15$, while $B$ (Boron), $Ga$ (Gallium), and $In$ (Indium) belong to group-$13$.
Therefore, adding $As$ to $Si$ creates an $n$-type semiconductor due to the presence of an extra electron.

(D) defect is called non-stoichiometric if the ratio of the number of cations to anions in the crystal becomes different from that indicated by its chemical formula.
This occurs due to the presence of metal excess or metal deficiency in the crystal lattice.

(C) Brass is an alloy of $Cu$ and $Zn$. In brass,$Zn$ atoms replace some of the $Cu$ atoms in the crystal lattice of copper. Since the host atoms are replaced by foreign atoms of similar size,this is classified as a substitutional impurity defect.

(A) Step $1$: Calculate the moles of solute $(n_{solute})$.
$n_{solute} = \frac{\text{mass}}{\text{molar mass}} = \frac{394 \ g}{342 \ g \ mol^{-1}} \approx 1.152 \ mol$.
Step $2$: Calculate the moles of solvent $(n_{solvent})$.
$n_{solvent} = \frac{622 \ g}{18 \ g \ mol^{-1}} \approx 34.556 \ mol$.
Step $3$: Calculate the mole fraction of solute $(X_{solute})$.
$X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{1.152}{1.152 + 34.556} = \frac{1.152}{35.708} \approx 0.032$.