MHT CET 2024 Chemistry Question Paper with Answer and Solution

(C) $\text{Percent atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants used in the reaction}} \times 100$
$\text{Percent atom economy} = \frac{65 \ u}{78 \ u} \times 100 = 83.33 \% \approx 83 \%$

(C) The molar mass of $NaOH$ is calculated as: $23 + 16 + 1 = 40 \ u$.
The percentage by mass of oxygen is given by the formula: $\frac{\text{Atomic mass of } O}{\text{Molar mass of } NaOH} \times 100$.
Substituting the values: $\frac{16}{40} \times 100 = 40 \%$.

(C) Given mass of $CO_2 = 8.8 \times 10^{-2} \ kg = 88 \ g$.
The molar mass of $CO_2$ is $44 \ g \ mol^{-1}$.
The number of moles $n$ is calculated as $n = \frac{\text{mass}}{\text{molar mass}} = \frac{88 \ g}{44 \ g \ mol^{-1}} = 2 \ mol$.
According to the ideal gas equation,$PV = nRT$.
Substituting $n = 2$,we get $PV = 2 \ RT$.

(B) Using the ideal gas equation: $PV = nRT$
Rearranging for $n$: $n = \frac{PV}{RT}$
Given values: $P = 3.18 \times 10^5 \ N \ m^{-2}$,$V = 8.314 \times 10^{-3} \ m^3$,$T = 318 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
Substituting the values: $n = \frac{(3.18 \times 10^5) \times (8.314 \times 10^{-3})}{8.314 \times 318}$
$n = \frac{3.18 \times 10^2}{318} = \frac{318}{318} = 1 \ mole$

(B) According to Boyle's law,for a fixed amount of an ideal gas at constant temperature,the pressure $P$ is inversely proportional to the volume $V$,i.e.,$P \propto \frac{1}{V}$.
This implies that $PV = k$,where $k$ is a constant.
Therefore,if we plot $PV$ on the y-axis and $P$ on the x-axis,the value of $PV$ remains constant regardless of the change in pressure $P$.
This results in a horizontal straight line parallel to the pressure axis.

(B) According to Boyle's law,at constant temperature and amount of gas:
$P_1 V_1 = P_2 V_2$
Given:
$P_1 = 1 \ atm$,$V_1 = 25 \ mL$
$P_2 = 1.25 \ atm$,$V_2 = ?$
Substituting the values:
$1 \ atm \times 25 \ mL = 1.25 \ atm \times V_2$
$V_2 = \frac{1 \ atm \times 25 \ mL}{1.25 \ atm} = 20 \ mL$

(A) According to Boyle's law,at constant temperature,the product of pressure and volume is constant.
$P_1 V_1 = P_2 V_2$
Given:
$P_1 = 105 \ kPa$,$V_1 = 11.2 \ dm^3$
$P_2 = 210 \ kPa$,$V_2 = ?$
Substituting the values:
$V_2 = \frac{P_1 V_1}{P_2} = \frac{105 \ kPa \times 11.2 \ dm^3}{210 \ kPa} = 5.6 \ dm^3$

(A) Normal temperature and pressure $(NTP)$ implies $P_1 = 1 \ atm = 1.01325 \times 10^5 \ Nm^{-2}$ and $V_1 = 1 \ dm^3$.
Given $P_2 = 1.032 \times 10^5 \ Nm^{-2}$.
Using Boyle's Law,$P_1 V_1 = P_2 V_2$.
$V_2 = \frac{P_1 V_1}{P_2} = \frac{1.01325 \times 10^5 \ Nm^{-2} \times 1 \ dm^3}{1.032 \times 10^5 \ Nm^{-2}}$.
$V_2 \approx 0.982 \ dm^3$.

(B) According to Dalton's Law of Partial Pressures,the partial pressure of a gas $(P_1)$ in a mixture is the product of its mole fraction $(x_1)$ and the total pressure $(P_{total})$ of the mixture.
$P_1 = x_1 \cdot P_{total}$
Rearranging this equation to solve for the mole fraction $(x_1)$ gives:
$x_1 = \frac{P_1}{P_{total}}$
Therefore,the correct relation is $x_1 = \frac{P_1}{P_{total}}$.

(A) Given: $n = 3.4 \ mol$,$T = 300 \ K$,$V = 68 \ mL = 68 \times 10^{-3} \ L = 0.068 \ dm^3$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation: $PV = nRT$.
$P = \frac{nRT}{V} = \frac{3.4 \ mol \times 8.314 \ J \ K^{-1} \ mol^{-1} \times 300 \ K}{0.068 \ dm^3}$.
Since $1 \ J = 1 \ Pa \cdot m^3$ and $1 \ dm^3 = 10^{-3} \ m^3$,the pressure $P = \frac{8480.28}{0.068 \times 10^{-3}} \ Pa = 124710000 \ Pa = 1.2471 \times 10^5 \ kPa$ is incorrect in the original prompt's unit conversion.
Recalculating: $P = \frac{3.4 \times 8.314 \times 300}{0.068 \times 10^{-3}} \ Pa = 1.2471 \times 10^8 \ Pa = 1.2471 \times 10^5 \ kPa$.

(C) Boyle's law states that at constant temperature,$PV = k_1$.
Charles's law states that at constant pressure,$\frac{V}{T} = k_2$.
Combining these two laws for a fixed amount of gas,we get the combined gas law: $\frac{PV}{T} = k$.
For a gas changing from state $1$ $(P_1, V_1, T_1)$ to state $2$ $(P_2, V_2, T_2)$,the relationship is expressed as:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.

(B) The ideal gas constant $R$ is a fundamental physical constant.
In the units of $L \ atm \ K^{-1} \ mol^{-1}$,the value of $R$ is approximately $0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Among the given options,$0.082$ is the closest approximation to this value.
Therefore,the correct option is $B$.

(A) Let the mass of each gas be $m = 1 \ g$.
The number of moles $n = \frac{m}{M}$,where $M$ is the molar mass.
Since $m$ is constant,$n \propto \frac{1}{M}$.
The molar masses are: $M(H_2) = 2 \ g/mol$,$M(N_2) = 28 \ g/mol$,$M(O_2) = 32 \ g/mol$,$M(Cl_2) = 71 \ g/mol$.
Thus,the number of moles follows the order: $n(H_2) > n(N_2) > n(O_2) > n(Cl_2)$.
From the ideal gas equation $PV = nRT$,at constant $T$ and $V$,$P \propto n$.
Therefore,the gas with the highest number of moles will exert the highest pressure.
Since $H_2$ has the lowest molar mass,it has the highest number of moles for a given mass,and thus exerts the highest pressure.

(B) Step $1$: Calculate the number of moles of each gas.
$n_{O_2} = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$
$n_{Ar} = \frac{80 \ g}{40 \ g/mol} = 2 \ mol$
$n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$
Step $2$: Calculate the total number of moles.
$n_{\text{total}} = n_{O_2} + n_{Ar} + n_{H_2} = 1 + 2 + 2 = 5 \ mol$
Step $3$: Calculate the mole fraction of dioxygen $(x_{O_2})$.
$x_{O_2} = \frac{n_{O_2}}{n_{\text{total}}} = \frac{1}{5} = 0.2$
Step $4$: Calculate the partial pressure of dioxygen $(P_{O_2})$.
$P_{O_2} = x_{O_2} \times P_{\text{total}} = 0.2 \times 10 \ bar = 2 \ bar$

(C) The compressibility factor,denoted by $Z$,is defined as the ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure.
For $n$ moles of a gas,the ideal gas equation is $PV = nRT$.
Therefore,the compressibility factor is given by the ratio $Z = \frac{PV}{nRT}$.
Thus,option $C$ is the correct formula.

(D) The ease of liquefaction of a gas depends on its critical temperature $(T_{c})$.
Gases with higher critical temperatures are easier to liquefy because they can be liquefied at higher temperatures.
The critical temperatures of the given gases are approximately: $SO_2$ $(430 \ K)$,$Cl_2$ $(417 \ K)$,$NH_3$ $(405 \ K)$,and $O_2$ $(154.6 \ K)$.
Since $O_2$ has the lowest critical temperature,it is the most difficult to liquefy among the given options.

(B) The number of protons $(Z)$ is calculated by subtracting the number of neutrons $(N)$ from the mass number $(A)$:
$Z = A - N = 40 - 21 = 19$.
Since the atomic number is $19$,the element is potassium $(K)$.
The standard representation of an element is ${ }_{Z}^{A} X$,where $A$ is the mass number and $Z$ is the atomic number.
Substituting the values,we get ${ }_{19}^{40} X$.
Therefore,the correct option is $(B)$.

(B) Isobars are defined as atoms of different elements that have the same mass number but different atomic numbers.
For example,${ }_{6}^{14}C$ and ${ }_{7}^{14}N$ both have a mass number of $14$ but different atomic numbers ($6$ and $7$ respectively).

(B) To determine the number of electrons in each species,we look at their atomic numbers $(Z)$:
$1$. For $K^{+}$: Atomic number of $K$ is $19$. Since it is a cation with a $+1$ charge,it has $19 - 1 = 18$ electrons.
$2$. For $Ca$: Atomic number of $Ca$ is $20$. As a neutral atom,it has $20$ electrons.
$3$. For $Mg$: Atomic number of $Mg$ is $12$. As a neutral atom,it has $12$ electrons.
$4$. For $Cl$: Atomic number of $Cl$ is $17$. As a neutral atom,it has $17$ electrons.
Therefore,the species containing $20$ electrons is $Ca$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$E_n = -R_H \times \frac{Z^2}{n^2} \ J$
For a hydrogen atom,the atomic number $Z = 1$.
For the fourth orbit,$n = 4$.
Substituting the values:
$E_4 = -2.18 \times 10^{-18} \times \frac{1^2}{4^2} \ J$
$E_4 = -2.18 \times 10^{-18} \times \frac{1}{16} \ J$
$E_4 = -0.13625 \times 10^{-18} \ J \approx -0.136 \times 10^{-18} \ J$

(D) For the first orbit of a monopositive $He^{+}$ ion,the atomic number $Z = 2$ and the orbit number $n = 1$.
The energy of an electron in the $n^{th}$ orbit is given by the formula:
$E_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \ J$
Substituting the values $Z = 2$ and $n = 1$:
$E_1 = -2.18 \times 10^{-18} \left( \frac{2^2}{1^2} \right) \ J$
$E_1 = -2.18 \times 10^{-18} \times 4 \ J$
$E_1 = -8.72 \times 10^{-18} \ J$

(D) For the first orbit of the monopositive $He^{+}$ ion,the atomic number $Z = 2$ and the principal quantum number $n = 1$.
The energy of an orbit is given by the formula: $E_n = -2.18 \times 10^{-18} \left(\frac{Z^2}{n^2}\right) \ J$.
Substituting the values: $E_1 = -2.18 \times 10^{-18} \left(\frac{2^2}{1^2}\right) \ J$.
$E_1 = -2.18 \times 10^{-18} \times 4 \ J = -8.72 \times 10^{-18} \ J$.

(B) Frequency is defined as the number of waves that pass through a given point in one second.
It is measured in Hertz $(Hz)$,where $1 \ Hz = 1 \ s^{-1}$.
- $(A)$ Wavelength: The distance between two consecutive crests or troughs.
- $(C)$ Wave number: The number of wavelengths per unit length $(1 / \lambda)$.
- $(D)$ Amplitude: The maximum displacement of the wave from its mean position.

(C) The radius of the $n^{\text{th}}$ orbit is given by the formula: $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For $He^{+}$, the atomic number $Z = 2$ and for the first orbit $n = 1$.
Substituting these values: $r_1 = \frac{52.9 \times (1)^2}{2} \ pm = 26.45 \ pm$.

(B) According to Bohr's model of the atom,the angular momentum $(L)$ of an electron in a stationary orbit is quantized and is given by the equation:
$L = mvr = \frac{nh}{2 \pi}$
where $m$ is the mass of the electron,$v$ is the velocity,$r$ is the radius of the orbit,$n$ is the principal quantum number $(n = 1, 2, 3, ...)$,and $h$ is Planck's constant.

(C) The Bohr model of the atom could not explain the splitting of spectral lines in the presence of a magnetic field,which is known as the Zeeman effect. Therefore,the statement that it explains the Zeeman effect is incorrect.

(C) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula: $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For the $Li^{2+}$ ion, the atomic number $Z = 3$.
For the first orbit, the principal quantum number $n = 1$.
Substituting these values into the formula: $r_1 = \frac{52.9 \times (1)^2}{3} \ pm = 17.63 \ pm$.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For $He^{+}$, the atomic number $Z = 2$ and for the first orbit, $n = 1$.
Substituting these values: $r_1 = \frac{52.9 \times (1)^2}{2} \ pm = 26.45 \ pm$.

(B) The lowest energy transition in the Balmer series occurs from $n_2 = 3$ to $n_1 = 2$.
Using the Rydberg formula for wave number: $\overline{v} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Substituting the values: $\overline{v} = R_H \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$.
$\overline{v} = R_H \left[ \frac{1}{4} - \frac{1}{9} \right] = R_H \left[ \frac{9-4}{36} \right] = R_H \left( \frac{5}{36} \right)$.

(C) The relationship between wavelength $(\lambda)$ and wave number $(\bar{\nu})$ is given by: $\lambda = \frac{1}{\bar{\nu}}$
Given $\bar{\nu} = 11516 \ cm^{-1}$.
$\lambda = \frac{1}{11516 \ cm^{-1}} \approx 8.6835 \times 10^{-5} \ cm$.
To convert this to nanometers $(nm)$,we use the conversion factor $1 \ nm = 10^{-7} \ cm$:
$\lambda = 8.6835 \times 10^{-5} \ cm \times \frac{1 \ nm}{10^{-7} \ cm} = 868.35 \ nm \approx 868 \ nm$.

(A) For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.
The lowest energy transition in the Lyman series corresponds to the transition from the nearest higher energy level,which is $n_2 = 2$.
The Rydberg formula for the wave number $\bar{v}$ is given by $\bar{v} = R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Substituting the values $n_1 = 1$ and $n_2 = 2$:
$\bar{v} = R_{H} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \ cm^{-1}$
$\bar{v} = R_{H} \left( 1 - \frac{1}{4} \right) \ cm^{-1}$
$\bar{v} = R_{H} \left( \frac{3}{4} \right) \ cm^{-1}$

(B) The wavenumber $(\bar{\nu})$ is defined as the reciprocal of the wavelength $(\lambda)$.
$\bar{\nu} = \frac{1}{\lambda}$
Given,$\lambda = 0.25 \mu m = 0.25 \times 10^{-6} \ m$.
Substituting the value in the formula:
$\bar{\nu} = \frac{1}{0.25 \times 10^{-6} \ m} = 4.0 \times 10^6 \ m^{-1}$.

(D) The $Balmer$ series refers to a series of emission spectral lines in the visible region of the electromagnetic spectrum for hydrogen.
These lines are produced when an electron transitions from a higher energy level $(n \geq 3)$ to the second energy level $(n=2)$ of the hydrogen atom.
The wavelengths of the lines in the $Balmer$ series fall within the visible light spectrum.
They correspond to the colors observed in hydrogen's emission spectrum,including red,blue,and violet lines.

(B) According to the de Broglie's equation,$\lambda = \frac{h}{p}$.
Therefore,the momentum $p$ is given by $p = \frac{h}{\lambda}$.
Given,$h = 6.63 \times 10^{-34} \ J \ s$ and $\lambda = 6.0 \ \mathring{A} = 6.0 \times 10^{-10} \ m$.
Substituting the values,$p = \frac{6.63 \times 10^{-34} \ J \ s}{6.0 \times 10^{-10} \ m} = 1.105 \times 10^{-24} \ kg \ m \ s^{-1} \approx 1.1 \times 10^{-24} \ kg \ m \ s^{-1}$.

(C) The frequency $\nu$ is calculated using the formula $\nu = \frac{c}{\lambda}$.
Given: Speed of light $c = 3 \times 10^8 \ m \ s^{-1}$ and wavelength $\lambda = 400 \ nm = 400 \times 10^{-9} \ m$.
Substituting the values: $\nu = \frac{3 \times 10^8 \ m \ s^{-1}}{400 \times 10^{-9} \ m} = 7.5 \times 10^{14} \ Hz$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 6.64 \times 10^{-27} \ kg$,and $v = 3 \times 10^3 \ ms^{-1}$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34}}{6.64 \times 10^{-27} \times 3 \times 10^3} \ m$.
$\lambda = \frac{6.63}{19.92} \times 10^{-34 - (-27) - 3} \ m$.
$\lambda \approx 0.333 \times 10^{-10} \ m = 0.0333 \times 10^{-9} \ m$.
Since $1 \ nm = 10^{-9} \ m$,we get $\lambda = 0.0333 \ nm$.

(C) The value of $(n+l)$ for the $3d$ orbital is calculated as follows:
For $3d$,$n=3$ and $l=2$,so $(n+l) = 3+2 = 5$.
Now,let us calculate the $(n+l)$ values for the given options:
$A) 4s$: $n=4, l=0 \implies (n+l) = 4+0 = 4$
$B) 3s$: $n=3, l=0 \implies (n+l) = 3+0 = 3$
$C) 4p$: $n=4, l=1 \implies (n+l) = 4+1 = 5$
$D) 2p$: $n=2, l=1 \implies (n+l) = 2+1 = 3$
Comparing these values,the $4p$ orbital has the same $(n+l)$ value of $5$ as the $3d$ orbital.

(C) The principal quantum number $n$ represents the shell number,and the azimuthal quantum number $l$ represents the subshell type.
For $l=0$,the subshell is $s$.
For $l=1$,the subshell is $p$.
For $l=2$,the subshell is $d$.
For $l=3$,the subshell is $f$.
Given $n=3$ and $l=2$,the orbital is $3d$.

(D) The $N$ shell corresponds to the principal quantum number $n = 4$.
For a given $n$,the possible values of the azimuthal quantum number $l$ range from $0$ to $n-1$.
Thus,for $n = 4$,$l = 0, 1, 2, 3$ (corresponding to $4s, 4p, 4d, 4f$ subshells).
The number of orbitals in a subshell is given by $2l + 1$.
For $l = 0$ $(4s)$: $2(0) + 1 = 1$ orbital.
For $l = 1$ $(4p)$: $2(1) + 1 = 3$ orbitals.
For $l = 2$ $(4d)$: $2(2) + 1 = 5$ orbitals.
For $l = 3$ $(4f)$: $2(3) + 1 = 7$ orbitals.
Total number of orbitals in $N$ shell $= 1 + 3 + 5 + 7 = 16$.
Alternatively,the total number of orbitals in a shell is given by $n^2 = 4^2 = 16$.

(D) The orbital designation is determined by the principal quantum number $n$ and the azimuthal quantum number $l$.
For $l=0$,the orbital is $s$.
For $l=1$,the orbital is $p$.
For $l=2$,the orbital is $d$.
For $l=3$,the orbital is $f$.
Given $n=4$ and $l=3$,the orbital is $4f$.

(C) According to the Heisenberg uncertainty principle,it is impossible to determine simultaneously the exact position and exact momentum of a microscopic particle like an electron.
Mathematically,it is expressed as $\Delta x \times \Delta p \geq \frac{h}{4\pi}$,where $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum.

(B) The electronic configuration of $N_2$ molecule ($14$ electrons) is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$.
The antibonding orbitals are those marked with an asterisk $(*)$.
These are $(\sigma^* 1s)^2$ and $(\sigma^* 2s)^2$.
Total number of electrons in antibonding orbitals $= 2 + 2 = 4$.

(A) An intensive property is independent of the amount of substance present in the system. Examples include viscosity,surface tension,and specific heat.
An extensive property depends on the amount of substance present in the system.
Internal energy $(U)$ is an extensive property because its value depends on the quantity of matter in the system.

(A) The correct answer is $A$.
Intensive properties are properties that do not depend on the amount or size of the substance and remain the same regardless of the quantity of the material present.
Extensive properties depend on the amount of substance.
$1$. Surface tension and viscosity are both intensive properties because they are independent of the amount of substance.
$2$. Mass,internal energy,and heat capacity are extensive properties as they depend on the amount of matter.
$3$. Temperature,boiling point,and specific heat are intensive properties.
Since both surface tension and viscosity are intensive,option $A$ is correct.

(B) The work done by the system during expansion is given by the formula: $W = -P_{ext} \Delta V$.
Given:
External pressure $P_{ext} = 2 \times 10^5 \ Nm^{-2}$
Change in volume $\Delta V = 200 \ cm^3 = 200 \times 10^{-6} \ m^3 = 2 \times 10^{-4} \ m^3$.
Substituting the values:
$W = -(2 \times 10^5 \ Nm^{-2}) \times (2 \times 10^{-4} \ m^3)$
$W = -40 \ J$.
Therefore,the work done by the system is $-40 \ J$.

(C) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = q + w$.
Here,heat is absorbed by the system,so $q = +605 \ J$.
Work is done by the system on the surroundings,so $w = -380 \ J$.
Substituting these values into the equation: $\Delta U = 605 \ J + (-380 \ J) = +225 \ J$.

(D) Given:
$q = 302.6 \ J$ (heat absorbed by the system is positive).
$P_{ext} = 2 \ Pa$.
$V_1 = 100 \ L$, $V_2 = 200 \ L$.
$\Delta V = V_2 - V_1 = 200 \ L - 100 \ L = 100 \ L$.
Since $1 \ L \cdot Pa = 1 \ J$ is not the standard conversion (usually $1 \ L \cdot atm = 101.3 \ J$), we use the given units directly:
Work done $W = -P_{ext} \Delta V = -(2 \ Pa) \times (100 \ L) = -200 \ J$.
According to the First Law of Thermodynamics:
$\Delta U = q + W$.
$\Delta U = 302.6 \ J + (-200 \ J) = 102.6 \ J$.
The closest value among the options is $100 \ J$.

(D) The first law of thermodynamics is a statement of the law of conservation of energy.
It states that energy cannot be created or destroyed; it can only be transferred or transformed from one form to another.
In the context of thermodynamics,the total energy of the universe (system + surroundings) remains constant.
Mathematically,for an isolated system,$\Delta U_{universe} = 0$,which implies that the energy of the universe is conserved.

(B) Heat is supplied to the system,so $q = 500 \ J$.
At constant volume,the change in volume $\Delta V = 0$.
Since work done $w = -P_{ext} \Delta V$,it follows that $w = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Substituting the values,$\Delta U = 500 \ J + 0 \ J = 500 \ J$.
Thus,$q = \Delta U = 500 \ J$ and $w = 0$.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
For a process at constant volume (isochoric process),the work done $w = -P_{ext} \cdot \Delta V = 0$.
Therefore,$\Delta U = q_v$,where $q_v$ is the heat of reaction at constant volume.

(A) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
$\Lambda_{0(NaBr)} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$
$\Lambda_{0(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{0(KBr)} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{0(KCl)} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$
To find $\Lambda_{0(NaBr)}$,we perform the operation:
$\Lambda_{0(NaBr)} = \Lambda_{0(NaCl)} + \Lambda_{0(KBr)} - \Lambda_{0(KCl)}$
$\Lambda_{0(NaBr)} = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

(A) The formula for molar conductivity is $\Lambda_{m} = \frac{1000 \times \kappa}{c}$.
Given,conductivity $\kappa = 6.07 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.005 \ M$.
Substituting the values:
$\Lambda_{m} = \frac{1000 \times 6.07 \times 10^{-4}}{0.005} \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\Lambda_{m} = \frac{0.607}{0.005} \ \Omega^{-1} \ cm^2 \ mol^{-1} = 121.4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(B) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\varrho)$. Thus,$\kappa = \frac{1}{\varrho}$.
For a given electrolytic cell,the relationship between conductivity $(\kappa)$ and conductance $(G)$ is given by $\kappa = G \cdot G^*$,where $G^*$ is the cell constant,defined as $\frac{l}{A}$.
Evaluating the options:
$A$: $\kappa = \frac{1}{\varrho}$ is the correct definition of conductivity.
$B$: $\kappa = G \cdot \frac{1}{a}$ is incorrect because the correct relation is $\kappa = G \cdot a$ (assuming $a$ is the cell constant).
$C$: $\kappa = \frac{1}{R} \cdot \frac{l}{A}$ is correct because $G = \frac{1}{R}$ and $\frac{l}{A}$ is the cell constant.
$D$: $\kappa = \Lambda_m \cdot c$ is correct based on the definition of molar conductivity,$\Lambda_m = \frac{\kappa}{c}$.
Therefore,the incorrect expression is $B$.

(B) The conductivity $(\kappa)$ is calculated using the formula: $\kappa = \frac{\text{Cell constant}}{R}$
Substituting the given values: $\kappa = \frac{1.1 \ cm^{-1}}{94.5 \ \Omega}$
$\kappa = 0.0116 \ \Omega^{-1} \ cm^{-1}$

(C) The molar conductivity of a weak electrolyte like $CH_3COOH$ does not vary linearly with the square root of concentration,as shown in the graph.
Unlike strong electrolytes,the curve for a weak electrolyte does not intersect the y-axis upon extrapolation.
Therefore,the molar conductivity at infinite dilution $(\Lambda_0)$ for weak electrolytes cannot be determined by the graphical extrapolation method.

(D) The cell constant $\left(\frac{l}{A}\right)$ is defined by the relationship: $\frac{l}{A} = \kappa \times R$
where $\kappa$ is the conductivity and $R$ is the resistance.
Given:
$\kappa = 1.90 \times 10^{-6} \ S \ cm^{-1}$
$R = 115 \ \Omega$
Calculation:
$\frac{l}{A} = (1.90 \times 10^{-6} \ S \ cm^{-1}) \times (115 \ \Omega) = 0.2185 \ cm^{-1}$
Rounding to three decimal places,we get $0.218 \ cm^{-1}$.
Thus,the correct option is $D$.

(C) $(1)$ Diamond: Diamond does not conduct electricity. It is an insulator because it has no free electrons or ions to carry charge. The carbon atoms in diamond form a strong covalent network,preventing the free flow of electrons.
$(2)$ Sulphur (solid): Sulphur in its solid form also does not conduct electricity. It is a non-metal and does not have free electrons or charged particles to facilitate conduction.
$(3)$ Molten $NaCl$: Molten $NaCl$ (sodium chloride in its liquid form) conducts electricity. When $NaCl$ is heated to its melting point,it dissociates into free sodium $(Na^+)$ and chloride ions $(Cl^-)$,which are able to move freely and carry an electric current.
$(4)$ Crystalline $NaCl$: Crystalline $NaCl$ (solid sodium chloride) does not conduct electricity. While it contains ions ($Na^+$ and $Cl^-$),they are not free to move in the solid state. For conduction,the ions need to be mobile,which happens only in the molten state.

(C) The conductivity $(\kappa)$ is calculated using the formula: $\kappa = \frac{\text{Cell constant}}{R}$.
Given: Cell constant = $1.32 \ cm^{-1}$,Resistance $(R)$ = $528 \ \Omega$.
$\kappa = \frac{1.32 \ cm^{-1}}{528 \ \Omega} = 0.0025 \ \Omega^{-1} \ cm^{-1}$.

(B) The relationship between conductivity $(k)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$k = \frac{G^*}{R}$
Therefore,the cell constant is:
$G^* = k \times R$
Given:
$k = 0.0012 \ \Omega^{-1} \ cm^{-1}$
$R = 600 \ \Omega$
Substituting the values:
$G^* = 0.0012 \ \Omega^{-1} \ cm^{-1} \times 600 \ \Omega = 0.72 \ cm^{-1}$

(D) $\because V = I R$ and $Q = I t$
Conductance $(G) = \frac{1}{R} = \frac{I}{V} = \frac{Q}{V t}$
Substituting with units,
$\therefore 1 \text{ Siemen} = \frac{1}{\Omega} = \Omega^{-1} = \frac{A}{V} = A V^{-1} = \frac{C}{V s} = C V^{-1} s^{-1}$
Since $\Omega$ is the unit of resistance,it is not equivalent to $1$ Siemen.

(A) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{1000 \times \kappa}{C}$
Given:
Conductivity $(\kappa)$ = $0.00250 \ \Omega^{-1} \ cm^{-1}$
Concentration $(C)$ = $0.02 \ M$
Substituting the values:
$\Lambda_m = \frac{1000 \times 0.00250}{0.02} \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\Lambda_m = \frac{2.5}{0.02} \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\Lambda_m = 125 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

(B) The formula for cell constant $(G^*)$ is given by:
$G^* = \kappa \times R$
Where:
- $\kappa$ is the conductivity $(S \ cm^{-1})$,
- $R$ is the resistance $(\Omega)$.
Given:
- $\kappa = 1.70 \times 10^{-4} \ S \ cm^{-1}$,
- $R = 100 \ \Omega$.
Calculation:
$G^* = (1.70 \times 10^{-4} \ S \ cm^{-1}) \times (100 \ \Omega) = 0.017 \ cm^{-1}$.
Therefore,the cell constant is $0.017 \ cm^{-1}$.
Hence,the correct option is $B$.

(D) saturated $KCl$ solution is not typically used as a standard solution for conductivity calibration.
The reason is that a saturated $KCl$ solution may have variable composition due to temperature changes,partial crystallization,and the difficulty in ensuring a uniform,reproducible concentration.
Thus,it does not have a precisely known or easily reproducible conductivity value compared to the other standard concentrations.
Conclusion: Out of the given options,the saturated $KCl$ solution (Option $D$) is not suitable for use as a standard solution in determining the cell constant because its concentration and thus its conductivity are not stable or easily reproducible.

(C) The conductivity $(\kappa)$ is given by the formula: $\kappa = \frac{\text{Cell constant}}{R}$
Given: Cell constant $= 0.9 \ cm^{-1}$ and $R = 6530 \ \Omega$.
Substituting the values: $\kappa = \frac{0.9 \ cm^{-1}}{6530 \ \Omega} = 1.378 \times 10^{-4} \ \Omega^{-1} \ cm^{-1} \approx 1.38 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$.

(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{1000 \times \kappa}{c}$
Given: $\Lambda_m = 101 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\kappa = 1.01 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$
Substituting the values: $101 = \frac{1000 \times 1.01 \times 10^{-2}}{c}$
$101 = \frac{10.1}{c}$
$c = \frac{10.1}{101} = 0.1 \ M$

(C) The formula for molar conductivity is $\Lambda_m = \frac{1000 \times \kappa}{C}$.
Given: $\kappa = 0.00216 \ \Omega^{-1} \ cm^{-1}$ and $C = 0.02 \ M$.
Substituting the values: $\Lambda_m = \frac{1000 \times 0.00216}{0.02} = 108.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(D) The formula for conductivity $(k)$ is given by: $k = \frac{1}{R} \times \frac{l}{A}$
Given: Resistance $(R)$ = $31.5 \ \Omega$ and Cell constant $(\frac{l}{A})$ = $0.315 \ cm^{-1}$
Substituting the values: $k = \frac{1}{31.5 \ \Omega} \times 0.315 \ cm^{-1}$
$k = 0.01 \ \Omega^{-1} \ cm^{-1}$

(B) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$\kappa = \frac{1}{R} \times G^*$
Given:
Resistance $(R)$ = $60 \ \Omega$
Conductivity $(\kappa)$ = $0.014 \ \Omega^{-1} \ cm^{-1}$
Substituting the values:
$0.014 \ \Omega^{-1} \ cm^{-1} = \frac{1}{60 \ \Omega} \times G^*$
$G^* = 0.014 \ \Omega^{-1} \ cm^{-1} \times 60 \ \Omega$
$G^* = 0.84 \ cm^{-1}$

(B) The correct answer is $B$. Dichlorodifluoromethane.
Freon-$12$,also known as dichlorodifluoromethane $(CCl_2F_2)$,is a commonly used refrigerant.
It was widely used in refrigeration and air conditioning systems for many years until its use was phased out due to concerns about ozone layer depletion.
Here is a quick explanation of the other options:
$-$ $A$. Monochloromonofluoromethane: This is not the correct structure for Freon-$12$.
$-$ $C$. Chlorodifluoromethane: This is Freon-$22$,not Freon-$12$.
$-$ $D$. Dichlorofluoromethane: This is not the correct chemical name for Freon-$12$.

(A) Nanowires are one-dimensional nanostructures in which two dimensions are in the nanoscale $(< 100 \ nm)$.
Quantum dots are zero-dimensional structures in which all three dimensions are in the nanoscale $(< 100 \ nm)$.
Microcapsules are generally larger than the nanoscale range.
Nanorings are typically considered zero-dimensional or specific nanostructures depending on their geometry,but nanowires are the standard example of having two dimensions in the nanoscale.

(A) The Scanning Electron Microscope $(SEM)$ is the instrument used to examine the surface structure of materials at high magnification.
It works by scanning a focused beam of electrons across the surface of a sample.
The electrons interact with the atoms on the surface,producing signals that are used to create an image of the surface's topography,morphology,and composition.

(A) One-dimensional nanostructures are materials that have two dimensions in the nanometer range and one dimension that is much larger, such as $Nanowires$.
$Nanoparticles$ are zero-dimensional, $Thin films$ are two-dimensional, and $Quantum dots$ are zero-dimensional nanostructures.

(C) $PS$ (Polystyrene) is believed to leach styrene,which is a possible human carcinogen,into food when used as a household plastic.

(B) An allylic alcohol is one where the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond $(C=C-C-OH)$.
$A$ $1^{\circ}$ allylic alcohol has the $-OH$ group attached to a primary carbon ($CH_2$ group) which is adjacent to a double bond.
$1.$ $CH_2=CH-CH(CH_3)-OH$: The $-OH$ is on a secondary carbon,so it is a $2^{\circ}$ allylic alcohol.
$2.$ $CH_2=CH-CH_2-OH$: The $-OH$ is on a primary carbon adjacent to the double bond,so it is a $1^{\circ}$ allylic alcohol.
$3.$ $CH_2=CH-C(CH_3)_2-OH$: The $-OH$ is on a tertiary carbon,so it is a $3^{\circ}$ allylic alcohol.
$4.$ $CH_3-CH=CH-CH_2-CH_2-OH$: The $-OH$ is on a carbon that is not adjacent to the double bond,so it is not an allylic alcohol.
Therefore,the correct option is $B$.

(D) Dihydric phenols are compounds containing two hydroxyl $(-OH)$ groups attached to a benzene ring.
$1$. Catechol has two $-OH$ groups at the $1,2$-positions.
$2$. Resorcinol has two $-OH$ groups at the $1,3$-positions.
$3$. Quinol (or hydroquinone) has two $-OH$ groups at the $1,4$-positions.
$4$. Phloroglucinol has three $-OH$ groups at the $1,3,5$-positions (trihydric phenol).
$5$. Pyrogallol has three $-OH$ groups at the $1,2,3$-positions (trihydric phenol).
Therefore,the pair of dihydric phenols is Catechol and Quinol.

(C) Hydroquinone is a dihydroxy benzene derivative where the two hydroxyl $(-OH)$ groups are attached at the $1$ and $4$ positions of the benzene ring.
According to $IUPAC$ nomenclature,this structure is named as Benzene-$1, 4$-diol.

(D) Cyclohexylamine consists of a cyclohexane ring $(C_6H_{11})$ attached to an amino group $(-NH_2)$.
Combining these,the molecular formula is $C_6H_{11} + NH_2 = C_6H_{13}N$.

(C) The classification of amines is based on the number of alkyl or aryl groups attached to the nitrogen atom.
Primary $(1^{\circ})$ amines contain the $-NH_2$ group.
Secondary $(2^{\circ})$ amines contain the $-NH-$ group.
Tertiary $(3^{\circ})$ amines contain the $-N-$ group (where the nitrogen is bonded to three carbon atoms).
In butan$-1$-amine $(CH_3CH_2CH_2CH_2NH_2)$,the nitrogen is attached to only one carbon atom,making it a primary $(1^{\circ})$ amine. Therefore,it is not a tertiary amine.

(C) In the carbinol system,alcohols are named as derivatives of methyl alcohol,which is referred to as $CH_3OH$ or carbinol.
For isobutyl alcohol,the structure is $(CH_3)_2CH-CH_2OH$.
This can be viewed as a carbinol group $(-CH_2OH)$ attached to an isopropyl group $((CH_3)_2CH-)$.
Therefore,the name according to the carbinol system is isopropyl carbinol.

(C) trihydric alcohol is an alcohol that contains three hydroxyl $(-OH)$ groups in its molecule.
$1$. Quinol (Benzene$-1,4-$diol) has two $-OH$ groups (dihydric).
$2$. Catechol (Benzene$-1,2-$diol) has two $-OH$ groups (dihydric).
$3$. Glycerol $(CH_2(OH)-CH(OH)-CH_2(OH))$ has three $-OH$ groups (trihydric).
$4$. Resorcinol (Benzene$-1,3-$diol) has two $-OH$ groups (dihydric).
Therefore,the correct answer is Glycerol.

(A) dihydric compound is one that contains two hydroxyl $(-OH)$ groups in its structure.
$1$. Crotonyl alcohol $(CH_3-CH=CH-CH_2OH)$ contains only one $-OH$ group,so it is a monohydric alcohol.
$2$. Resorcinol $(C_6H_4(OH)_2)$ contains two $-OH$ groups attached to the benzene ring,so it is a dihydric phenol.
$3$. Ethylene glycol $(HO-CH_2-CH_2-OH)$ contains two $-OH$ groups,so it is a dihydric alcohol.
$4$. Quinol $(C_6H_4(OH)_2)$ contains two $-OH$ groups attached to the benzene ring,so it is a dihydric phenol.
Therefore,Crotonyl alcohol is not a dihydric compound.

(C) The structures of the given acids are as follows:
$1$. Adipic acid: $HOOC-(CH_2)_4-COOH$ (Dicarboxylic acid)
$2$. Glutaric acid: $HOOC-(CH_2)_3-COOH$ (Dicarboxylic acid)
$3$. Valeric acid: $CH_3-CH_2-CH_2-CH_2-COOH$ (Monocarboxylic acid)
$4$. Malonic acid: $HOOC-CH_2-COOH$ (Dicarboxylic acid)
Therefore,Valeric acid is not a dicarboxylic acid.

(C) In a secondary benzylic alcohol,the $-OH$ group is attached to an $sp^3$ hybridized carbon atom,which is directly bonded to an aromatic ring and two other carbon atoms (or one carbon and one hydrogen atom,specifically a secondary carbon).
$1$. In option $A$,the $-OH$ group is attached to a primary carbon $(C_6H_5CH_2OH)$,making it a primary benzylic alcohol.
$2$. In option $B$,the $-OH$ group is attached to a tertiary carbon $(C_6H_5C(CH_3)_2OH)$,making it a tertiary benzylic alcohol.
$3$. In option $C$,the $-OH$ group is attached to a secondary carbon $(C_6H_5CH(OH)CH_3)$,which is bonded to the aromatic ring,making it a secondary benzylic alcohol.
$4$. In option $D$,the $-OH$ group is attached to a tertiary carbon $(C_6H_5C(OH)(CH_3)CH_2CH_3)$,making it a tertiary benzylic alcohol.
Therefore,the correct answer is $C$.

(B) Aldehydes are organic compounds containing a functional group with the structure $-CHO$.
The general molecular formula for saturated aliphatic aldehydes is $C_n H_{2n} O$,where $n$ is the number of carbon atoms $(n \ge 1)$.
For example,if $n=1$,the formula is $CH_2O$ (Formaldehyde,$HCHO$).
If $n=2$,the formula is $C_2H_4O$ (Acetaldehyde,$CH_3CHO$).

(D) An allylic halide is a compound in which the halogen atom is attached to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C-C-X)$.
In option $A$,$B$,and $C$,the halogen atom $(X)$ is attached to the carbon atom directly adjacent to the double bond.
In option $D$,the structure is $CH_3-CH=CH-CH_2-CH_2-X$. Here,the halogen atom is attached to a carbon atom that is two positions away from the double bond,making it a homoallylic halide,not an allylic halide.
Therefore,the correct option is $D$.

(D) tricarboxylic acid is an organic compound that contains three carboxylic acid $(-COOH)$ functional groups.
$1$. Propionic acid $(CH_3CH_2COOH)$ is a monocarboxylic acid.
$2$. Oxalic acid $(HOOC-COOH)$ is a dicarboxylic acid.
$3$. Malonic acid $(HOOC-CH_2-COOH)$ is a dicarboxylic acid.
$4$. Citric acid $(HOOC-CH_2-C(OH)(COOH)-CH_2-COOH)$ contains three $-COOH$ groups,making it a tricarboxylic acid.
Therefore,the correct option is $D$.

(D) molecule is chiral if it contains a chiral center,which is a carbon atom bonded to four different groups.
$1$. $2-$Chloropropane: $CH_3-CHCl-CH_3$ (The central carbon is bonded to two identical methyl groups,so it is achiral).
$2$. $2-$Chloro$-2-$methylbutane: $CH_3-CCl(CH_3)-CH_2-CH_3$ (The central carbon is bonded to two identical methyl groups,so it is achiral).
$3$. $3-$Chloro$-3-$methylbutane: $CH_3-CH_2-CCl(CH_3)-CH_3$ (The central carbon is bonded to two identical methyl groups,so it is achiral).
$4$. $2-$Chloropentane: $CH_3-CHCl-CH_2-CH_2-CH_3$ (The $C-2$ carbon is bonded to four different groups: $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$,so it is chiral).
Therefore,the correct option is $D$.

(C) chiral carbon atom is defined as a carbon atom that is bonded to four different groups or atoms.
Let us analyze the structures:
$A$. $2-$Iodopropane: $CH_3-CH(I)-CH_3$. The central carbon is attached to two identical $-CH_3$ groups,so it is achiral.
$B$. $2-$Iodo$-2-$methylbutane: $CH_3-C(I)(CH_3)-CH_2-CH_3$. The central carbon is attached to two identical $-CH_3$ groups,so it is achiral.
$C$. $2-$Iodo$-3-$methylbutane: $CH_3-CH(I)-CH(CH_3)_2$. The carbon at position $2$ is attached to four different groups: $-H$,$-I$,$-CH_3$,and $-CH(CH_3)_2$. Thus,it is a chiral molecule.
$D$. $3-$Iodopentane: $CH_3-CH_2-CH(I)-CH_2-CH_3$. The carbon at position $3$ is attached to two identical $-CH_2CH_3$ groups,so it is achiral.
Therefore,the correct option is $C$.

(B) The structure of threose is $CHO-CH(OH)-CH(OH)-CH_2OH$.
$A$ chiral carbon atom is a carbon atom bonded to four different groups.
In the structure of threose,the two central carbon atoms (at positions $2$ and $3$) are each bonded to four different groups: $-H$,$-OH$,$-CHO$ (or $-CH_2OH$),and the other chiral carbon.
Therefore,there are $2$ chiral carbon atoms in the threose molecule,as indicated by the asterisks $(*)$ in the structure.

(B) chiral molecule is one that contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$n$-Butyl bromide: $CH_3-CH_2-CH_2-CH_2Br$ (No chiral carbon).
$sec$-Butyl bromide: $CH_3-CH(Br)-CH_2-CH_3$. The carbon atom marked with an asterisk $({}^*)$ is bonded to four different groups: $-H, -CH_3, -Br, -CH_2CH_3$. Thus,it is chiral.
Isobutyl bromide: $(CH_3)_2CH-CH_2Br$ (No chiral carbon).
$tert$-Butyl bromide: $(CH_3)_3CBr$ (No chiral carbon).
Therefore,$sec$-butyl bromide is the chiral molecule.

(C) molecule is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$A$. $2-$Bromopropane: $CH_3-CH(Br)-CH_3$. The central carbon is bonded to two identical methyl groups. It is achiral.
$B$. $2-$Bromo$-2-$methylbutane: $CH_3-C(Br)(CH_3)-CH_2-CH_3$. The central carbon is bonded to two identical methyl groups. It is achiral.
$C$. $2-$Bromo$-3-$methylbutane: $CH_3-CH(Br)-CH(CH_3)_2$. The carbon at position $2$ is bonded to four different groups: $-H$,$-Br$,$-CH_3$,and $-CH(CH_3)_2$. Thus,it is chiral.
$D$. $3-$Bromopentane: $CH_3-CH_2-CH(Br)-CH_2-CH_3$. The central carbon is bonded to two identical ethyl groups. It is achiral.
Therefore,the correct option is $C$.

(D) The chemical formula for copper pyrites is $CuFeS_2$.
Thus,the elements present in copper pyrites are Copper $(Cu)$,Iron $(Fe)$,and Sulfur $(S)$.

(B) The minerals and their chemical formulas are as follows:
$A$. Siderite: $FeCO_3$
$B$. Calamine: $ZnCO_3$
$C$. Chalcocite: $Cu_2S$
$D$. Limonite: $2Fe_2O_3 \cdot 3H_2O$
Among the given options,Calamine $(ZnCO_3)$ is a mineral of zinc.

(A) $Cleveite$ is a variety of uraninite and it contains uranium,which is a radioactive element.

(A) The mineral $CuFeS_2$ is known as Chalcopyrite,which is a major ore of copper.
$Zincite$ is an ore of zinc $(ZnO)$.
$Limonite$ is an ore of iron $(Fe_2O_3 \cdot 3H_2O)$.
$Siderite$ is an ore of iron $(FeCO_3)$.

(D) The Swartz reaction is a method used to prepare alkyl fluorides by heating alkyl chlorides or alkyl bromides in the presence of metallic fluorides such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
The general reaction is: $R-X + AgF \rightarrow R-F + AgX$ (where $X = Cl, Br$).
Thus,the product obtained is an alkyl fluoride.

(D) For a given alkyl group,the boiling point of alkyl halides increases as the size and mass of the halogen atom increase.
This is due to the increase in the magnitude of van der Waals forces with an increase in the size and mass of the halogen.
Thus,the boiling point follows the order: $CH_3I > CH_3Br > CH_3Cl > CH_3F$.
Therefore,$CH_3I$ has the highest boiling point.

(C) Alcohols react with phosphorus pentachloride $(PCl_5)$ to produce alkyl chlorides,phosphorus oxychloride $(POCl_3)$,and hydrogen chloride $(HCl)$.
The balanced chemical equation is:
$R-OH + PCl_5 \longrightarrow R-Cl + POCl_3 + HCl$

(D) is methyl cyanide $(CH_3-CN)$ formed by the nucleophilic substitution of $CH_3-I$ with $KCN$.
The reduction of $CH_3-CN$ using $Na / C_2H_5OH$ is known as the Mendius reduction,which produces ethylamine $(CH_3-CH_2-NH_2)$ as product $B$.
$CH_3-I + KCN \rightarrow CH_3-CN (A) + KI$
$CH_3-CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3-CH_2-NH_2 (B)$

(B) Alkyl fluorides are prepared by heating alkyl chlorides or bromides with metal fluorides such as $AgF$,$Hg_2F_2$,$AsF_3$,$SbF_3$,etc. This reaction is known as the Swartz reaction.
$R-Cl + AgF \longrightarrow R-F + AgCl \downarrow$