MHT CET 2021 Chemistry Question Paper with Answer and Solution

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

(D) According to Charles's Law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ at constant pressure.
Given: $V_1 = 2 \ dm^3$,$T_1 = 0^{\circ} C = 273 \ K$.
The temperature is decreased by $272^{\circ} C$,so the new temperature $T_2 = 273 \ K - 272 \ K = 1 \ K$.
Substituting the values: $\frac{2}{273} = \frac{V_2}{1}$.
Therefore,$V_2 = \frac{2}{273} \ dm^3$.

(C) Given: $P = 5 \ atm$,$T = 300 \ K$,$mass = 22 \ g$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Number of moles $(n)$ of $CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{22 \ g}{44 \ g \ mol^{-1}} = 0.5 \ mol$.
Using the ideal gas equation: $PV = nRT$.
$V = \frac{nRT}{P} = \frac{0.5 \times 0.0821 \times 300}{5}$.
$V = \frac{12.315}{5} = 2.463 \ L$.
Since $1 \ L = 1 \ dm^3$,the volume is $2.46 \ dm^3$.

(B) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 1.013 \times 10^5 \ Nm^{-2}$,$V_1 = 2.27 \ L$.
Final volume $V_2 = 4540 \ mL = 4.540 \ L$.
Substituting the values: $1.013 \times 10^5 \times 2.27 = P_2 \times 4.540$.
$P_2 = \frac{1.013 \times 10^5 \times 2.27}{4.540} = 0.5065 \times 10^5 \ Nm^{-2} = 5.065 \times 10^4 \ Nm^{-2}$.

(A) According to Charles's Law,for a fixed amount of gas at constant pressure,the volume is directly proportional to the temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 3.4 \ L$,$T_1 = 298 \ K$,$V_2 = 6.8 \ L$.
Substituting the values: $\frac{3.4}{298} = \frac{6.8}{T_2}$.
Solving for $T_2$: $T_2 = \frac{6.8 \times 298}{3.4} = 2 \times 298 = 596 \ K$.

(C) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 2 \ atm$,$V_1 = 250 \ mL$,$P_2 = 2.5 \ atm$.
Substituting the values: $2 \times 250 = 2.5 \times V_2$.
$V_2 = \frac{2 \times 250}{2.5} = \frac{500}{2.5} = 200 \ mL$.

(A) The critical temperature of water is approximately $647 \ K$ $(647.096 \ K)$.

(A) According to Bohr's postulate,the frequency $(v)$ of the emitted or absorbed radiation during a transition between two stationary states with energy difference $\Delta E$ is given by the equation:
$h v = \Delta E$
Where $h$ is Planck's constant.
Rearranging for frequency:
$v = \frac{\Delta E}{h}$

(B) The energy of an electron in a stationary state is given by the formula: $E_n = -R_H \times \frac{1}{n^2}$
Where $R_H$ is the Rydberg constant,which is $2.18 \times 10^{-18} \ J$.
For $n=2$:
$E_2 = -2.18 \times 10^{-18} \times \frac{1}{2^2}$
$E_2 = -2.18 \times 10^{-18} \times \frac{1}{4}$
$E_2 = -0.545 \times 10^{-18} \ J$

(B) The relationship between frequency $(v)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $v = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^8 \ m/s$ and $\lambda = 460 \ nm = 460 \times 10^{-9} \ m$.
Substituting the values: $v = \frac{3 \times 10^8 \ m/s}{460 \times 10^{-9} \ m} = 6.52 \times 10^{14} \ Hz$.
Thus,the frequency is approximately $6.5 \times 10^{14} \ Hz$.

(C) The relationship between speed of light $(c)$,frequency $(\nu)$,and wavelength $(\lambda)$ is given by $c = \nu \times \lambda$.
Given: $c = 3 \times 10^8 \ m/s$ and $\nu = 50 \ Hz$.
Rearranging the formula for wavelength: $\lambda = \frac{c}{\nu}$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{50} = 0.06 \times 10^8 \ m = 6 \times 10^6 \ m$.

(C) The frequency $v$ is given by the formula $v = \frac{c}{\lambda}$.
Given wavelength $\lambda = 480 \ nm = 480 \times 10^{-9} \ m$.
The speed of light $c = 3 \times 10^8 \ m/s$.
Substituting the values: $v = \frac{3 \times 10^8 \ m/s}{480 \times 10^{-9} \ m} = 6.25 \times 10^{14} \ Hz$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$.
Given $E_n = -3.4 \ eV$,we have $-3.4 = -13.6 / n^2$,which gives $n^2 = 4$,so $n = 2$.
According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{n h}{2 \pi}$.
Substituting $n = 2$,we get $L = \frac{2 h}{2 \pi} = \frac{h}{\pi}$.

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.
Since $p = mv$ and kinetic energy $E_K = \frac{p^2}{2m}$,we have $p = \sqrt{2mE_K}$.
Thus,$\lambda = \frac{h}{\sqrt{2mE_K}}$.
For an electron in a hydrogen atom,the velocity $v$ in the $n^{\text{th}}$ orbit is given by $v_n = \frac{v_0 Z}{n}$.
Since $p = mv$,the momentum $p_n$ is proportional to $\frac{1}{n}$.
Therefore,$\lambda_n = \frac{h}{p_n} \propto n$.
When the electron jumps from the third excited state $(n = 4)$ to the ground state $(n = 1)$,the ratio of the wavelengths is $\frac{\lambda_{n=1}}{\lambda_{n=4}} = \frac{1}{4}$.

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.
Since the kinetic energy $K = qV = \frac{p^2}{2m}$,the momentum is $p = \sqrt{2mqV}$.
Thus,$\lambda = \frac{h}{\sqrt{2mqV}}$.
For the electron: $\lambda = \frac{h}{\sqrt{2mVq}}$.
For the proton of mass $M$ accelerated through $9 \ V$: $\lambda_{p} = \frac{h}{\sqrt{2M(9 \ V)q}} = \frac{h}{3\sqrt{2MVq}}$.
Dividing the two expressions: $\frac{\lambda}{\lambda_{p}} = \frac{\sqrt{2M(9 \ V)q}}{\sqrt{2mVq}} = 3 \sqrt{\frac{M}{m}}$.
Therefore,$\lambda_{p} = \frac{\lambda}{3} \sqrt{\frac{m}{M}}$.

(A) The principal quantum number $n=4$ indicates the $4th$ shell.
The azimuthal quantum number $\ell=3$ corresponds to the $f$-subshell.
Therefore,the orbital is $4f$.

(B) For a given azimuthal quantum number $\ell$,the magnetic quantum number $m_\ell$ can take values ranging from $-\ell$ to $+\ell$,including zero.
Therefore,the total number of possible values for the magnetic quantum number is $(2 \ell + 1)$.

(D) According to the Pauli exclusion principle,no two electrons in an atom can have the same set of all four quantum numbers.
For two electrons in the same orbital,the principal $(n)$,azimuthal $(l)$,and magnetic $(m_l)$ quantum numbers are identical.
Therefore,they must be distinguished by their spin quantum number $(m_s)$,which can be $+\frac{1}{2}$ or $-\frac{1}{2}$.

(A) The azimuthal quantum number $\ell=2$ corresponds to the $d$-subshell.
Each orbital can hold a maximum of $2$ electrons.
The number of orbitals in a subshell is given by $(2\ell+1)$.
For $\ell=2$,the number of orbitals $= 2(2)+1 = 5$.
Therefore,the maximum number of electrons $= 5 \times 2 = 10$ electrons.

(C) Extensive properties are those properties of a system that depend on the amount of matter (size and mass) present in the system.
$Mass$,$Volume$,and $Internal \ energy$ are extensive properties because they depend on the quantity of matter.
$Pressure$ is an intensive property because it is independent of the amount of matter present in the system.

(B) Intensive properties are those that are independent of the amount of matter present in the system.
Extensive properties depend on the quantity of matter present.
$1$. Surface tension: Intensive property.
$2$. Heat capacity: Extensive property (as it depends on the mass of the substance).
$3$. Viscosity: Intensive property.
$4$. Temperature: Intensive property.
Therefore,heat capacity is $NOT$ an intensive property.

(A) Internal energy $(U)$ is a state function.
Therefore,the change in internal energy $(\Delta U)$ depends only on the initial and final states of the system,not on the path taken.

(C) Given: $V_1 = 10 \ L$,$V_2 = 5 \ L$,$W = 1730 \ J$,$T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
For isothermal reversible compression,the work done on the system is given by the formula:
$W = -2.303 \ nRT \log_{10} \left(\frac{V_2}{V_1}\right)$
Since work is done on the gas,$W$ is positive in this context $(W = 1730 \ J)$:
$1730 = -2.303 \times n \times 8.314 \times 300 \times \log_{10} \left(\frac{5}{10}\right)$
$1730 = -2.303 \times n \times 8.314 \times 300 \times \log_{10} (0.5)$
$1730 = -2.303 \times n \times 8.314 \times 300 \times (-0.3010)$
$1730 = n \times 5744.14 \times 0.3010$
$1730 = n \times 1729$
$n = \frac{1730}{1729} \approx 1 \ mol$.

(A) Work of compression occurs when the volume of the system decreases during a reaction,which corresponds to a negative change in the number of moles of gas $(\Delta n_g < 0)$.
In option $A$: $NH_{3(g)} + HCl_{(g)} \longrightarrow NH_4Cl_{(s)}$,the change in moles of gas is $\Delta n_g = 0 - (1 + 1) = -2$. Since the volume decreases significantly as gaseous reactants form a solid,this reaction shows work of compression.
In option $B$: $\Delta n_g = 3 - 1 = +2$ (Expansion).
In option $C$: $\Delta n_g = (2 + 1) - 2 = +1$ (Expansion).
In option $D$: $\Delta n_g = 1 - 0 = +1$ (Expansion).
Therefore,the correct reaction is $NH_{3(g)} + HCl_{(g)} \longrightarrow NH_4Cl_{(s)}$.

(D) For an insulated container,the process is adiabatic,so $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
The work done during expansion against a constant external pressure is given by $w = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 2.5 \ atm$,$V_1 = 2.5 \ L$,and $V_2 = 4.5 \ L$.
$\Delta V = V_2 - V_1 = 4.5 \ L - 2.5 \ L = 2.0 \ L$.
$w = -2.5 \ atm \times 2.0 \ L = -5.0 \ atm \cdot L$.
Since $1 \ atm \cdot L = 101.3 \ J$,we have $w = -5.0 \times 101.3 \ J = -506.5 \ J$.
Therefore,$\Delta U = 0 + (-506.5 \ J) = -506.5 \ J$.

(D) The formula for work done during compression is $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
Given $P_{ext} = 4.05 \ bar$,$V_1 = 2.5 \times 10^{-2} \ m^3$,and $V_2 = 1.3 \times 10^{-2} \ m^3$.
Substituting the values: $W = -4.05 \ bar \times (1.3 \times 10^{-2} - 2.5 \times 10^{-2}) \ m^3$.
$W = -4.05 \times (-1.2 \times 10^{-2}) \ bar \cdot m^3 = 4.86 \times 10^{-2} \ bar \cdot m^3$.
Since $1 \ bar = 10^5 \ Pa$ and $1 \ Pa \cdot m^3 = 1 \ J$,we convert the units:
$W = 4.86 \times 10^{-2} \times 10^5 \ J = 4860 \ J$.

(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \cdot \Delta V$.
Given $W = -6 \ dm^3 \ bar$ and $\Delta V = V_2 - V_1 = (20 - 15) \ dm^3 = 5 \ dm^3$.
Substituting the values: $-6 \ dm^3 \ bar = -P_{ext} \cdot (5 \ dm^3)$.
$P_{ext} = \frac{6}{5} \ bar = 1.2 \ bar$.
Since $1 \ bar = 10^5 \ Pa$,the external pressure is $1.2 \times 10^5 \ Pa$.

(A) The work done by a gas is given by $W = -P_{ext} \cdot \Delta V$.
Since the process occurs at constant volume,$\Delta V = 0$,which implies $W = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that the heat supplied $Q = x \ J$ and $W = 0$,we get $\Delta U = x \ J$.
Therefore,$Q = \Delta U = x \ J$ and $W = 0$.

(D) For an isothermal reversible expansion process,the work done is given by the formula:
$W = -2.303 \ nRT \log \frac{P_1}{P_2}$
Given:
$n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$P_1 = 210 \ kPa$,$P_2 = 105 \ kPa$
Substituting the values:
$W = -2.303 \times 1 \times 8.314 \times 300 \times \log \frac{210}{105}$
$W = -2.303 \times 8.314 \times 300 \times \log 2$
Using $\log 2 \approx 0.3010$:
$W = -2.303 \times 8.314 \times 300 \times 0.3010$
$W \approx -1729 \ J$
The magnitude of work done is $|W| = 1729 \ J$.

(A) The expansion of a gas against no external pressure (opposing force),such as expansion into a vacuum,is defined as $free \ expansion$.

(A) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
For a monoatomic ideal gas,the degrees of freedom $f = 3$.
The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.
Thus,$\gamma - 1 = \frac{2}{3}$.
Since the volume $V$ is proportional to the length $L$ of the gas column ($V = A \times L$,where $A$ is the constant cross-sectional area),we have $\frac{V_1}{V_2} = \frac{L_1}{L_2}$.
Substituting this into the adiabatic relation: $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = \left(\frac{L_1}{L_2}\right)^{2/3}$.

(A) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that heat $Q$ is provided to the system,$Q = +x \ kJ = 1000x \ J$.
Since work $W$ is done on the system,$W = +y \ J$.
Therefore,the change in internal energy is $\Delta U = 1000x + y \ J$.

(A) The work done during expansion is given by the formula $W = -P_{ext} \Delta V$.
Since the gas expands,work is done by the system,so $W = -5 \ bar \ dm^3$.
Given $P_{ext} = 2.5 \ bar$,$V_2 = 4.5 \ L$,and $V_1 = x \ L$.
Since $1 \ L = 1 \ dm^3$,we have:
$-5 \ bar \ dm^3 = -2.5 \ bar \times (4.5 \ L - x \ L)$.
Dividing both sides by $-2.5 \ bar$:
$2 = 4.5 - x$.
Rearranging for $x$:
$x = 4.5 - 2 = 2.5 \ L$.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
Heat absorbed by the system is positive,so $q = +54 \ J$.
Work done by the system on the surroundings is negative,so $w = -238 \ J$.
Substituting these values into the equation:
$\Delta U = 54 \ J + (-238 \ J) = -184 \ J$.
Thus,the change in internal energy of the system is $-184 \ J$.

(C) According to the first law of thermodynamics,$\Delta U = q + W$.
Here,the heat absorbed by the system is $q = +701 \ J$.
The work done by the system on the surroundings is $W = -394 \ J$.
Substituting these values into the equation:
$\Delta U = 701 \ J + (-394 \ J) = 307 \ J$.
Therefore,the change in internal energy of the system is $307 \ J$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Since the system gives out heat,$q = -x \ J$.
Since the system does work on its surroundings,$w = -y \ J$.
Therefore,$\Delta U = (-x) + (-y) = -x - y \ J$.

(A) The formula for work done during expansion against constant external pressure is $W = -P_{\text{ext}} \Delta V$.
Given: $W = -5.09 \ kJ = -5090 \ J$,$V_1 = 2 \times 10^{-2} \ m^3$,$V_2 = 3 \times 10^{-2} \ m^3$.
Change in volume $\Delta V = V_2 - V_1 = (3 - 2) \times 10^{-2} \ m^3 = 1 \times 10^{-2} \ m^3$.
Substituting the values: $-5090 = -P_{\text{ext}} \times (1 \times 10^{-2})$.
$P_{\text{ext}} = \frac{5090}{10^{-2}} = 509000 \ Nm^{-2} = 5.09 \times 10^5 \ Nm^{-2}$.

(A) According to the $I^{st}$ law of thermodynamics,$\Delta U = q + w$.
Given: Heat absorbed by the system,$q = +710 \ J$.
Change in internal energy,$\Delta U = +460 \ J$.
Substituting these values into the equation: $460 \ J = 710 \ J + w$.
$w = 460 \ J - 710 \ J$.
$w = -250 \ J$.
Since the work done $w$ is negative,it indicates that work is done by the system.

(A) For an isochoric process,the volume remains constant,so $\Delta V = 0$.
Since the work done is given by $W = -P \Delta V$,for an isochoric process,$W = 0$.
The first law of thermodynamics is expressed as $\Delta U = Q + W$.
Substituting $W = 0$ and $Q = Q_v$ (heat at constant volume),we get $\Delta U = Q_v$.

(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
Given: $P_{ext} = 2.40 \times 10^5 \ Pa$,$V_2 = 2.2 \times 10^{-3} \ m^3$,$W = -0.048 \ kJ = -48 \ J$.
Substituting the values: $-48 \ J = -2.4 \times 10^5 \ Pa \times (2.2 \times 10^{-3} \ m^3 - V_1)$.
Dividing both sides by $-2.4 \times 10^5 \ Pa$: $\frac{-48}{-2.4 \times 10^5} = 2.2 \times 10^{-3} - V_1$.
$20 \times 10^{-5} = 2.2 \times 10^{-3} - V_1$.
$0.2 \times 10^{-3} = 2.2 \times 10^{-3} - V_1$.
$V_1 = 2.2 \times 10^{-3} - 0.2 \times 10^{-3} = 2.0 \times 10^{-3} \ m^3$.

(D) According to the first law of thermodynamics,$\Delta U = q + W$.
Work done on the system is positive,so $W = +62 \ J$.
Heat transferred to the surrounding is negative,so $q = -128 \ J$.
Substituting these values into the equation:
$\Delta U = -128 \ J + 62 \ J = -66 \ J$.
Therefore,the internal energy change is $-66 \ J$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
Therefore,the difference is $\Delta H - \Delta U = \Delta n_g RT$.
For the reaction: $C_2H_{6(g)} + 3.5O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,the number of moles of gaseous products is $2$ and gaseous reactants is $1 + 3.5 = 4.5$.
$\Delta n_g = n_{g(products)} - n_{g(reactants)} = 2 - 4.5 = -2.5$.
Temperature $T = 25 + 273 = 298 \ K$.
$\Delta H - \Delta U = -2.5 \times 8.314 \times 298 = -6193.93 \ J$.
Converting to $kJ$: $-6193.93 \ J = -6.19 \ kJ \approx -6.2 \ kJ$.

(D) The given reaction is: $H_{2(g)} + Cl_{2(g)} \longrightarrow 2 HCl_{(g)}, \Delta H = -194 \ kJ$
The heat of formation $(\Delta H_f)$ is defined as the enthalpy change when $1 \ mole$ of a substance is formed from its constituent elements in their standard states.
For the reaction: $\Delta H_{reaction} = 2 \Delta H_{f(HCl)} - [\Delta H_{f(H_2)} + \Delta H_{f(Cl_2)}]$
Since $H_{2(g)}$ and $Cl_{2(g)}$ are elements in their standard states,their $\Delta H_f = 0$.
$-194 \ kJ = 2 \Delta H_{f(HCl)} - 0 - 0$
$\Delta H_{f(HCl)} = -194 / 2 \ kJ \ mol^{-1} = -97 \ kJ \ mol^{-1}$

(C) The combustion reaction for acetylene is: $C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \longrightarrow 2CO_{2(g)} + H_2O_{(\ell)}$; $\Delta_{c}H^{\circ} = -1300 \ kJ \ mol^{-1}$.
The molar mass of acetylene $(C_2H_2)$ is $(2 \times 12) + (2 \times 1) = 26 \ g \ mol^{-1}$.
Since $26 \ g$ of acetylene releases $1300 \ kJ$ of energy upon complete combustion,the enthalpy change for $39 \ g$ is calculated as:
$\Delta H = \left( \frac{-1300 \ kJ \ mol^{-1}}{26 \ g \ mol^{-1}} \right) \times 39 \ g = -50 \times 39 = -1950 \ kJ$.

(C) The balanced chemical equation for the formation of $2$ moles of ammonia is:
$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$
We know the relation $\Delta H - \Delta U = \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species:
$\Delta n_g = (\text{Sum of stoichiometric coefficients of gaseous products}) - (\text{Sum of stoichiometric coefficients of gaseous reactants})$
$\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting the value of $\Delta n_g$ in the relation:
$\Delta H - \Delta U = -2 RT$.

(A) The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g / mol$.
Number of moles of $CH_4$ in $24 \ g$ is calculated as $n = \frac{24 \ g}{16 \ g / mol} = 1.5 \ mol$.
The enthalpy of formation for $1 \ mol$ of $CH_4$ is given as $-75 \ kJ / mol$.
Therefore,the enthalpy change for the formation of $1.5 \ mol$ of $CH_4$ is $\Delta H = 1.5 \ mol \times (-75 \ kJ / mol) = -112.5 \ kJ$.

(C) The formation reaction for $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
The enthalpy of formation is calculated using bond enthalpies as: $\Delta H_f = \sum BE_{\text{reactants}} - \sum BE_{\text{products}}$
$\Delta H_f = [\frac{1}{2} BE_{(N \equiv N)} + \frac{3}{2} BE_{(H-H)}] - [3 BE_{(N-H)}]$
Substituting the given values:
$\Delta H_f = [\frac{1}{2} \times 941 + \frac{3}{2} \times 436] - [3 \times 389]$
$\Delta H_f = [470.5 + 654] - 1167$
$\Delta H_f = 1124.5 - 1167 = -42.5 \ kJ/mol$

(D) The chemical equation for the formation of $NO_{2(g)}$ is: $\frac{1}{2} N_{2(g)} + O_{2(g)} \longrightarrow NO_{2(g)}$.
In an endothermic process,heat is absorbed from the surroundings during the reaction.
By definition,for an endothermic reaction,the change in enthalpy is positive,i.e.,$\Delta H > 0$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H - \Delta U = \Delta n_g RT$
Here,$\Delta n_g$ is the change in the number of moles of gaseous species.
For the reaction: $2 \ C_6H_{6(\ell)} + 15 \ O_{2(g)} \rightarrow 12 \ CO_{2(g)} + 6 \ H_2O_{(\ell)}$
$\Delta n_g = (n_{g, \text{products}}) - (n_{g, \text{reactants}}) = 12 - 15 = -3$
Now,substituting the values: $\Delta H - \Delta U = -3 \times 8.314 \ J \ K^{-1} \ mol^{-1} \times 298 \ K$
$= -7432.7 \ J = -7.432 \ kJ$

(A) The volume $V$ of a cylinder is given by the formula $V = \pi r^2 h$,where $r$ is the radius of the base and $h$ is the height of the water level.
Given that the radius $r = 3 \ m$ is constant,we differentiate the volume with respect to time $t$:
$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$
We are given the rate of change of volume $\frac{dV}{dt} = 36 \ m^3 / min$.
Substituting the known values into the equation:
$36 = \pi (3)^2 \frac{dh}{dt}$
$36 = 9 \pi \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{36}{9 \pi} = \frac{4}{\pi} \ m / min$
Thus,the water level is rising at the rate of $\frac{4}{\pi} \ m / min$.

(A) The formula for molar conductivity is $\Lambda_m = \frac{1000 \times \kappa}{C}$,where $\kappa$ is the conductivity and $C$ is the molar concentration.
Rearranging for conductivity: $\kappa = \frac{\Lambda_m \times C}{1000}$.
Given: $\Lambda_m = 106 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $C = 0.1 \ M$.
Substituting the values: $\kappa = \frac{106 \times 0.1}{1000} \ \Omega^{-1} \ cm^{-1}$.
$\kappa = \frac{10.6}{1000} \ \Omega^{-1} \ cm^{-1} = 1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$.

(A) Specific conductance or conductivity $(k)$ is defined as the reciprocal of specific resistance or resistivity $(\rho)$.
Mathematically,$k = \frac{1}{\rho}$.
Since resistivity $(\rho)$ is the resistance offered by a unit volume of the conductor,conductivity represents the ease with which current flows through that unit volume.
Therefore,conductivity is inversely proportional to resistivity.

(A) The formula for molar conductivity is $\Lambda_m = \frac{k \times 1000}{M}$.
Given $\Lambda_m = 2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $M = 0.4 \ M$.
Substituting the values: $2.5 \times 10^2 = \frac{k \times 1000}{0.4}$.
$k = \frac{2.5 \times 10^2 \times 0.4}{1000} = \frac{100}{1000} = 0.1 \ \Omega^{-1} \ cm^{-1}$.
Resistivity $\rho$ is the reciprocal of conductivity $k$: $\rho = \frac{1}{k} = \frac{1}{0.1} = 10 \ \Omega \ cm$.

(D) The relationship between molar conductivity $(\Lambda_m)$,conductivity $(k)$,and molarity $(M)$ is given by the formula: $\Lambda_m = \frac{k \times 1000}{M}$.
Given: $\Lambda_m = 230 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $k = 0.0115 \ \Omega^{-1} \ cm^{-1}$.
Substituting the values into the formula: $230 = \frac{0.0115 \times 1000}{M}$.
Rearranging for $M$: $M = \frac{0.0115 \times 1000}{230}$.
$M = \frac{11.5}{230} = 0.05 \ M$.

(B) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$\kappa = \frac{1}{R} \times G^*$
Given:
$\kappa = 200 \ \Omega^{-1} \ cm^{-1}$
$G^* = 1 \ cm^{-1}$
Substituting the values:
$200 = \frac{1}{R} \times 1$
$R = \frac{1}{200} \ \Omega$
$R = 0.005 \ \Omega = 5 \times 10^{-3} \ \Omega$

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(k)$ is given by the formula: $\Lambda_m = \frac{k \times 1000}{M}$.
Given: $\Lambda_m = 230 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $M = 0.04 \ M$.
Substituting the values: $230 = \frac{k \times 1000}{0.04}$.
Solving for $k$: $k = \frac{230 \times 0.04}{1000}$.
$k = \frac{9.2}{1000} = 9.2 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

(C) Kohlrausch's law of independent migration of ions is primarily used to determine the limiting molar conductivity $(\Lambda^0_m)$ of weak electrolytes,which cannot be obtained by extrapolation of the $\Lambda_m$ versus $\sqrt{C}$ plot.
$CH_3COOH$ is a weak electrolyte.
$KCl$,$Na_2SO_4$,and $HCl$ are strong electrolytes whose limiting molar conductivities can be determined directly by extrapolation.

(B) The formula for conductivity $(\kappa)$ is given by: $\kappa = \frac{1}{R} \times \frac{\ell}{A}$
Where $\frac{\ell}{A}$ is the cell constant.
Given: $R = 645 \ \Omega$ and $\text{Cell constant} = 1.29 \ cm^{-1}$.
Substituting the values: $\kappa = \frac{1}{645} \times 1.29$
$\kappa = 0.002 \ \Omega^{-1} \ cm^{-1}$
$\kappa = 2 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(C) Specific conductance or Conductivity $(\kappa)$ is defined as the conductance of $1 \ cm^3$ of an electrolytic solution.
Mathematically,$\kappa = \frac{1}{\rho}$,where $\rho$ is the resistivity.
Since $R = \rho \frac{\ell}{A}$,we have $\frac{1}{\rho} = \frac{1}{R} \cdot \frac{\ell}{A}$,which implies $\kappa = G \times G^*$,where $G$ is conductance and $G^*$ is the cell constant.
The $SI$ unit of conductivity is $S \ cm^{-1}$ or $ohm^{-1} \ cm^{-1}$.

(C) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^*)$ is given by: $\kappa = \frac{1}{R} \times G^*$.
Rearranging for the cell constant: $G^* = \kappa \times R$.
Given: $\kappa = 1.2 \ S \ m^{-1}$ and $R = 30 \ \Omega$.
$G^* = 1.2 \ S \ m^{-1} \times 30 \ \Omega = 36 \ m^{-1}$.
Since $1 \ m = 100 \ cm$,then $1 \ m^{-1} = 10^{-2} \ cm^{-1}$.
$G^* = 36 \times 10^{-2} \ cm^{-1} = 0.36 \ cm^{-1}$.

(D) The formula for molar conductivity is $\Lambda_{m} = \frac{k \times 1000}{M}$.
Given:
Conductivity $(k) = 0.0112 \ \Omega^{-1} \ cm^{-1}$
Molarity $(M) = 0.04 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0112 \times 1000}{0.04} = \frac{11.2}{0.04} = 280 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(C) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given:
Conductivity $\kappa = 0.0627 \ S \ cm^{-1}$
Molarity $M = 0.3 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0627 \times 1000}{0.3}$
$\Lambda_{m} = \frac{62.7}{0.3} = 209 \ S \ cm^2 \ mol^{-1}$.

(D) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given conductivity $\kappa = 2.67 \times 10^{-4} \ S \ cm^{-1}$ and molarity $M = 0.012 \ M$.
Substituting the values: $\Lambda_{m} = \frac{2.67 \times 10^{-4} \times 1000}{0.012}$.
$\Lambda_{m} = \frac{0.267}{0.012} = 22.25 \ S \ cm^2 \ mol^{-1}$.
Rounding to the nearest provided option,the correct answer is $22.2 \ S \ cm^2 \ mol^{-1}$.

(D) Specific conductance or conductivity $(\kappa)$ is defined as the reciprocal of specific resistance $(\rho)$: $\kappa = \frac{1}{\rho}$.
Since $R = \rho \frac{\ell}{A}$,we have $\frac{1}{\rho} = \frac{1}{R} \times \frac{\ell}{A}$.
Thus,$\kappa = G \times G^*$,where $G$ is conductance and $G^*$ is the cell constant.
Therefore,the conductivity of a solution is defined as the conductance offered by $1 \ cm^3$ of an electrolytic solution.
The $SI$ unit of $\kappa$ is $S \ cm^{-1}$ or $ohm^{-1} \ cm^{-1}$.

(D) The relationship between molar conductivity $(\Lambda_{m})$ and conductivity $(k)$ is given by the formula:
$\Lambda_{m} = \frac{k \times 1000}{M}$
Given:
$\Lambda_{m} = 412.3 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$
$M = 0.02 \ M$
Substituting the values:
$412.3 = \frac{k \times 1000}{0.02}$
Solving for $k$:
$k = \frac{412.3 \times 0.02}{1000}$
$k = 8.246 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(A) The structure of $Benzene-1,2-diol$ consists of a benzene ring with two hydroxyl $(-OH)$ groups attached at adjacent positions ($1$ and $2$).
This compound is commonly known as $Catechol$ (or $Pyrocatechol$).

(B) primary $(1^{\circ})$ amine is a compound where the nitrogen atom is attached to only one alkyl or aryl group,represented by the general formula $R-NH_2$.
$1$. Ethyl methyl propyl amine: $CH_3-CH_2-N(CH_3)-CH_2-CH_2-CH_3$. The nitrogen is attached to three alkyl groups,so it is a $3^{\circ}$ amine.
$2$. Hexamethylene diamine: $H_2N-(CH_2)_6-NH_2$. Each nitrogen atom is attached to only one carbon atom,so it is a $1^{\circ}$ amine.
$3$. Diphenyl amine: $Ph-NH-Ph$. The nitrogen is attached to two phenyl groups,so it is a $2^{\circ}$ amine.
$4$. $N,N$-Dimethyl aniline: $Ph-N(CH_3)_2$. The nitrogen is attached to one phenyl group and two methyl groups,so it is a $3^{\circ}$ amine.
Therefore,Hexamethylene diamine is the primary amine.

(B) An allylic alcohol is one where the $-OH$ group is attached to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C)$.
$1$. $A$ secondary allylic alcohol is one where the carbon atom attached to the $-OH$ group is bonded to two other carbon atoms.
$2$. Let us analyze the options:
- Option $A$: $CH_2=CH-C(CH_3)_2-OH$ is a tertiary allylic alcohol.
- Option $B$: $CH_2=CH-CH(CH_3)-OH$ is a secondary allylic alcohol because the carbon attached to $-OH$ is bonded to one $-CH_3$ group and one $-CH=CH_2$ group.
- Option $C$: $CH_2=CH-CH_2-OH$ is a primary allylic alcohol.
- Option $D$: $CH_3-CH=CH-CH_2-OH$ is a primary allylic alcohol.
Therefore,the correct option is $B$.

(B) Catechol is a common name for a dihydroxybenzene derivative where the two hydroxyl $(-OH)$ groups are attached to adjacent carbon atoms on the benzene ring.
According to $IUPAC$ nomenclature,the benzene ring is numbered such that the substituents receive the lowest possible locants.
For catechol,the $-OH$ groups are at positions $1$ and $2$.
Therefore,the $IUPAC$ name is Benzene-$1,2$-diol.

(A) The boiling point of haloalkanes increases with an increase in the molecular mass due to the greater magnitude of $Van \ der \ Waals$ forces of attraction.
Since $CH_3F$ (Fluoromethane) has the lowest molecular mass among the given compounds,it experiences the weakest $Van \ der \ Waals$ forces and thus has the lowest boiling point.

(A) dicarboxylic acid is an organic compound containing two carboxylic acid functional groups $(-COOH)$.
Glutaric acid $(HOOC-(CH_2)_3-COOH)$ and Malonic acid $(HOOC-CH_2-COOH)$ are both dicarboxylic acids.
Valeric acid,Caproic acid,and Propionic acid are monocarboxylic acids.
Therefore,the correct pair is Glutaric acid and Malonic acid.

(D) benzylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom that is directly attached to an aromatic ring.
In $4-$Iodotoluene,the iodine is attached directly to the benzene ring (aryl halide).
In $1-$Iodo$-2-$phenylethane,the iodine is attached to a carbon atom that is separated from the ring by another $CH_2$ group.
In Iodobenzene,the iodine is attached directly to the benzene ring (aryl halide).
In Iodophenylmethane (also known as benzyl iodide),the iodine is attached to a $CH_2$ group which is directly attached to the benzene ring. Thus,it is a benzylic halide.

(B) benzylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom that is directly attached to an aromatic ring (benzene ring).
In $1-$chloro$-2-$phenylbutane $(CH_3-CH_2-CH(C_6H_5)-CH_2Cl)$,the chlorine atom is attached to a carbon atom that is not directly bonded to the benzene ring.
Therefore,it is not a benzylic halide.

(C) tertiary amine is symmetrical if the nitrogen atom is attached to three identical alkyl or aryl groups.
In $N(CH_3)_3$ (trimethylamine),the nitrogen is attached to three methyl groups,but this is not provided as an option.
Looking at the provided images:
Option $A$ is not a tertiary amine.
Option $B$ is $N,N$-dimethylaniline (asymmetric).
Option $C$ is triphenylamine,where the nitrogen atom is attached to three identical phenyl groups. Therefore,it is a symmetrical tertiary amine.
Option $D$ is a primary amine.

(C) haloalkyne is an organic compound where a halogen atom is directly bonded to an $sp$ hybridized carbon atom of an alkyne group.
Example: $CH \equiv C-Cl$ (chloroethyne).
In this structure,the carbon atom bonded to the chlorine is $sp$ hybridized.

(D) primary $(1^{\circ})$ amine is an amine where the nitrogen atom is attached to only one alkyl or aryl group,represented by the general formula $R-NH_2$.
$1$. $N$-methyl aniline $(C_6H_5NHCH_3)$ is a secondary amine.
$2$. $N$-phenylbenzenamine $((C_6H_5)_2NH)$ is a secondary amine.
$3$. Methyl phenylamine $(C_6H_5NHCH_3)$ is a secondary amine.
$4$. Isopropyl amine $((CH_3)_2CHNH_2)$ has the $-NH_2$ group attached to a carbon atom,which is bonded to two other carbon atoms. Although it is a secondary carbon,the amine itself is a primary amine because the nitrogen is attached to only one carbon atom ($R-NH_2$ structure).
Therefore,the correct option is $D$.

(C) The structure of $Crotonyl$ alcohol is $CH_3-CH=CH-CH_2-OH$.
In this molecule,the $-OH$ group is attached to a $sp^3$ hybridized carbon atom,which is adjacent to a carbon-carbon double bond $(C=C)$.
Compounds in which the $-OH$ group is attached to a $sp^3$ hybridized carbon atom next to a double bond are classified as allylic alcohols.
Therefore,$Crotonyl$ alcohol is an example of an allylic alcohol.

(C) dihydric alcohol (or diol) contains two hydroxyl $(-OH)$ groups attached to the carbon chain.
$1$. Catechol is $1,2$-dihydroxybenzene (dihydric).
$2$. Hydroquinone is $1,4$-dihydroxybenzene (dihydric).
$3$. Resorcinol is $1,3$-dihydroxybenzene (dihydric).
$4$. Phloroglucinol is $1,3,5$-trihydroxybenzene,which is a trihydric phenol,not a dihydric alcohol.
Therefore,the correct answer is $C$.

(C) secondary $(2^{\circ})$ amine is a compound where the nitrogen atom is bonded to two carbon atoms and one hydrogen atom $(R_2NH)$.
$1$. Hexane-$1,6$-diamine is a primary amine $(R-NH_2)$.
$2$. $N,N$-Dimethylbenzenamine is a tertiary amine $(R_3N)$.
$3$. $N$-methylbenzenamine $(C_6H_5-NH-CH_3)$ has the nitrogen atom attached to a phenyl group and a methyl group,making it a secondary amine.
$4$. Prop-$2$-en-$1$-amine is a primary amine $(R-NH_2)$.
Therefore,$N$-methylbenzenamine is the correct answer.

(C) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C-C-X)$.
In $3$-chloropropene $(CH_2=CH-CH_2Cl)$,the chlorine atom is attached to the $sp^3$ hybridized carbon atom which is adjacent to the $C=C$ double bond.
Therefore,$3$-chloropropene is an allylic halide.

(B) An amino group is defined as the primary amine functional group,represented by the formula $-NH_2$.
$A$. $N$-Phenylbenzenamine is a secondary amine $(Ph_2NH)$.
$B$. $4$-Bromoaniline is a primary aromatic amine $(Br-C_6H_4-NH_2)$,which contains the $-NH_2$ group.
$C$. $N,N$-Dimethylbenzenamine is a tertiary amine $(Ph-N(CH_3)_2)$.
$D$. $N$-Methylmethanamine is a secondary amine $(CH_3-NH-CH_3)$.
Therefore,$4$-Bromoaniline contains the amino group.

(D) vinylic halide is defined as a compound in which the halogen atom is directly bonded to a carbon atom that is part of a carbon-carbon double bond $(C=C)$.
This carbon atom is $sp^2$ hybridized.
For example,in vinyl chloride $(CH_2=CHCl)$,the chlorine atom is attached to an $sp^2$ hybridized carbon atom of an aliphatic chain.

(B) The acidic strength of carboxylic acids is directly proportional to the $-I$ (inductive) effect of the substituent attached to the alpha-carbon.
$Acidic \text{ } strength \propto -I \text{ } effect$.
The order of $-I$ effect for halogens is $F > Cl > Br > I$ due to the difference in electronegativity.
Therefore,the order of acidic strength is $Fluoroacetic \text{ } acid > Chloroacetic \text{ } acid > Bromoacetic \text{ } acid > Iodoacetic \text{ } acid$.
Thus,Fluoroacetic acid is the strongest acid.

(C) The stability of the conjugate acid (ammonium ion) in aqueous solution depends on the extent of hydrogen bonding with water molecules.
More hydrogen atoms attached to the nitrogen atom allow for more extensive hydrogen bonding with water,leading to greater stabilization through solvation.
In $NH_4^+$,there are $4$ hydrogen atoms available for hydrogen bonding.
In $RNH_3^+$,there are $3$ hydrogen atoms.
In $R_2NH_2^+$,there are $2$ hydrogen atoms.
In $R_3NH^+$,there is only $1$ hydrogen atom.
Therefore,$NH_4^+$ is stabilized to the greatest extent due to solvation.

(B) The basicity of amines in the aqueous phase depends on the combined effect of inductive effect,solvation effect,and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
However,among the given options,$(C_2H_5)N(CH_3)_2$ is a tertiary amine with an ethyl group,which provides a stronger inductive effect ($+I$ effect) compared to methyl groups,making it more basic than the others listed.

(B) The basicity of amines is inversely proportional to their $pK_{b}$ values.
In the gas phase or non-polar solvents,the basicity order is $3^{\circ} > 2^{\circ} > 1^{\circ} > NH_3$.
However,in aqueous solution,the basicity is determined by a combination of inductive effect,solvation effect,and steric hindrance.
For ethylamines in aqueous solution,the order of basicity is $(CH_3CH_2)_2NH > CH_3CH_2NH_2 > (CH_3CH_2)_3N > C_6H_5NH_2$.
Since $(CH_3CH_2)_2NH$ is the most basic,it will have the lowest $pK_{b}$ value.

(C) The basic strength of amines in the aqueous phase depends on the inductive effect,solvation effect,and steric hindrance.
For aliphatic amines,the secondary amine $(CH_3)_2NH$ exhibits the highest basic strength due to the combined effect of the $+I$ effect of two methyl groups and favorable solvation,which outweighs the steric hindrance compared to tertiary amines.

(C) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In $Phenylmethanamine$ $(C_6H_5CH_2NH_2)$,the lone pair is localized on the nitrogen atom as it is not in conjugation with the benzene ring.
In $Benzenamine$ $(C_6H_5NH_2)$,$N$-Methylaniline,and $N, N$-Dimethylaniline,the lone pair on the nitrogen atom is involved in resonance with the benzene ring,which decreases its availability for protonation.
Among the aromatic amines,the presence of electron-donating alkyl groups on the nitrogen atom increases the electron density on the nitrogen,making them stronger bases than $Benzenamine$.
Therefore,$Benzenamine$ has the least available lone pair due to resonance and the absence of electron-donating groups,making it the weakest base among the given options.

(D) In $Aniline$ $(C_6H_5NH_2)$,the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring.
This delocalization of the lone pair makes it less available for protonation,thereby making $Aniline$ the weakest base among the given options.

(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 $C$.

(A) compound is optically inactive if it does not contain a chiral center (an asymmetric carbon atom bonded to four different groups).
$1.$ $2-$Chloro-$2-$methylbutane: The $C-2$ carbon is bonded to two identical methyl groups,so it is not chiral. Thus,it is optically inactive.
$2.$ $3-$Chlorohexane: The $C-3$ carbon is bonded to $-H$,$-Cl$,$-CH_2CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.
$3.$ $2-$Chloro-$3-$methylbutane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH(CH_3)_2$. Since all four groups are different,it is chiral and optically active.
$4.$ $2-$Chloropentane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.
Therefore,$2-$Chloro-$2-$methylbutane is the optically inactive compound.

(B) $Baryte$ is $BaSO_4$.
$Cryolite$ is $Na_3 AlF_6$.
$Galena$ is $PbS$.
$Epsom$ salt is $MgSO_4 \cdot 7 H_2 O$.
Therefore,the correct pair is $Cryolite$ - $Na_3 AlF_6$.

(A) The $S_{N}1$ reaction is a unimolecular nucleophilic substitution reaction where the rate-determining step is the formation of a carbocation intermediate.
Since the nucleophile attacks only after the formation of the carbocation,the rate of the $S_{N}1$ reaction is independent of the concentration and strength of the nucleophile.
$A$ more powerful nucleophile actually favours the $S_{N}2$ mechanism,where the nucleophile attacks the substrate simultaneously with the leaving group departure.
Therefore,the statement that a more powerful nucleophile favours $S_{N}1$ is incorrect.

(B) The reaction of alkyl halides with metallic fluorides like $AgF$ is known as the Swarts reaction.
In this reaction,the halogen atom (in this case,$Br$) is replaced by a fluorine atom.
$tert$-butyl bromide is $(CH_3)_3C-Br$.
When it reacts with $AgF$,the $Br$ atom is replaced by $F$ to form $2$-Fluoro-$2$-methylpropane,which is $(CH_3)_3C-F$.

(B) The reaction of haloalkanes with $AgNO_2$ (silver nitrite) yields nitroalkanes as the major product because $AgNO_2$ is a covalent compound,and the nitrogen atom acts as the nucleophilic center.
In contrast,ionic nitrites like $NaNO_2$ or $KNO_2$ primarily yield alkyl nitrites $(R-ONO)$ because the oxygen atom is more nucleophilic in these ionic species.
Therefore,for the conversion of chloroethane $(CH_3CH_2Cl)$ to nitroethane $(CH_3CH_2NO_2)$,the reagent $A$ is silver nitrite $(AgNO_2)$.

(B) Chlorobenzene undergoes electrophilic aromatic substitution (chlorination) in the presence of anhydrous $FeCl_3$ as a Lewis acid catalyst.
The chlorine atom on the benzene ring is ortho- and para-directing due to the resonance effect.
However,the para-isomer ($1,4-$dichlorobenzene) is the major product due to less steric hindrance compared to the ortho-isomer ($1,2-$dichlorobenzene).

(B) The reaction of $2-$Chlorobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction,specifically $E2$).
According to $Saytzeff's$ rule,the major product is the more substituted alkene.
$CH_3-CH_2-CHCl-CH_3 \xrightarrow{Alc. KOH} CH_3-CH=CH-CH_3$ ($But-2-ene$,major product) $+ CH_3-CH_2-CH=CH_2$ ($But-1-ene$,minor product).
Therefore,$But-2-ene$ is the major product.

(B) The reaction proceeds as follows:
$1$. $CH_3Br$ reacts with $AgNO_2$ to form nitromethane $(A)$ as the major product: $CH_3Br + AgNO_2 \longrightarrow CH_3NO_2 + AgBr$.
$2$. Nitromethane $(CH_3NO_2)$ is then reduced by $Sn / HCl$ (a reducing agent) to form methylamine $(B)$: $CH_3NO_2 + 6[H] \xrightarrow{Sn / HCl} CH_3NH_2 + 2H_2O$.
Therefore,product $B$ is $CH_3NH_2$.

(A) The $S_N1$ mechanism proceeds via the formation of a planar carbocation intermediate.
Because the carbocation is planar,the nucleophile can attack from either the front side or the back side.
This leads to both inversion and retention of configuration,resulting in a racemic mixture if the substrate is chiral.
Backside attack is a characteristic feature of $S_N2$ mechanisms,not $S_N1$.
Therefore,option $A$ is $NOT$ a feature of $S_N1$ mechanisms.

(A) The $Wurtz-Fittig$ reaction involves the coupling of an aryl halide $(C_6H_5Cl)$ and an alkyl halide $(CH_3Cl)$ in the presence of sodium metal in dry ether to form an alkylbenzene $(C_6H_5-CH_3)$.
The reaction is: $C_6H_5Cl + CH_3Cl + 2 \ Na \xrightarrow[dry]{ether} C_6H_5-CH_3 + 2 \ NaCl$.
Option $A$ represents this reaction.