AIEEE 2010 Chemistry Question Paper with Answer and Solution

(D) The potential energy function is $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$.
At equilibrium,the force $F = -\frac{dU}{dx} = 0$.
$F = -(\frac{-12a}{x^{13}} + \frac{6b}{x^7}) = \frac{12a}{x^{13}} - \frac{6b}{x^7} = 0$.
$\frac{12a}{x^{13}} = \frac{6b}{x^7} \Rightarrow x^6 = \frac{2a}{b} \Rightarrow x = (\frac{2a}{b})^{1/6}$.
At $x = \infty$,$U(\infty) = 0$.
At equilibrium,$U_{\text{equilibrium}} = \frac{a}{(2a/b)^2} - \frac{b}{(2a/b)} = \frac{ab^2}{4a^2} - \frac{b^2}{2a} = \frac{b^2}{4a} - \frac{2b^2}{4a} = -\frac{b^2}{4a}$.
The dissociation energy $D = U(\infty) - U_{\text{equilibrium}} = 0 - (-\frac{b^2}{4a}) = \frac{b^2}{4a}$.

(C) The initial nucleus is represented as $_{Z}X^{A}$.
When an $\alpha$-particle $(_{2}He^{4})$ is emitted,the mass number decreases by $4$ and the atomic number decreases by $2$.
For $3$ $\alpha$-particles,the change is: $\Delta A = 3 \times 4 = 12$ and $\Delta Z = 3 \times 2 = 6$.
After $3$ $\alpha$ emissions,the nucleus becomes $_{Z-6}Y^{A-12}$.
When a positron $(_{+1}e^{0})$ is emitted,the mass number remains unchanged and the atomic number decreases by $1$.
For $2$ positrons,the change is: $\Delta A = 2 \times 0 = 0$ and $\Delta Z = 2 \times 1 = 2$.
After $2$ positron emissions,the final nucleus is $_{Z-6-2}Y^{A-12} = _{Z-8}Y^{A-12}$.
Number of protons $(P)$ = $Z - 8$.
Number of neutrons $(N)$ = Mass number - Atomic number = $(A - 12) - (Z - 8) = A - Z - 4$.
The ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 4}{Z - 8}$.

(D) Given,linear charge density $\lambda = \frac{q}{\pi r}$.
Consider a small element of the ring subtending an angle $d\theta$ at the center. The charge on this element is $dq = \lambda (r d\theta) = \frac{q}{\pi r} (r d\theta) = \frac{q}{\pi} d\theta$.
The electric field due to this element at the center $O$ is $dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q d\theta}{\pi r^2}$.
Due to symmetry,the horizontal components of the electric field cancel out,and the vertical components add up.
The vertical component of the field is $dE_y = dE \cos\theta$ (directed along $-\hat{j}$).
Integrating from $-\pi/2$ to $\pi/2$:
$E = \int_{-\pi/2}^{\pi/2} \frac{1}{4\pi\varepsilon_0} \frac{q d\theta}{\pi r^2} \cos\theta (-\hat{j})$
$E = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} \int_{-\pi/2}^{\pi/2} \cos\theta d\theta$
$E = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} [\sin\theta]_{-\pi/2}^{\pi/2}$
$E = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} [1 - (-1)] = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} (2) = -\frac{q}{2\pi^2\varepsilon_0 r^2} \hat{j}$.

(C) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.
Given $V_1 = V$ and $V_2 = 32V$,we have $T_1 V^{\gamma - 1} = T_2 (32V)^{\gamma - 1}$.
This simplifies to $T_1 = (32)^{\gamma - 1} T_2$.
For a diatomic gas,the adiabatic index $\gamma = 7/5$,so $\gamma - 1 = 2/5$.
Substituting this,$T_1 = (32)^{2/5} T_2 = (2^5)^{2/5} T_2 = 2^2 T_2 = 4 T_2$.
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Substituting $T_1 = 4T_2$,we get $\eta = 1 - \frac{T_2}{4T_2} = 1 - 0.25 = 0.75$.

(D) Energy required to break one mole of $Cl_2$ molecules is $242 \, kJ \, mol^{-1} = 242 \times 10^3 \, J \, mol^{-1}.$
Energy required to break a single $Cl-Cl$ bond is $E = \frac{242 \times 10^3}{6.02 \times 10^{23}} \, J.$
Using the relation $E = \frac{hc}{\lambda}$,the wavelength is $\lambda = \frac{hc}{E}.$
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34} \, J \cdot s \times 3 \times 10^8 \, m \cdot s^{-1} \times 6.02 \times 10^{23} \, mol^{-1}}{242 \times 10^3 \, J \cdot mol^{-1}}.$
$\lambda \approx 4.94 \times 10^{-7} \, m = 494 \times 10^{-9} \, m = 494 \, nm.$

(B) Ionisation energy $(IE)$ is the energy required to move an electron from the ground state to infinity.
$IE = E_{\infty} - E_{1} = 0 - E_{1} = -E_{1}$
Therefore,$E_{1}$ of $He^{+} = -19.6 \times 10^{-18} \, J \, atom^{-1}$.
Energy of a hydrogen-like species at state $n$ is given by $(E_{n})_{species} = (E_{1})_{H} \times \frac{Z^{2}}{n^{2}}$.
For $He^{+}$ $(Z=2, n=1)$: $(E_{1})_{He^{+}} = (E_{1})_{H} \times 2^{2} = -19.6 \times 10^{-18} \, J \, atom^{-1}$.
So,$(E_{1})_{H} = \frac{-19.6 \times 10^{-18}}{4} \, J \, atom^{-1}$.
For $Li^{2+}$ $(Z=3, n=1)$: $(E_{1})_{Li^{2+}} = (E_{1})_{H} \times 3^{2} = \frac{-19.6 \times 10^{-18}}{4} \times 9 = -4.41 \times 10^{-17} \, J \, atom^{-1}$.

(D) All the given ions $(O^{2-}, F^{-}, Na^{+}, Mg^{2+}, Al^{3+})$ are isoelectronic,meaning they all have $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
$O^{2-}$ $(Z=8)$,$F^{-}$ $(Z=9)$,$Na^{+}$ $(Z=11)$,$Mg^{2+}$ $(Z=12)$,$Al^{3+}$ $(Z=13)$.
Since the nuclear charge increases from $O^{2-}$ to $Al^{3+}$,the ionic radius decreases in the order: $O^{2-} > F^{-} > Na^{+} > Mg^{2+} > Al^{3+}$.

(D) At equilibrium,the water vapour exerts a pressure equal to its saturated vapour pressure.
Using the ideal gas equation,$PV = nRT$,where $P = 3170 \ Pa$,$V = 1.0 \ dm^3 = 1.0 \times 10^{-3} \ m^3$,$T = 300 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$n = \frac{PV}{RT} = \frac{3170 \times 1.0 \times 10^{-3}}{8.314 \times 300} \ mol$.
$n = \frac{3.17}{2494.2} \ mol \approx 1.27 \times 10^{-3} \ mol$.

(B) The reaction for the formation of $2 \ mol$ of $NH_3$ is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
The enthalpy change for this reaction is $\Delta H = 2 \times \Delta_f H^{\circ}(NH_3) = 2 \times (-46.0) = -92 \ kJ \ mol^{-1}$.
Using bond enthalpies,$\Delta H = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products}$.
Given bond energies (as enthalpy of formation from atoms is negative of bond dissociation energy): $BE(N \equiv N) = 712 \ kJ \ mol^{-1}$ and $BE(H-H) = 436 \ kJ \ mol^{-1}$.
Let $x$ be the bond enthalpy of $N-H$ bond. There are $6$ $N-H$ bonds in $2 \ mol$ of $NH_3$.
$-92 = [BE(N \equiv N) + 3 \times BE(H-H)] - 6x$.
$-92 = [712 + 3 \times 436] - 6x$.
$-92 = [712 + 1308] - 6x$.
$-92 = 2020 - 6x$.
$6x = 2020 + 92 = 2112$.
$x = 2112 / 6 = +352 \ kJ \ mol^{-1}$.

(B) At equilibrium,the Gibbs free energy change $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium we have $\Delta H - T_e \Delta S = 0$,which gives $T_e = \frac{\Delta H}{\Delta S}$.
For a reaction to be spontaneous,the condition is $\Delta G < 0$.
Substituting the expression for $\Delta G$,we get $\Delta H - T \Delta S < 0$,which implies $\Delta H < T \Delta S$.
Dividing by $\Delta S$ (since $\Delta S > 0$),we get $T > \frac{\Delta H}{\Delta S}$.
Since $T_e = \frac{\Delta H}{\Delta S}$,the condition for spontaneity is $T > T_e$.

(A) An acid is a substance that donates a proton $(H^{+})$.
In reaction $(i)$,$H_2PO_4^-$ is formed as a product,so it is not acting as an acid.
In reaction $(ii)$,$H_2PO_4^- + H_2O \rightarrow HPO_4^{2-} + H_3O^{+}$,$H_2PO_4^-$ donates a proton to $H_2O$,thus it acts as an acid.
In reaction $(iii)$,$H_2PO_4^- + OH^{-} \rightarrow H_2O + HPO_4^{2-}$,$H_2PO_4^-$ donates a proton to $OH^{-}$,thus it also acts as an acid.
Therefore,$H_2PO_4^-$ acts as an acid in both $(ii)$ and $(iii)$.

(C) For the first dissociation step: $H_2CO_3(aq) \rightleftharpoons HCO_3^-(aq) + H^+(aq)$.
Given $K_1 = 4.2 \times 10^{-7}$ and initial concentration $C = 0.034 \ M$.
Since $K_1$ is very small,$[H^+] \approx [HCO_3^-] = \sqrt{K_1 \times C} = \sqrt{4.2 \times 10^{-7} \times 0.034} \approx 1.195 \times 10^{-4} \ M$.
For the second dissociation step: $HCO_3^-(aq) \rightleftharpoons CO_3^{2-}(aq) + H^+(aq)$.
$K_2 = \frac{[CO_3^{2-}][H^+]}{[HCO_3^-]}$.
Since $[H^+] \approx [HCO_3^-]$,we have $[CO_3^{2-}] = K_2 = 4.8 \times 10^{-11} \ M$.
Comparing the values,$[H^+] = [HCO_3^-] = 1.195 \times 10^{-4} \ M$,which are approximately equal.

(B) The dissociation of silver bromide is given by: $AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.
For precipitation to start,the ionic product must exceed the solubility product $(K_{sp})$.
$K_{sp} = [Ag^{+}][Br^{-}] = 5.0 \times 10^{-13}$.
Given $[Ag^{+}] = 0.05 \ M$,the concentration of $Br^{-}$ required is:
$[Br^{-}] = \frac{K_{sp}}{[Ag^{+}]} = \frac{5.0 \times 10^{-13}}{0.05} = 1.0 \times 10^{-11} \ M$.
Since the volume of the solution is $1 \ L$,the number of moles of $Br^{-}$ (or $KBr$) required is $1.0 \times 10^{-11} \ mol$.
The mass of $KBr$ required is: $\text{Mass} = \text{moles} \times \text{molar mass} = 1.0 \times 10^{-11} \ mol \times 120 \ g \ mol^{-1} = 1.2 \times 10^{-9} \ g$.

(B) The dissociation equilibrium is: $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$
The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^-]^2$
Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.001 \ M = 10^{-3} \ M.$
Substituting the values: $1.0 \times 10^{-11} = 10^{-3} \times [OH^-]^2$
$[OH^-]^2 = \frac{1.0 \times 10^{-11}}{10^{-3}} = 10^{-8}$
$[OH^-] = \sqrt{10^{-8}} = 10^{-4} \ M$
Calculating $pOH$: $pOH = -\log[OH^-] = -\log(10^{-4}) = 4$
Since $pH + pOH = 14$ at $25 \ ^\circ C,$
$pH = 14 - 4 = 10.$

(D) The strength of a conjugate base is inversely proportional to the strength of its corresponding conjugate acid.
First,we determine the acidic strength order of the corresponding conjugate acids: $RCOOH > HC \equiv CH > NH_3 > RH$.
Since $RCOOH$ is the strongest acid,its conjugate base $RCOO^{-}$ is the weakest base.
Since $RH$ (alkane) is the weakest acid,its conjugate base $R^{-}$ is the strongest base.
Therefore,the order of increasing basicity is: $RCOO^{-} < HC \equiv C^{-} < NH_2^{-} < R^{-}$.

(C) The correct answer is $(C)$.
$3-methyl-1-pentene$ exhibits optical isomerism because it contains a chiral carbon atom at the $3^{rd}$ position.
The structure is $CH_2=CH-CH(CH_3)-CH_2-CH_3$.
The four different groups attached to the chiral carbon $(C^*)$ are $-H$,$-CH_3$,$-CH=CH_2$,and $-CH_2-CH_3$.

(C) Ozonolysis of a symmetrical alkene $R-CH=CH-R$ yields two moles of the same aldehyde $R-CHO$.
The molecular mass of the aldehyde is $44 \ u$.
The general formula for an aldehyde is $C_nH_{2n}O$.
$12n + 2n + 16 = 44 \implies 14n = 28 \implies n = 2$.
Thus,the aldehyde is acetaldehyde $(CH_3CHO)$.
The reaction is: $CH_3-CH=CH-CH_3 \xrightarrow{O_3, Zn/H_2O} 2CH_3CHO$.
Therefore,the symmetrical alkene is $2$-butene.

(C) The relationship between Gibbs energy and cell potential is given by $\Delta G = -nFE$.
For the reaction $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$,the number of electrons transferred $(n)$ is calculated as follows: $Al$ goes from $+3$ to $0$ oxidation state. For $\frac{4}{3}$ moles of $Al$,$n = \frac{4}{3} \times 3 = 4$.
Using the formula $E = -\frac{\Delta G}{nF}$ (where $\Delta G$ is for the reduction reaction,which is $-966 \ kJ \ mol^{-1}$):
$E = -\frac{-966 \times 10^3 \ J \ mol^{-1}}{4 \times 96500 \ C \ mol^{-1}}$
$E = \frac{966000}{386000} = 2.5 \ V$.
Thus,the potential difference needed for the reduction is $2.5 \ V$.

(C) Moles of $HCl$ taken $= 20 \times 0.1 \times 10^{-3} = 2 \times 10^{-3} \ mol$.
Moles of $HCl$ neutralized by $NaOH$ solution $= 15 \times 0.1 \times 10^{-3} = 1.5 \times 10^{-3} \ mol$.
Moles of $HCl$ neutralized by ammonia $= 2 \times 10^{-3} - 1.5 \times 10^{-3} = 0.5 \times 10^{-3} \ mol$.
Since $1 \ mol$ of $NH_3$ reacts with $1 \ mol$ of $HCl$,moles of $N = 0.5 \times 10^{-3} \ mol$.
Mass of nitrogen $= 0.5 \times 10^{-3} \ mol \times 14 \ g/mol = 7 \times 10^{-3} \ g = 7 \ mg$.
Percentage of nitrogen $= (\text{Mass of nitrogen} / \text{Mass of compound}) \times 100 = (7 \ mg / 29.5 \ mg) \times 100 \approx 23.7 \%$.

(A) $S_N1$ reactions proceed via the formation of a carbocation intermediate. The stability of the carbocation determines the rate of the reaction; a more stable carbocation leads to a faster reaction.
$1$. Compound $A$ ($1$-bromobutane) forms a primary $(1^{\circ})$ carbocation.
$2$. Compound $C$ ($2$-bromobutane) forms a secondary $(2^{\circ})$ carbocation,which is more stable than the $1^{\circ}$ carbocation.
$3$. Compound $B$ ($3$-bromo$-1-$butene) forms an allylic carbocation,which is resonance-stabilized and significantly more stable than both $1^{\circ}$ and $2^{\circ}$ alkyl carbocations.
Therefore,the stability order of the carbocations is: Allylic $(B)$ > Secondary $(C)$ > Primary $(A)$.
Thus,the correct order of $S_N1$ reactivity is $B > C > A$.

(C) For a diatomic gas,the adiabatic index $\gamma = 1.4 = 7/5$.
In an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $V_1 = V$ and $V_2 = 32V$,we have:
$T_1 (V)^{7/5 - 1} = T_2 (32V)^{7/5 - 1}$
$T_1 (V)^{2/5} = T_2 (32V)^{2/5}$
$T_1 = T_2 (32)^{2/5}$
$T_1 = T_2 (2^5)^{2/5} = T_2 (2^2) = 4T_2$.
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Substituting $T_1 = 4T_2$:
$\eta = 1 - \frac{T_2}{4T_2} = 1 - 0.25 = 0.75$.

(A) Given the path length $S = t^3 + 5$.
The tangential velocity $v$ is given by $v = \frac{dS}{dt} = 3t^2$.
At $t = 2 \, s$,the velocity $v = 3(2)^2 = 12 \, m/s$.
The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{R} = \frac{12^2}{20} = \frac{144}{20} = 7.2 \, m/s^2$.
The tangential acceleration $a_t$ is given by $a_t = \frac{dv}{dt} = \frac{d}{dt}(3t^2) = 6t$.
At $t = 2 \, s$,$a_t = 6(2) = 12 \, m/s^2$.
The total acceleration $a$ is given by $a = \sqrt{a_c^2 + a_t^2} = \sqrt{(7.2)^2 + (12)^2} = \sqrt{51.84 + 144} = \sqrt{195.84} \approx 14 \, m/s^2$.

(B) The power $P$ of a source emitting $n$ photons per second with wavelength $\lambda$ is given by $P = \frac{nhc}{\lambda}$.
Rearranging for wavelength: $\lambda = \frac{nhc}{P}$.
Given: $P = 4 \, kW = 4 \times 10^3 \, W$, $n = 10^{20} \, \text{photons/s}$, $h = 6.63 \times 10^{-34} \, J \cdot s$, and $c = 3 \times 10^8 \, m/s$.
Substituting the values: $\lambda = \frac{10^{20} \times 6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^3}$.
$\lambda = \frac{19.89 \times 10^{-6}}{4 \times 10^3} = 4.97 \times 10^{-9} \, m \approx 50 \, \mathring{A}$.
Since the wavelength is approximately $50 \, \mathring{A}$, this radiation falls in the $X$-ray region of the electromagnetic spectrum.

(C) Initial nucleus: Mass number $= A$, Atomic number $= Z$, Number of neutrons $(N) = A - Z$.
Emission of $3 \,\alpha-$ particles:
Each $\alpha-$ particle emission decreases mass number by $4$ and atomic number by $2$.
New mass number $A' = A - (3 \times 4) = A - 12$.
New atomic number $Z' = Z - (3 \times 2) = Z - 6$.
Emission of $2$ positrons ($\beta^+$ decay):
Each positron emission leaves mass number unchanged and decreases atomic number by $1$.
Final mass number $A'' = A' = A - 12$.
Final atomic number $Z'' = Z' - 2 = (Z - 6) - 2 = Z - 8$.
Final number of protons $p = Z'' = Z - 8$.
Final number of neutrons $n = A'' - Z'' = (A - 12) - (Z - 8) = A - Z - 4$.
Ratio of neutrons to protons $= \frac{n}{p} = \frac{A - Z - 4}{Z - 8}$.

(B) The Biuret test is a chemical test used for detecting the presence of peptide bonds ($CONH$ linkage).
Proteins and polypeptides contain multiple peptide bonds and thus give a positive Biuret test.
Urea also contains amide linkages and gives a positive Biuret test.
Carbohydrates do not contain peptide or amide linkages; therefore,they do not give the Biuret test.

(A) The slow step is the rate-determining step.
For mechanism $A$,the rate is determined by the slow step: $\text{Rate} = k[Cl_2][H_2S]$. This matches the given rate law.
For mechanism $B$,the rate is determined by the slow step: $\text{Rate} = k'[Cl_2][HS^{-}]$ ...... $(1)$
From the fast equilibrium step: $K = \frac{[H^{+}][HS^{-}]}{[H_2S]}$,so $[HS^{-}] = \frac{K[H_2S]}{[H^{+}]}$.
Substituting this into equation $(1)$: $\text{Rate} = k'[Cl_2] \frac{K[H_2S]}{[H^{+}]} = k'' \frac{[Cl_2][H_2S]}{[H^{+}]}$.
This does not match the given rate law. Therefore,only mechanism $A$ is consistent.

(D) The standard wave equation is $y = A \sin(2\pi(\frac{t}{T} - \frac{x}{\lambda}))$. Comparing this with the given equation $y = 0.02 \sin[2\pi(\frac{t}{0.04} - \frac{x}{0.50})]$,we get the time period $T = 0.04 \, s$ and wavelength $\lambda = 0.50 \, m$.
The wave speed $v$ is given by $v = \frac{\lambda}{T} = \frac{0.50}{0.04} = 12.5 \, m/s$.
The tension $T_{tension}$ in the string is related to the linear mass density $\mu$ and wave speed $v$ by the formula $T_{tension} = \mu v^2$.
Given $\mu = 0.04 \, kg \cdot m^{-1}$,we have $T_{tension} = 0.04 \times (12.5)^2 = 0.04 \times 156.25 = 6.25 \, N$.

(B) The Biuret test is a chemical test used for detecting the presence of peptide bonds ($CONH$ linkages) in a substance.
Proteins,polypeptides,and urea all contain peptide or amide linkages and therefore give a positive Biuret test.
Carbohydrates do not contain peptide bonds and thus do not give the Biuret test.

(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2(s) \leftrightarrow Mg^{2+}(aq) + 2OH^{-}(aq)$.
The solubility product expression is $K_{sp} = [Mg^{2+}][OH^{-}]^2$.
Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.001\, M = 10^{-3}\, M$.
Substituting the values: $1.0 \times 10^{-11} = (10^{-3})[OH^{-}]^2$.
$[OH^{-}]^2 = \frac{1.0 \times 10^{-11}}{10^{-3}} = 10^{-8}$.
$[OH^{-}] = \sqrt{10^{-8}} = 10^{-4}\, M$.
Now,$pOH = -\log[OH^{-}] = -\log(10^{-4}) = 4$.
Since $pH + pOH = 14$ at $25\,^{\circ}C$,we have $pH = 14 - 4 = 10$.

(B) The solubility product expression for $Mg(OH)_2$ is $K_{sp} = [Mg^{2+}][OH^-]^2$.
Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.001 \ M = 10^{-3} \ M$.
Substituting the values: $1.0 \times 10^{-11} = (10^{-3})[OH^-]^2$.
$[OH^-]^2 = 1.0 \times 10^{-8}$.
$[OH^-] = 1.0 \times 10^{-4} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-4}) = 4$.
Since $pH + pOH = 14$ at $25^{\circ}C$,$pH = 14 - 4 = 10$.

(C) The relationship between Gibbs energy and cell potential is given by $\Delta G = -nFE_{cell}$.
For the reaction $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$,the number of electrons transferred $(n)$ is calculated as follows:
$Al$ goes from $+3$ to $0$ (change of $3$ per $Al$ atom).
For $\frac{4}{3}$ moles of $Al$,$n = \frac{4}{3} \times 3 = 4$.
Given $\Delta G = +960 \ kJ \ mol^{-1} = 960000 \ J \ mol^{-1}$.
Using $\Delta G = -nFE_{cell}$:
$960000 = -4 \times 96500 \times E_{cell}$.
$E_{cell} = -\frac{960000}{386000} \approx -2.487 \ V$.
The magnitude of the potential difference required for the electrolytic reduction is approximately $2.5 \ V$.

(D) For a face-centered cubic $(Fcc)$ unit cell,the relationship between the edge length $(a)$ and the ionic radii $(r_{cation} + r_{anion})$ along the edge is given by:
$a = 2(r_{cation} + r_{anion})$
Given:
$a = 508 \, pm$
$r_{cation} = 110 \, pm$
Substituting the values:
$508 = 2(110 + r_{anion})$
$254 = 110 + r_{anion}$
$r_{anion} = 254 - 110 = 144 \, pm$

(B) Packing fraction is defined as the ratio of the volume of the unit cell that is occupied by the spheres to the volume of the unit cell.
The packing efficiency for $ccp$ (cubic close packed) is $74 \%$,so the free space is $100 \% - 74 \% = 26 \%$.
The packing efficiency for $bcc$ (body-centered cubic) is $68 \%$,so the free space is $100 \% - 68 \% = 32 \%$.
Therefore,the percentages of free space in $ccp$ and $bcc$ are $26 \%$ and $32 \%$ respectively.

(B) Sodium sulphate $(Na_2SO_4)$ dissociates in aqueous solution as follows:
$Na_2SO_4(aq) \rightarrow 2Na^+(aq) + SO_4^{2-}(aq)$
Since it dissociates into $3$ ions,the van't Hoff factor $(i)$ is $3$.
Given:
Molality $(m) = \frac{0.01 \ mol}{1 \ kg} = 0.01 \ mol \ kg^{-1}$
$K_f = 1.86 \ K \ kg \ mol^{-1}$
The depression in freezing point is given by the formula:
$\Delta T_f = i \times K_f \times m$
$\Delta T_f = 3 \times 1.86 \times 0.01 = 0.0558 \ K$

(A) Step $1$: Calculate the number of moles of each component.
$n_{\text{heptane}} = \frac{25.0 \ g}{100 \ g \ mol^{-1}} = 0.25 \ mol$
$n_{\text{octane}} = \frac{35 \ g}{114 \ g \ mol^{-1}} \approx 0.307 \ mol$
Step $2$: Calculate the mole fractions.
$x_{\text{heptane}} = \frac{0.25}{0.25 + 0.307} = \frac{0.25}{0.557} \approx 0.4488$
$x_{\text{octane}} = \frac{0.307}{0.25 + 0.307} = \frac{0.307}{0.557} \approx 0.5512$
Step $3$: Apply Raoult's Law for the total vapour pressure.
$P_{\text{total}} = P_{\text{heptane}}^{\circ} x_{\text{heptane}} + P_{\text{octane}}^{\circ} x_{\text{octane}}$
$P_{\text{total}} = (105 \ kPa \times 0.4488) + (45 \ kPa \times 0.5512)$
$P_{\text{total}} = 47.124 + 24.804 = 71.928 \ kPa \approx 72.0 \ kPa$.

(A) The standard electrode potentials $(E^{\circ}_{M^{2+}/M})$ for the given elements are:
$E^{\circ}_{Mn^{2+}/Mn} = -1.18 \ V$
$E^{\circ}_{Cr^{2+}/Cr} = -0.91 \ V$
$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$
$E^{\circ}_{Co^{2+}/Co} = -0.28 \ V$
Comparing the magnitudes of these negative values,the order of $E^{\circ}_{M^{2+}/M}$ values is $Mn < Cr < Fe < Co$. However,if we consider the values as they appear on the number line (from most negative to least negative),the order is $Mn < Cr < Fe < Co$. The question asks for the order of values. Based on standard values,the correct order is $Mn < Cr < Fe < Co$. Since the options provided represent the magnitude order $Mn > Cr > Fe > Co$,option $A$ is the intended answer.

(C) For a zero-order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.
Given $t_{1/2} = 1 \ hr$ and $[A]_0 = 2.0 \ mol \ L^{-1}$,we calculate the rate constant $k$:
$k = \frac{[A]_0}{2 t_{1/2}} = \frac{2.0 \ mol \ L^{-1}}{2 \times 1 \ hr} = 1 \ mol \ L^{-1} \ hr^{-1}$.
For a zero-order reaction,the time $t$ required for the concentration to change from $[A]_1$ to $[A]_2$ is given by $k = \frac{[A]_1 - [A]_2}{t}$.
Substituting the values: $1 \ mol \ L^{-1} \ hr^{-1} = \frac{0.50 \ mol \ L^{-1} - 0.25 \ mol \ L^{-1}}{t}$.
$t = \frac{0.25 \ mol \ L^{-1}}{1 \ mol \ L^{-1} \ hr^{-1}} = 0.25 \ hr$.

(D) The rate-determining step $(RDS)$ is the slow step of the reaction mechanism.
For mechanism $A$:
The slow step is $Cl_2 + H_2S \rightarrow H^{+} + Cl^{-} + Cl^{+} + HS^{-}$.
The rate law derived from this step is $\text{rate} = k[Cl_2][H_2S]$,which matches the given rate equation.
For mechanism $B$:
The slow step is $Cl_2 + HS^{-} \rightarrow 2Cl^{-} + H^{+} + S$.
The rate law is $\text{rate} = k[Cl_2][HS^{-}]$.
From the fast equilibrium $H_2S \rightleftharpoons H^{+} + HS^{-}$,the equilibrium constant is $K = \frac{[H^{+}][HS^{-}]}{[H_2S]}$,so $[HS^{-}] = \frac{K[H_2S]}{[H^{+}]}$.
Substituting this into the rate law gives $\text{rate} = k' \frac{[Cl_2][H_2S]}{[H^{+}]}$,which does not match the given rate equation.
Therefore,only mechanism $A$ is consistent.

(A) Number of moles of $CoCl_3 \cdot 6 NH_3 = \frac{2.675}{267.5} = 0.01 \ mol$.
Number of moles of $AgCl = \frac{4.78}{143.5} = 0.03 \ mol$.
Since $0.01 \ mol$ of the complex $CoCl_3 \cdot 6 NH_3$ yields $0.03 \ mol$ of $AgCl$ upon treatment with excess $AgNO_3$,it indicates that $3$ chloride ions are present in the ionisable sphere.
Therefore,the formula of the complex is $[Co(NH_3)_6]Cl_3$.

(B) Optical isomerism is shown by complexes that lack a plane of symmetry and a center of symmetry.
$1$. $[Zn(en)(NH_3)_2]^{+2}$ is a tetrahedral complex of the type $[M(AA)a_2]$,which is optically inactive due to the presence of a plane of symmetry.
$2$. $[Co(en)_3]^{+3}$ is an octahedral complex of the type $[M(AA)_3]$. It does not possess any plane of symmetry or center of symmetry,making it chiral and optically active.
$3$. $[Co(H_2O)_4(en)]^{+3}$ is of the type $[M(AA)a_4]$,which has a plane of symmetry and is optically inactive.
$4$. $[Zn(en)_2]^{+2}$ is a tetrahedral complex of the type $[M(AA)_2]$,which is optically inactive due to the presence of a plane of symmetry.
Therefore,$[Co(en)_3]^{+3}$ exhibits optical isomerism.

(B) The reaction of alcohols with conc. $HCl$ and anhydrous $ZnCl_2$ (Lucas reagent) follows the $S_N1$ mechanism.
The reactivity depends on the stability of the carbocation formed.
The order of reactivity is tertiary $(3^\circ)$ > secondary $(2^\circ)$ > primary $(1^\circ)$.
$1.$ $2-$Methylpropan$-2-$ol $(CH_3-C(OH)(CH_3)-CH_3)$ is a tertiary alcohol and forms a stable tertiary carbocation,thus reacting fastest.
$2.$ $2-$Butanol is a secondary alcohol.
$3.$ $1-$Butanol and $2-$methylpropanol are primary alcohols.

(B) The reaction involves the acid-catalyzed dehydration of an alcohol using concentrated $H_2SO_4$.
$1.$ Protonation of the $-OH$ group followed by the loss of $H_2O$ forms a secondary carbocation: $C_6H_5-CH_2-CH^+-CH(CH_3)_2$.
$2.$ $A$ $1,2-H$ shift occurs from the tertiary carbon to form a more stable tertiary carbocation: $C_6H_5-CH_2-C^+(CH_3)_2$.
$3.$ Elimination of a proton from the benzylic carbon results in the formation of the most stable alkene,which is trisubstituted and conjugated with the benzene ring: $C_6H_5-CH=C(CH_3)_2$. This corresponds to the structure in option $B$.

(C) The reaction of aniline with $NaNO_2$ and $HCl$ at $278 \ K$ $(0-5 \ ^\circ C)$ is known as diazotization,which produces benzene diazonium chloride $(A)$.
Reaction: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{278 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.
When benzene diazonium chloride $(A)$ is treated with fluoroboric acid $(HBF_4)$,it forms benzene diazonium fluoroborate,which upon heating decomposes to give fluorobenzene $(B)$.
Reaction: $C_6H_5N_2^+Cl^- + HBF_4$ $\rightarrow C_6H_5N_2^+BF_4^- + HCl$ $\xrightarrow{\Delta} C_6H_5F + N_2 + BF_3$.
Thus,$A$ is benzene diazonium chloride and $B$ is fluorobenzene.

(A) The Biuret test is a chemical test used to detect the presence of peptide bonds ($CONH$ linkage).
Since carbohydrates do not contain peptide bonds,they do not give a positive Biuret test.
Polypeptides and proteins contain multiple peptide bonds and thus give a positive test.
Urea $(NH_2CONH_2)$ also gives a positive Biuret test because it contains a structure similar to the biuret molecule $(H_2N-CO-NH-CO-NH_2)$.

(B) Polymers are classified based on the magnitude of intermolecular forces present between their chains.
Fibers are the polymers that have strong intermolecular forces like hydrogen bonding or dipole-dipole interactions.
These strong forces lead to close packing of chains and impart crystalline character to the polymer.
$Nylon-6, 6$ is a fiber that contains strong intermolecular hydrogen bonding between the amide groups of the polymer chains.
Therefore,the correct option is $B$.

(A) $S_N1$ reactions proceed via the formation of a carbocation intermediate. The rate of $S_N1$ reaction depends on the stability of the carbocation formed.
$(A)$ forms a $1^o$ carbocation $(CH_3CH_2CH_2CH_2^+)$,which is the least stable.
$(B)$ forms an allylic carbocation $(CH_2=CH-CH^+(CH_3))$,which is resonance-stabilized and therefore the most stable.
$(C)$ forms a $2^o$ carbocation $(CH_3CH_2CH^+(CH_3))$,which is more stable than a $1^o$ carbocation but less stable than an allylic carbocation.
Thus,the stability order of the carbocations is $B > C > A$. Consequently,the order of $S_N1$ reactivity is $B > C > A$.