AIPMT 1992 Chemistry Question Paper with Answer and Solution

(C) The kinetic energy of the particle is given by $K = (1/2)MV^2$. Since the speed $V$ is uniform,the kinetic energy remains constant throughout the motion. Therefore,the change in kinetic energy is $0$.
Let the initial velocity at one point be $\vec{v}_1 = V\hat{i}$.
At the diametrically opposite point,the velocity will be $\vec{v}_2 = -V\hat{i}$.
The change in momentum $\Delta \vec{p}$ is given by $\Delta \vec{p} = M\vec{v}_2 - M\vec{v}_1$.
Substituting the values,$\Delta \vec{p} = M(-V\hat{i}) - M(V\hat{i}) = -2MV\hat{i}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2MV$.

(C) We are given the relation $\frac{R}{C_V} = 0.67$.
Since $C_V = \frac{f}{2} R$,where $f$ is the degree of freedom,we can write $\frac{R}{C_V} = \frac{2}{f}$.
Equating the two,we get $\frac{2}{f} = 0.67$.
Solving for $f$,we get $f = \frac{2}{0.67} \approx 2.985 \approx 3$.
Since the degree of freedom $f = 3$,the gas is monoatomic.

(C) According to Faraday's law of electromagnetic induction, the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$.
From Ohm's law, the induced current is $i = \frac{e}{R} = -\frac{1}{R} \frac{d\phi}{dt}$, where $R$ is the resistance of the loop.
Since current is the rate of flow of charge, $i = \frac{dq}{dt}$.
Equating the two expressions for current: $\frac{dq}{dt} = -\frac{1}{R} \frac{d\phi}{dt}$.
Integrating both sides, we get $q = -\frac{1}{R} \Delta \phi$.
Therefore, the total charge induced $(q)$ depends on the total change in magnetic flux $(\Delta \phi)$ and the resistance of the loop $(R)$, and is independent of the time taken or the rate of change of flux.
Thus, the correct option is $(C)$.

(C) In the $N_2$ molecule,the two nitrogen atoms are connected by a triple bond.
Each nitrogen atom contributes $3$ electrons to the bond formation.
Therefore,the total number of electrons involved in the bond formation is $3 + 3 = 6$ electrons.

(D) coordinate or dative bond is a type of covalent bond in which both electrons of the shared pair are contributed by only one of the two participating atoms.
For the formation of a coordinate bond,the donor atom must possess at least one lone pair of electrons (an unshared electron pair) to donate to the acceptor atom.

(C) $CO_2$ has $sp$ hybridization and a linear geometry with a bond angle of $180^{\circ}$. The central carbon atom is bonded to two oxygen atoms with no lone pairs,resulting in a linear shape.

(C) The strength of a hydrogen bond depends on the electronegativity of the atom to which the hydrogen atom is bonded.
Since fluorine $(F)$ is the most electronegative element,the $H-F$ bond is highly polar,resulting in the strongest hydrogen bonding in $HF$ compared to $H_2O$,$NH_3$,or acetic acid.

(C) The ideal gas equation is given by $PV = nRT$.
In this equation:
$P$ represents the pressure of the gas.
$V$ represents the volume occupied by $n$ moles of the gas.
$n$ represents the number of moles of the gas.
$R$ is the universal gas constant.
$T$ is the absolute temperature.
Therefore,$V$ is the volume occupied by $n$ moles of the gas at pressure $P$ and temperature $T$. Thus,the statement '$n$ moles of the gas have a volume $V$' is correct.

(B) The universal gas constant $R$ is defined by the ideal gas equation $PV = nRT$.
Its value depends on the units used for pressure,volume,and temperature.
In units of liters,atmospheres,Kelvin,and moles,the value is approximately $0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.

(A) Using the ideal gas law $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is mass,$M$ is molar mass).
Since $V$,$R$,and $M$ are constant,we have $\frac{P_1 V}{m_1 T_1} = \frac{P_2 V}{m_2 T_2}$.
Rearranging for $T_2$: $T_2 = T_1 \times \frac{P_2}{P_1} \times \frac{m_1}{m_2}$.
Given: $P_1 = 1 \ atm$,$T_1 = 300 \ K$,$m_1 = 2 \ g$,$P_2 = 0.75 \ atm$,$m_2 = 1 \ g$.
$T_2 = 300 \times \frac{0.75}{1} \times \frac{2}{1} = 300 \times 1.5 = 450 \ K$.

(C) From the ideal gas equation,$PV = nRT = (\frac{m}{M})RT$.
Since density $d = \frac{m}{V}$,we have $P = \frac{dRT}{M}$,which implies $d = \frac{PM}{RT}$.
At constant pressure $P$ and molar mass $M$,$d \propto \frac{1}{T}$.
Therefore,$\frac{d_1}{d_2} = \frac{T_2}{T_1}$.
Given $T_1 = 27 + 273 = 300\, K$,$d_1 = d$,and $d_2 = 0.75\, d$.
Substituting the values: $\frac{d}{0.75\, d} = \frac{T_2}{300\, K}$.
$T_2 = \frac{300}{0.75} = 400\, K$.

(D) The correct answer is $D$. An ideal gas cannot be liquefied because there are no intermolecular forces of attraction between the molecules of an ideal gas. Liquefaction requires the presence of attractive forces to bring molecules closer together to form a liquid phase.

(B) The equilibrium constant $K_c$ for the reaction $C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g)$ is given by the expression:
$K_c = \frac{[C_2H_6]}{[C_2H_4][H_2]}$
Substituting the units of concentration (molarity,$mole \, litre^{-1}$):
$K_c = \frac{mole \, litre^{-1}}{(mole \, litre^{-1})(mole \, litre^{-1})} = \frac{1}{mole \, litre^{-1}} = litre \, mole^{-1}$
Therefore,the correct unit is $litre \, mole^{-1}$.

(A) When equal volumes are mixed,the concentration of each ion is halved. Let the initial concentrations be $C_1$ and $C_2$. The new concentration of each ion becomes $C/2$.
For precipitation to occur,the ionic product $Q_{sp}$ must be greater than $K_{sp}$ $(Q_{sp} > 1.8 \times 10^{-10})$.
For option $A$: $[Ag^{+}] = 10^{-4}/2 = 5 \times 10^{-5} \ M$ and $[Cl^{-}] = 10^{-4}/2 = 5 \times 10^{-5} \ M$.
$Q_{sp} = [Ag^{+}][Cl^{-}] = (5 \times 10^{-5}) \times (5 \times 10^{-5}) = 25 \times 10^{-10} = 2.5 \times 10^{-9}$.
Since $2.5 \times 10^{-9} > 1.8 \times 10^{-10}$,precipitation will occur.

(A) Let the given equations be:
$(i) H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H_1 = 131 \ kJ$
$(ii) CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H_2 = -282 \ kJ$
$(iii) H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H_3 = -242 \ kJ$
Adding equations $(i)$,$(ii)$,and $(iii)$:
$(H_2O_{(g)} + C_{(s)}) + (CO_{(g)} + \frac{1}{2}O_{2(g)}) + (H_{2(g)} + \frac{1}{2}O_{2(g)}) \to (CO_{(g)} + H_{2(g)}) + CO_{2(g)} + H_2O_{(g)}$
Canceling common terms on both sides:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
Therefore,$X = \Delta H_1 + \Delta H_2 + \Delta H_3 = 131 + (-282) + (-242) = -393 \ kJ$.

(B) The reaction of $BaO_2$ with dilute $H_2SO_4$ is given by:
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$
In the products,the most electronegative element is oxygen.
In $H_2O_2$,oxygen is in the peroxide state,so its oxidation state is $-1$.
In $BaSO_4$,oxygen is in the oxide state,so its oxidation state is $-2$.
Therefore,the oxidation states are $-1$ and $-2$.

(D) The valence shell electronic configuration of $ns^2 np^5$ corresponds to elements having $7$ electrons in their outermost shell.
These elements belong to group $17$ of the periodic table,which are known as the halogens.

(D) The ionisation energy is the energy required to remove an electron from a gaseous atom or ion.
$A$ large jump in successive ionisation energies occurs when an electron is removed from a stable,noble gas-like configuration.
If the jump occurs between the second and third ionisation energies,it means the first two electrons are removed easily,but the third electron is removed from a stable,fully filled shell.
This implies the element has $2$ valence electrons.
The configuration $1s^2, 2s^2p^6, 3s^2$ corresponds to Magnesium $(Mg)$,which has $2$ valence electrons in the $3s$ orbital.
Removing the first two electrons leaves a stable $2s^2p^6$ configuration; removing the third electron requires significantly more energy.

(C) The correct answer is $(C)$.
Anhydrous $CaCl_2$ is a common drying agent used in the laboratory for the fast drying of neutral gases.

(B) $1$. Identify the principal functional group: The aldehyde group $(-CHO)$ has the highest priority,so the parent chain must include the carbon of the $-CHO$ group and it is assigned position $1$.
$2$. Select the longest carbon chain containing the principal functional group: The chain has $5$ carbons,so the parent alkane is pentane. The suffix for aldehyde is $-al$,so the parent name is pentanal.
$3$. Number the chain: The carbon of the $-CHO$ group is $C-1$. The methyl group $(-CH_3)$ is at $C-4$ and the hydroxyl group $(-OH)$ is at $C-4$.
$4$. Assemble the name: The substituents are $4-$hydroxy and $4-$methyl. Alphabetically,'h' comes before 'm'. Thus,the name is $4-$hydroxy$-4-$methylpentanal. However,re-evaluating the structure $CH_3-CH(OH)-CH_2-CH(CH_3)-CHO$: The chain is $C_1(CHO)-C_2(H)(CH_3)-C_3(H_2)-C_4(H)(OH)-C_5(H_3)$. The substituents are at position $4$ (hydroxy) and position $2$ (methyl). Therefore,the correct name is $4-$hydroxy$-2-$methylpentanal.

(A) The $IUPAC$ name $2$-methyl-$2$-butene indicates a $4$-carbon chain (but-ene) with a double bond at the $2^{nd}$ position and a methyl group at the $2^{nd}$ position.
The structure is $CH_3-C(CH_3)=CH-CH_3$.

(D) The given hybridisation sequence is $sp^3, sp^2, sp^2, sp^3, sp^2, sp^2, sp, sp$.
Counting the carbons:
$1(sp^3): CH_3-$
$2(sp^2): -CH=$
$3(sp^2): =CH-$
$4(sp^3): -CH_2-$
$5(sp^2): -CH=$
$6(sp^2): =CH-$
$7(sp): -C\equiv$
$8(sp): \equiv CH$
Combining these,we get $CH_3-CH=CH-CH_2-CH=CH-C\equiv CH$.
This matches option $D$.

(A) Geometrical isomerism is shown by compounds that have restricted rotation around a double bond and where each carbon atom of the double bond is attached to two different groups.
In $2-$butene $(CH_3-CH=CH-CH_3)$,the carbon atoms of the double bond are each bonded to a hydrogen atom $(H)$ and a methyl group $(CH_3)$.
Thus,it exists as cis-$2-$butene and trans-$2-$butene,which are geometrical isomers.

(A) Baeyer's reagent is an alkaline solution of cold $KMnO_4$.
It is primarily used for the detection of unsaturation (double or triple bonds) in a molecule,as it causes the purple color of $KMnO_4$ to disappear.

(C) In a benzene molecule,all $C-C$ bond lengths are equal due to resonance.
Benzene does not readily undergo addition reactions because it is highly stable due to aromaticity.
There is a cyclic delocalisation of $\pi$ electrons in the benzene ring.
Monosubstitution of benzene gives only a single isomeric product because all six positions in the benzene ring are equivalent.

(C) catalyst is a substance that increases the rate of both the forward and backward reactions to the same extent by lowering the activation energy.
Consequently,it helps the system reach the equilibrium state in a shorter amount of time without altering the position of the equilibrium or the enthalpy change of the reaction.
Therefore,the correct option is $C$.

(B) Resin is a sticky substance collected from the stem or wood of $Pinus$ trees.
Through the process of distillation,this resin is separated into two main products: turpentine (a volatile oil) and rosin (a solid residue).
Therefore,$Pinus$ is the correct source for these commercial products.

(A) The common Indian bullfrog is scientifically known as $Rana$ $tigrina$.
Although the nomenclature has been revised to $Hoplobatrachus$ $tigerinus$ (formerly $Rana$ $tigerina$) by taxonomists,the name $Rana$ $tigrina$ remains the most widely recognized and used name in biological literature and textbooks.

(C) The $Columella auris$ is a bone found in the middle ear of amphibians, reptiles, and birds. It is homologous to the $stapes$ bone found in the middle ear of mammals. Embryologically and evolutionarily, the $Columella auris$ is derived from the $hyomandibular$ bone of the fish skull. Therefore, it is a modified $hyomandibular$.

(D) In both photosynthesis and cellular respiration,the synthesis of $ATP$ occurs via a process called chemiosmosis,which is driven by an electron transport chain $(ETC)$.
This process involves a series of oxidation-reduction (redox) reactions where electrons are transferred from one carrier to another.
As electrons move through the $ETC$,they release energy,which is used to pump protons $(H^+)$ across a membrane,creating a proton gradient.
The energy stored in this electrochemical gradient is then used by the enzyme $ATP$ synthase to synthesize $ATP$ from $ADP$ and inorganic phosphate.
Therefore,the fundamental source of energy for $ATP$ synthesis in these processes is the energy released during the transfer of electrons.

(C) Amoebiasis,also known as amoebic dysentery,is a protozoan disease.
It is caused by the parasite $Entamoeba \ histolytica$.
The infection is transmitted through the ingestion of contaminated food and water containing the quadrinucleate cysts of the parasite.
These cysts enter the human intestine,where they excyst and cause damage to the intestinal mucosa,leading to symptoms like abdominal pain,cramps,and stool with excess mucus and blood.

(B) Short-day plants $(SDP)$ are those that flower when the photoperiod is shorter than a critical duration. For these plants to flower,they require a photoperiod shorter than the critical day length and,crucially,an uninterrupted long night. If the long night is interrupted by a flash of light,the flowering process is inhibited.

(D) Vernalization is the process by which flowering is promoted by exposure of plants to a period of low temperature. This ensures that plants flower during the favorable season,allowing them to complete their life cycle before the onset of unfavorable conditions. Therefore,the correct option is $D$.

(D) In the electron transport system $(ETS)$ of cellular respiration,the final complex is Complex $IV$,also known as cytochrome $c$ oxidase.
This complex contains cytochromes $a$ and $a_3$,and two copper centers.
Cytochrome $a_3$ is the terminal cytochrome that directly transfers electrons to the final electron acceptor,which is molecular oxygen $(O_2)$,to form water $(H_2O)$.

(A) Apical dominance is primarily caused by the hormone auxin.
$IAA$ (Indole$-3-$acetic acid) is the most common naturally occurring auxin in plants.
It promotes the growth of the apical bud while inhibiting the growth of lateral buds.

(C) $C_4$ plants are those that utilize the $C_4$ photosynthetic pathway to fix carbon dioxide.
Examples of $C_4$ plants include maize (corn),sugarcane,sorghum,and certain grasses.
Papaya,potato,and pea are examples of $C_3$ plants,which follow the Calvin cycle for carbon fixation.
Therefore,maize is the correct answer.

(A) The photoelectric current $I$ is directly proportional to the number of photoelectrons emitted per unit time.
The number of photoelectrons emitted per unit time depends on the intensity of the incident light and the number of incident photons,provided the frequency of incident light $v$ is greater than the threshold frequency $(hv > W)$.
Since the work function $W$ does not affect the number of photoelectrons emitted (it only affects the kinetic energy of the emitted electrons),the photoelectric current remains unchanged as long as the intensity of the incident light is constant.
Therefore,$I_1 = I_2$.

(A) The acidity of carboxylic acids is determined by the electron-withdrawing effect ($-I$ effect) of the substituents attached to the alpha-carbon.
Stronger electron-withdrawing groups stabilize the carboxylate anion more effectively,thereby increasing acidity.
The electronegativity of the halogens follows the order: $F > Cl > Br$.
Therefore,the $-I$ effect is strongest for the trifluoromethyl group $(-CF_3)$ and weakest for the methyl group $(-CH_3)$.
The order of acidic strength is: $CF_3COOH > CCl_3COOH > CBr_3COOH > CH_3COOH$.
Thus,$CF_3COOH$ is the strongest acid.

(D) The polarity of a covalent bond depends on the difference in electronegativity between the two bonded atoms.
Greater the difference in electronegativity,higher is the polarity of the bond.
The electronegativity values of the atoms involved are: $C = 2.55$,$O = 3.44$,$Br = 2.96$,$S = 2.58$,and $F = 3.98$.
The differences in electronegativity are:
$C-O: |3.44 - 2.55| = 0.89$
$C-Br: |2.96 - 2.55| = 0.41$
$C-S: |2.58 - 2.55| = 0.03$
$C-F: |3.98 - 2.55| = 1.43$
Since the difference in electronegativity is highest for the $C-F$ bond,it has the most polar character.
Therefore,the correct option is $(D)$.

(B) The solubility of $AgBr$ is governed by its solubility product constant $(K_{sp})$.
In the presence of $NaBr$ or $HBr$,the common ion effect $(Br^-)$ significantly decreases the solubility of $AgBr$.
In pure water,$AgBr$ has a low solubility.
In $NH_4OH$,$Ag^+$ ions react with $NH_3$ to form a stable complex ion,$[Ag(NH_3)_2]^+$,according to the reaction: $AgBr(s) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^+(aq) + Br^-(aq)$.
This complex formation removes $Ag^+$ ions from the solution,shifting the equilibrium to the right and significantly increasing the solubility of $AgBr$ compared to the other solvents.

(C) The presence of an electron-releasing group like $-CH_3$,$-OH$,etc.,increases the electron density at the $o/p$ positions and thus makes the benzene ring more reactive towards electrophiles.
On the other hand,electron-withdrawing groups like $-COOH$,$-NO_2$,etc.,reduce the electron density of the benzene ring,thereby decreasing its reactivity towards electrophilic substitution.
Comparing the given compounds:
$1$. $-NO_2$ (in nitrobenzene) is a strong electron-withdrawing group.
$2$. $-COOH$ (in benzoic acid) is an electron-withdrawing group.
$3$. $-H$ (in benzene) is the reference.
$4$. $-CH_3$ (in toluene) is an electron-releasing group due to the $+I$ effect and hyperconjugation.
Therefore,the order of reactivity towards electrophilic nitration is: $\text{nitrobenzene} < \text{benzoic acid} < \text{benzene} < \text{toluene}$.
Thus,toluene is the most reactive.

(B) Glycogen is a multi-branched polysaccharide of glucose that serves as a form of energy storage in animals,fungi,and bacteria.
It is a homopolymer,meaning it is composed of repeating units of a single type of monomer,which is $D-glucose$.
The glucose units are linked together by $\alpha-1,4-glycosidic$ bonds in linear chains and $\alpha-1,6-glycosidic$ bonds at branch points.

(A) Given equations:
$(1) \ H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$(2) \ CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
Adding equations $(1)$,$(2)$,and $(3)$:
$(H_2O_{(g)} + C_{(s)} + CO_{(g)} + \frac{1}{2}O_{2(g)} + H_{2(g)} + \frac{1}{2}O_{2(g)}) \to (CO_{(g)} + H_{2(g)} + CO_{2(g)} + H_2O_{(g)})$
Canceling common terms on both sides:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
Therefore,$\Delta H = 131 + (-282) + (-242) = X$
$X = 131 - 524 = -393 \ kJ$.

(C) In an electromagnetic wave,the electric field vector $\overrightarrow{E}$ and the magnetic field vector $\overrightarrow{B}$ oscillate in phase and are mutually perpendicular to each other.
According to Maxwell's equations and the properties of electromagnetic waves,the direction of propagation of the wave is given by the direction of the Poynting vector $\overrightarrow{S}$,which is defined as $\overrightarrow{S} = \frac{1}{\mu_0} (\overrightarrow{E} \times \overrightarrow{B})$.
Therefore,the direction of propagation of the electromagnetic wave is along the direction of $\overrightarrow{E} \times \overrightarrow{B}$.

(A) The photoelectric current in a cell depends on the number of photoelectrons emitted per unit time.
This number is directly proportional to the number of incident photons per unit time (intensity of light),provided the energy of each incident photon $(hv)$ is greater than the work function $(W)$ of the metal surface.
Since the intensity of the incident light remains unchanged and the condition $hv > W_2$ (which implies $hv > W_1$ as well) is satisfied,the rate of emission of photoelectrons remains the same.
Therefore,the photoelectric current remains unchanged,i.e.,$I_1 = I_2$.

(A) The photoelectric current in a cell depends on the number of photoelectrons emitted per unit time.
This number is directly proportional to the intensity of the incident light and the quantum efficiency of the cathode material.
The work function $W$ determines the threshold frequency,but it does not affect the number of photoelectrons emitted for a given intensity of incident light,provided the energy of the incident photons $h\nu$ is greater than the work function $(h\nu > W)$.
Since the intensity of the incident light and all other experimental conditions remain unchanged,the number of photoelectrons emitted per second remains the same.
Therefore,the photoelectric current remains constant,i.e.,$I_1 = I_2$.

(C) In uniform circular motion,the speed $v$ of the particle remains constant,so the kinetic energy $K = \frac{1}{2} M v^{2}$ remains constant. Thus,the change in kinetic energy is $0$.
At diametrically opposite points,the velocity vectors are in opposite directions. Let the initial velocity be $\vec{v}_1 = v \hat{i}$ and the final velocity be $\vec{v}_2 = -v \hat{i}$.
The change in momentum $\Delta \vec{p}$ is given by $\Delta \vec{p} = M \vec{v}_2 - M \vec{v}_1$.
Substituting the values,$\Delta \vec{p} = M(-v \hat{i}) - M(v \hat{i}) = -2 M v \hat{i}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2 M v$.

(C) The induced current $I$ in a loop is given by $I = \frac{\varepsilon}{R} = \frac{1}{R} \left( -\frac{d\phi}{dt} \right)$.
The total charge $q$ induced is the integral of current over time: $q = \int I dt$.
Substituting the expression for $I$: $q = \int \frac{1}{R} \left( \frac{d\phi}{dt} \right) dt = \frac{1}{R} \int d\phi = \frac{\Delta \phi}{R}$.
Thus,the total charge induced depends on the total change in magnetic flux $\Delta \phi$ and the resistance $R$ of the loop.

(A) The photoelectric current depends on the number of photoelectrons emitted per unit time.
This number is directly proportional to the intensity of the incident light,provided the energy of the incident photons $(hv)$ is greater than the work function $(W)$ of the metal surface.
Since the condition $hv > W_2$ is satisfied (which implies $hv > W_1$ as well),every incident photon that was capable of ejecting an electron before the change is still capable of doing so after the change.
Because the intensity of the incident light remains unchanged,the rate of emission of photoelectrons remains the same.
Therefore,the photoelectric current remains constant,i.e.,$I_1 = I_2$.

(A) In $DNA$,the nitrogenous base pairs are held together by hydrogen bonds.
Adenine $(A)$ pairs with Thymine $(T)$ via $2$ hydrogen bonds,and Guanine $(G)$ pairs with Cytosine $(C)$ via $3$ hydrogen bonds.
These hydrogen bonds provide stability to the double-helical structure of $DNA$.

(C) Halogens have the highest electronegativity and the smallest size in their respective periods,allowing them to gain an electron easily to achieve a stable noble gas configuration. Therefore,they form anions most readily.

(D) . The $N \equiv N$ bond dissociation energy is very high,approximately $945 \ kJ \ mol^{-1}$,which makes the $N_2$ molecule chemically inert under normal conditions.

(C) The reaction of chlorine gas with dry slaked lime $(Ca(OH)_2)$ at room temperature produces bleaching powder $(CaOCl_2)$.
The chemical equation is:
$Ca(OH)_2 + Cl_2 \to CaOCl_2 + H_2O$
Thus,the main product is $CaOCl_2$.

(B) In the extraction of bromine from sea water,the concentrated mother liquor containing bromide ions $(Br^-)$ is treated with chlorine gas $(Cl_2)$.
Chlorine is a stronger oxidizing agent than bromine and displaces bromine from the bromide solution.
The chemical reaction is: $MgBr_2 + Cl_2 \to MgCl_2 + Br_2$.

(A) The correct answer is $(A)$. Alkaline solution of pyrogallol is a well-known reagent used in gas analysis to quickly and quantitatively absorb oxygen from a gas mixture.

(C) The chemical formula $H_3PO_2$ corresponds to hypophosphorus acid (also known as phosphinic acid).
In its structure,there is only one $P-OH$ group,which means only one hydrogen atom is ionizable.
Therefore,the basicity of $H_3PO_2$ is $1$.

(C) primary alcohol is defined by the structure $R-CH_2-OH$. The isomers of $C_5H_{11}OH$ that are primary alcohols are:
$1$. $CH_3-CH_2-CH_2-CH_2-CH_2-OH$ (pentan-$1$-ol)
$2$. $CH_3-CH(CH_3)-CH_2-CH_2-OH$ ($3$-methylbutan-$1$-ol)
$3$. $CH_3-CH_2-CH(CH_3)-CH_2-OH$ ($2$-methylbutan-$1$-ol)
$4$. $CH_3-C(CH_3)_2-CH_2-OH$ ($2,2$-dimethylpropan-$1$-ol)
Thus,there are $4$ primary alcohol isomers.

(B) Molality is defined as the number of moles of solute per $1 \ kg$ of solvent.
Since mass does not change with temperature,molality is independent of temperature.
In contrast,molarity,formality,and normality involve volume,which changes with temperature.

(A) Colligative properties are those properties of solutions that depend only on the number of solute particles and not on their nature. The four main colligative properties are:
$1$. Relative lowering of vapour pressure
$2$. Elevation in boiling point
$3$. Depression in freezing point
$4$. Osmotic pressure
Therefore,$(A)$ Osmotic pressure is a colligative property.

(A) The osmotic pressure $(\pi)$ of a solution is given by the formula $\pi = iCRT$.
Since both glucose and urea are non-electrolytes,their van't Hoff factor $(i)$ is $1$.
Both solutions have the same molar concentration $(C = 0.1 \ M)$ and are at the same temperature $(T)$.
Therefore,both solutions have the same osmotic pressure.
Since the osmotic pressures are equal,there is no net movement of the solvent across the semipermeable membrane.

(C) During the electrolysis of dilute $H_2SO_4$ using platinum electrodes,the following reactions occur:
At the cathode: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$
At the anode: $2H_2O (l) \rightarrow O_2 (g) + 4H^+ (aq) + 4e^-$
Thus,$O_2$ gas is evolved at the anode.

(C) Park's process is a metallurgical method used for the desilverization of lead. In this process,zinc is added to molten lead containing silver. Since silver is more soluble in molten zinc than in lead,it forms an alloy with zinc,which rises to the surface and is skimmed off. Thus,it is used in the extraction/purification of $Silver$.

(B) The Carbylamine reaction is a diagnostic test for primary amines.
In this reaction,a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$.
The general reaction is: $R-NH_2 + CHCl_3 + 3KOH(alc.) \to R-NC + 3KCl + 3H_2O$.
Since chloroform is a trihalogen methane,option $B$ is the correct description.

(C) The oxidation of secondary alcohols yields ketones. $CH_3COCH_3$ (acetone) is a ketone.
Isopropyl alcohol is a secondary alcohol with the structure $CH_3CH(OH)CH_3$.
Upon oxidation with oxidizing agents like $K_2Cr_2O_7/H^+$,isopropyl alcohol undergoes dehydrogenation to form $CH_3COCH_3$.

(A) . $CH_3-CO-CH_3$ (acetone) is a ketone.
Ketones,upon vigorous oxidation,undergo $C-C$ bond cleavage,resulting in carboxylic acids with a fewer number of carbon atoms than the parent ketone.
In contrast,primary alcohols $(CH_3-CH_2-CH_2-OH)$ and aldehydes ($CH_3-CH_2-CHO$ and $CCl_3-CH_2-CHO$) yield carboxylic acids with the same number of carbon atoms upon oxidation.

(D) Benzoic acid $(C_6H_5COOH)$ undergoes decarboxylation when heated with sodalime $(NaOH + CaO)$ to form benzene $(C_6H_6)$. Thus,$X$ is sodalime.
Phenol $(C_6H_5OH)$ undergoes reduction when heated with zinc dust $(Zn)$ to form benzene $(C_6H_6)$ and zinc oxide $(ZnO)$. Thus,$Y$ is zinc dust.
Therefore,$X$ is sodalime and $Y$ is zinc dust. The correct option is $(D)$.

(C) In aqueous solution,the basicity of aliphatic amines depends on the inductive effect,solvation effect,and steric hindrance.
For ethyl-substituted amines,the order of basicity is $(C_2H_5)_2NH > C_2H_5NH_2 > (C_2H_5)_3N$.
$NH_3$ is less basic than these aliphatic amines.
$C_6H_5NH_2$ (aniline) is the least basic due to the delocalization of the lone pair of electrons on the nitrogen atom into the benzene ring.
Therefore,the correct increasing order is $C_6H_5NH_2 < NH_3 < (C_2H_5)_3N < C_2H_5NH_2 < (C_2H_5)_2NH$.

(A) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups $(EWG)$ stabilize the carboxylate ion through the inductive effect ($-I$ effect),thereby increasing acidity.
The strength of the $-I$ effect depends on the electronegativity of the substituent.
Fluorine $(F)$ is more electronegative than Chlorine $(Cl)$ and Bromine $(Br)$.
Therefore,the $-I$ effect follows the order: $CF_3 > CCl_3 > CBr_3 > CH_3$.
Thus,$CF_3COOH$ is the strongest acid among the given options.