KCET 2009 Chemistry Question Paper with Answer and Solution

(C) The general trend of ionization energy in a period increases from left to right due to an increase in effective nuclear charge.
The expected order is $C < N < O < F$.
However,$N$ $(1s^2 2s^2 2p^3)$ has a stable half-filled $2p$ sub-shell,which makes it harder to remove an electron compared to $O$ $(1s^2 2s^2 2p^4)$.
Therefore,the ionization energy of $N$ is higher than that of $O$.
The correct order is $C < O < N < F$.

(C) For output $A$: The inputs to the $NAND$ gate are $1$ and $1$,so its output is $0$. This $0$ passes through a $NOT$ gate,becoming $1$. The other input to the $OR$ gate comes from a $1$ passing through a $NOT$ gate,becoming $0$. Thus,the $OR$ gate receives inputs $1$ and $0$,resulting in output $A = 1$.
For output $B$: The input $0$ passes through a $NOT$ gate to become $1$. The input $1$ passes through a $NOT$ gate to become $0$. These are inputs to a $NAND$ gate. Since $1$ and $0$ are inputs to $NAND$,the output is $1$. Thus,$B = 1$.
For output $C$: The inputs to the $NOR$ gate are $1$ and $1$,so its output is $0$. The $AND$ gate receives this $0$ and the original input $1$. Since $0$ and $1$ are inputs to $AND$,the output is $0$. Thus,$C = 0$.
Therefore,the outputs are $A = 1, B = 1, C = 0$.

(C) First,find the equivalent capacitance of the parallel combination of the $4\,\mu F$ and $2\,\mu F$ capacitors:
$C_p = 4\,\mu F + 2\,\mu F = 6\,\mu F$.
Now,the circuit consists of this equivalent capacitor $C_p$ in series with the $6\,\mu F$ capacitor and the $12\,V$ battery.
The total equivalent capacitance $C_{eq}$ of the circuit is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{6\,\mu F} = \frac{1}{6\,\mu F} + \frac{1}{6\,\mu F} = \frac{2}{6\,\mu F} = \frac{1}{3\,\mu F}$.
Thus,$C_{eq} = 3\,\mu F$.
The total charge $Q_{total}$ supplied by the battery is:
$Q_{total} = C_{eq} \times V = 3\,\mu F \times 12\,V = 36\,\mu C$.
Since the $6\,\mu F$ capacitor and the parallel combination $(C_p)$ are in series,they both carry the same charge $Q_{total} = 36\,\mu C$.
The voltage across the parallel combination $C_p$ is:
$V_p = \frac{Q_{total}}{C_p} = \frac{36\,\mu C}{6\,\mu F} = 6\,V$.
Since the $4\,\mu F$ and $2\,\mu F$ capacitors are in parallel,the voltage across the $4\,\mu F$ capacitor is also $6\,V$.
The charge $Q$ on the $4\,\mu F$ capacitor is:
$Q = C \times V = 4\,\mu F \times 6\,V = 24\,\mu C = 24 \times 10^{-6}\,C$.

(B) For combination $(i)$:
The inputs to the $NAND$ gate are $1$ and $1$,so its output is $\overline{1 \cdot 1} = 0$. This $0$ passes through a $NOT$ gate,becoming $1$. The inputs to the $NOR$ gate are $1$ and $0$. The output of a $NOR$ gate is $1$ only if both inputs are $0$. Since one input is $1$,the output $A = 0$.
For combination $(ii)$:
The inputs $0$ and $1$ pass through $NOT$ gates,becoming $1$ and $0$ respectively. These are the inputs to a $NAND$ gate. The output is $\overline{1 \cdot 0} = \overline{0} = 1$. Thus,$B = 1$.
For combination $(iii)$:
The inputs $1$ and $1$ pass through a $NOR$ gate,giving an output of $\overline{1 + 1} = 0$. This $0$ and the input $1$ are fed into an $AND$ gate. The output is $0 \cdot 1 = 0$. Thus,$C = 0$.
Therefore,the outputs are $A = 0, B = 1, C = 0$.

(C) Let the initial number of atoms at time $t=0$ be $N_{0}$.
Let the number of atoms at any instant of time $t$ be $N$.
According to the radioactive decay law,$N = N_{0} e^{-\lambda t}$.
Mean life is defined as $\tau = 1/\lambda$,where $\lambda$ is the decay constant.
Given that the time $t = \tau$,we substitute this into the decay equation:
$N = N_{0} e^{-\lambda \tau} = N_{0} e^{-\lambda (1/\lambda)} = N_{0} e^{-1}$.
Therefore,$N = N_{0}/e$.
The ratio of the initial number of atoms to the number of atoms present at time $t = \tau$ is:
$N_{0}/N = N_{0} / (N_{0}/e) = e$.

(A) Synthetic auxins are artificial compounds that induce physiological responses similar to those of $IAA$ (Indole$-3-$acetic acid).
$2,4-D$ ($2,4$-dichlorophenoxyacetic acid) is a synthetic auxin that does not occur naturally in plants and is widely used as a herbicide or weedicide.
In contrast,$IAA$,$GA$ (Gibberellic acid),and $ABA$ (Abscisic acid) are naturally occurring plant hormones.

(D) The plasma fluid that filters out from glomerular capillaries into Bowman's capsule of nephrons is called glomerular filtrate.
It is a non-colloidal part and contains urea,water,glucose,amino acids,vitamins,fatty acids,uric acid,creatinine,and salts.
$RBCs$,$WBCs$,platelets,and plasma proteins are the colloidal components of the blood and do not filter out from the glomerulus due to their large molecular size.
Thus,plasma proteins are present in higher concentration in the blood compared to the glomerular filtrate,where they are essentially absent.

(D) The $Hypothalamus$ acts as the primary control centre for various homeostatic functions in the human body.
It regulates body temperature,hunger,thirst,sleep,fatigue,and emotional responses such as anger,pleasure,love,and hate.
Since the patient is experiencing an abnormally low body temperature,loss of appetite,and extreme thirst,these symptoms indicate a dysfunction in the $Hypothalamus$.
Therefore,a brain scan would likely reveal a tumour in the $Hypothalamus$.

(A) During the growth phase of oogenesis,an egg nest forms an ovarian follicle (Graafian follicle).
One central oogonium grows and functions as a primary oocyte.
The surrounding follicular cells provide nourishment to the primary oocyte.
Yolk is deposited in the oocyte during this state.
This phenomenon of yolk deposition is called vitellogenesis.

(C) $DNA$ fingerprinting is a technique used to identify an individual based on the specificity of their $DNA$.
This technique relies on the fact that an individual's $DNA$ contains specific nucleotide sequences that are unique to them.
For $DNA$ profiling,cells must contain a nucleus to provide the necessary genetic material.
Among the given options,leucocytes (white blood cells) are nucleated and thus contain $DNA$,whereas erythrocytes (red blood cells) in humans are enucleated (lack a nucleus) and platelets are cell fragments lacking a nucleus.
Therefore,leucocytes are the ideal choice for identifying the criminal.

(C) Pollen grains are haploid $(n)$ structures formed after meiosis.
Given that the diploid number of the plant is $2n = 28$,the haploid number $(n)$ is $28 / 2 = 14$.
When pollen grains are cultured using the tissue culture method,the resulting callus consists of cells derived from the haploid pollen grains.
Therefore,the number of chromosomes in the cells of the callus will be $n = 14$.

(C) Genetic diversity refers to the variety in the number and types of genes and chromosomes present in different species,as well as variations in genes and their alleles within the same species. The introduction of high-yielding varieties (HYVs) is considered the greatest threat to genetic diversity in agricultural crops because it leads to the replacement of traditional,locally adapted landraces with a few genetically uniform,high-productivity strains.

(B) The compound $NH_4Cl$ (ammonium chloride) consists of an ammonium ion $(NH_4^+)$ and a chloride ion $(Cl^-)$.
$1$. The bond between $NH_4^+$ and $Cl^-$ is an ionic bond.
$2$. Within the $NH_4^+$ ion,there are three $N-H$ covalent bonds.
$3$. The fourth $N-H$ bond is formed by the donation of a lone pair of electrons from the nitrogen atom to the $H^+$ ion,which is a coordinate (dative) covalent bond.
Therefore,$NH_4Cl$ contains all three types of linkages.

(D) The pyramidal structure of the covalent molecule $AB_{3}$ is shown below:
In this structure,the central atom $A$ is bonded to three $B$ atoms,forming $3$ bond pairs.
There is $1$ lone pair of electrons present on the central atom $A$.
Therefore,the number of lone pairs is $1$ and the number of bond pairs is $3$.

(D) $H_{2}^{+}$: Electronic configuration is $\sigma 1s^{1}$,$\sigma^{*} 1s^{0}$.
Bond order $= \frac{1-0}{2} = 0.5$.
$H_{2}^{-}$: Electronic configuration is $\sigma 1s^{2}$,$\sigma^{*} 1s^{1}$.
Bond order $= \frac{2-1}{2} = 0.5$.
Although the bond orders of $H_{2}^{+}$ and $H_{2}^{-}$ are the same,$H_{2}^{+}$ is more stable than $H_{2}^{-}$.
This is because $H_{2}^{-}$ contains one electron in the antibonding molecular orbital $(\sigma^{*} 1s)$,which decreases its stability compared to $H_{2}^{+}$.

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$.
For $O_{2}^{+}$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $\frac{1}{2}(10 - 5) = 2.5$.
For $O_{2}$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $\frac{1}{2}(10 - 6) = 2.0$.
For $O_{2}^{-}$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $\frac{1}{2}(10 - 7) = 1.5$.
For $O_{2}^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $\frac{1}{2}(10 - 8) = 1.0$.
Thus,the increasing order of bond order is: $O_{2}^{2-} < O_{2}^{-} < O_{2} < O_{2}^{+}$.

(A) The carbon-carbon bond length in benzene is $1.39 \ \mathring{A}$.
In ethane $(C_{2}H_{6})$,the $C-C$ single bond length is $1.54 \ \mathring{A}$.
In ethene $(C_{2}H_{4})$,the $C=C$ double bond length is $1.34 \ \mathring{A}$.
Since $1.34 \ \mathring{A} < 1.39 \ \mathring{A} < 1.54 \ \mathring{A}$,the bond length in benzene lies between that of $C_{2}H_{6}$ and $C_{2}H_{4}$.

(A) The equilibrium constant $(K_c)$ of a chemical reaction depends only on the temperature of the system. It remains unchanged by the addition or removal of reactants or products,provided the temperature remains constant at $300 \ K$. Therefore,the value of the equilibrium constant remains $6.4$.

(B) In a pressure cooker,the pressure inside is significantly higher than the atmospheric pressure. According to the relationship between vapor pressure and boiling point,as the pressure increases,the boiling point of water also increases. Consequently,water boils at a higher temperature (above $100 \ ^{\circ}C$),which provides more thermal energy to the food,leading to faster cooking.

(B) On moving from left to right in a period,ionisation energy generally increases due to an increase in effective nuclear charge.
Thus,the expected order of ionisation energy for $C$,$N$,$O$,and $F$ is $C < N < O < F$.
However,the ionisation energy of $N$ $(1s^2 2s^2 2p^3)$ is greater than that of $O$ $(1s^2 2s^2 2p^4)$.
This is because $N$ has a stable half-filled $p$-orbital configuration,which requires more energy to remove an electron.
Therefore,the correct order is $C < O < N < F$.

(D) Let the sides be $b = 2a$ and $a$. The angles opposite to these sides are $B$ and $A$ respectively. Given $B - A = 60^{\circ}$.
By the sine rule,$\frac{b}{\sin B} = \frac{a}{\sin A}$ $\Rightarrow \frac{2a}{\sin B} = \frac{a}{\sin A}$ $\Rightarrow \sin B = 2 \sin A$.
Substitute $B = A + 60^{\circ}$:
$\sin(A + 60^{\circ}) = 2 \sin A$
$\sin A \cos 60^{\circ} + \cos A \sin 60^{\circ} = 2 \sin A$
$\frac{1}{2} \sin A + \frac{\sqrt{3}}{2} \cos A = 2 \sin A$
$\frac{\sqrt{3}}{2} \cos A = \frac{3}{2} \sin A$
$\tan A = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$
$A = 30^{\circ}$.
Then $B = 30^{\circ} + 60^{\circ} = 90^{\circ}$.
Since one angle is $90^{\circ}$,the triangle is a right-angled triangle.

(C) In a ruby laser,the active medium is a ruby crystal ($Al_2O_3$ doped with $Cr^{3+}$ ions).
When the ruby crystal is pumped with high-intensity light,the $Cr^{3+}$ ions are excited to higher energy levels.
These ions quickly decay to a metastable state,which has a relatively long lifetime.
Stimulated emission occurs when these ions transition from the metastable state to the ground state,releasing photons of a specific wavelength $(694.3 \ nm)$.
Therefore,the stimulated emission in a ruby laser is due to the transition from the metastable state to the ground state.

(C) The given compound is $(CH_3)_3 CCl$.
To find the $IUPAC$ name,we first identify the longest carbon chain containing the functional group.
The longest chain has $3$ carbon atoms,so the parent alkane is propane.
Numbering the chain from the end that gives the lowest number to the substituent,we get the chlorine atom and the methyl group at the $2^{nd}$ position.
Thus,the $IUPAC$ name is $2$-chloro-$2$-methylpropane.

(A) The nitro group $(-NO_2)$ is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
It withdraws electron density from the benzene ring,specifically from the ortho $(o)$ and para $(p)$ positions through resonance.
As a result,the electron density at the $o$ and $p$ positions decreases significantly compared to the meta $(m)$ positions.
Since the electrophile is electron-deficient,it prefers to attack the position with relatively higher electron density,which is the meta position.
Therefore,the nitro group is meta-directing.

(D) The electrophile involved in the sulphonation of benzene is $SO_{3}$.
During the reaction,concentrated sulfuric acid acts as both a reagent and a catalyst to generate the electrophile.
The reaction is: $2 H_{2}SO_{4} \longrightarrow SO_{3} + H_{3}O^{+} + HSO_{4}^{-}$

(B) Total milliequivalents of $H^{+}$ ions from $HCl$ and $HNO_{3}$:
$= (30 \times \frac{1}{3}) + (20 \times \frac{1}{2}) = 10 + 10 = 20 \ mEq$.
Total milliequivalents of $OH^{-}$ ions from $NaOH$:
$= 40 \times \frac{1}{4} = 10 \ mEq$.
Since $H^{+}$ reacts with $OH^{-}$ to form water,the remaining milliequivalents of $H^{+}$:
$= 20 - 10 = 10 \ mEq$.
The final volume of the solution is $1 \ dm^{3} = 1000 \ mL$.
Concentration of $H^{+}$ ions $[H^{+}] = \frac{10 \ mEq}{1000 \ mL} = 10^{-2} \ M$.
$pH = -\log[H^{+}] = -\log(10^{-2}) = 2$.

(A) Initial concentration of $[OH^-] = 10^{-6} \ M$.
Upon dilution by $100$ times,the concentration becomes $[OH^-] = \frac{10^{-6}}{100} = 10^{-8} \ M$.
Since this concentration is very low,we must consider the contribution of $[OH^-]$ from the auto-ionization of water,which is $10^{-7} \ M$.
Total $[OH^-] = 10^{-8} + 10^{-7} = 10^{-8} (1 + 10) = 11 \times 10^{-8} \ M$.
$pOH = -\log(11 \times 10^{-8}) = 8 - \log(11) \approx 8 - 1.0414 = 6.9586$.
$pH = 14 - pOH = 14 - 6.9586 = 7.0414$.
Thus,the $pH$ lies between $7$ and $8$.

(A) When an organic compound is heated with $CuO$,carbon is oxidized to $CO_{2}$ and hydrogen is oxidized to $H_{2}O$.
Since the compound produces $CO_{2}$,it must contain carbon.
Since it does not produce water,it does not contain hydrogen.
Among the given options,$CCl_{4}$ (carbon tetrachloride) is the only compound that contains carbon but no hydrogen.
Therefore,the correct option is $A$.

(D) The atomic mass of a metal is given by the formula: $\text{Atomic mass} = \text{Equivalent mass} \times \text{Valency}$.
Given,$\text{Equivalent mass} = 32$ and the metal is bivalent (valency $= 2$).
Therefore,$\text{Atomic mass} = 32 \times 2 = 64$.
The formula of the metal nitrate for a bivalent metal $M$ is $M(NO_3)_2$.
The molecular mass of $M(NO_3)_2 = \text{Atomic mass of } M + 2 \times (\text{Atomic mass of } N + 3 \times \text{Atomic mass of } O)$.
Substituting the values: $64 + 2 \times (14 + 3 \times 16) = 64 + 2 \times (14 + 48) = 64 + 2 \times 62 = 64 + 124 = 188$.

(C) The electronic configuration $1s^{2} 2s^{2} 2p^{6} 3s^{1}$ corresponds to the element Sodium ($Na$,atomic number $11$).
Sodium is an alkali metal belonging to Group $1$ of the periodic table.
Alkali metals react with oxygen to form oxides of the general formula $M_{2}O$ (e.g.,$Na_{2}O$).
These oxides are strongly basic in nature as they react with water to form hydroxides.

(A) Alkali metals have low ionisation energy due to their large atomic size.
They possess the minimum value of ionisation energy in their respective periods.
Therefore,the statement that they have high ionisation energy is incorrect.

(C) The reaction of $CO_{2}$ with $Ca(OH)_{2}$ in excess leads to the formation of $CaCO_{3}$ and water: $Ca(OH)_{2} + CO_{2} \rightarrow CaCO_{3} + H_{2}O$.
Millimoles of $Ca(OH)_{2} = 50 \ mL \times 0.5 \ M = 25 \ mmol$.
Since the stoichiometry is $1:1$,millimoles of $CaCO_{3}$ formed $= 25 \ mmol$.
The reaction of $CaCO_{3}$ with $HCl$ is: $CaCO_{3} + 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}$.
The milliequivalents of $CaCO_{3}$ $= \text{millimoles} \times n\text{-factor} = 25 \times 2 = 50 \ meq$.
At neutralization,milliequivalents of $HCl = \text{milliequivalents of } CaCO_{3} = 50 \ meq$.
Volume of $0.1 \ N \ HCl = \frac{50 \ meq}{0.1 \ N} = 500 \ cm^{3}$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.
The molar masses are: $H_{2} = 2 \text{ g/mol}$,$CH_{4} = 16 \text{ g/mol}$,$SO_{2} = 64 \text{ g/mol}$.
Since $M(H_{2}) < M(CH_{4}) < M(SO_{2})$,the rate of diffusion is $r(H_{2}) > r(CH_{4}) > r(SO_{2})$.
Therefore,the amount of gas remaining in the container after time $t$ will be in the reverse order of their diffusion rates.
The amount left is in the order: $SO_{2} > CH_{4} > H_{2}$.
Since partial pressure $p$ is directly proportional to the number of moles $n$ $(p = \frac{nRT}{V})$,the order of partial pressures is $pSO_{2} > pCH_{4} > pH_{2}$.

(D) The reaction is: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$
Initial moles: $SO_2 = 5, O_2 = 5, SO_3 = 0$
$60\%$ of $SO_2$ used up $= 5 \times 0.6 = 3 \ moles$.
At equilibrium:
$SO_2 = 5 - 3 = 2 \ moles$
$O_2 = 5 - (3/2) = 5 - 1.5 = 3.5 \ moles$
$SO_3 = 3 \ moles$
Total moles at equilibrium $= 2 + 3.5 + 3 = 8.5 \ moles$.
Partial pressure of $O_2 = (\text{moles of } O_2 / \text{total moles}) \times \text{Total Pressure}$
$pO_2 = (3.5 / 8.5) \times 1 \ atm = 0.41 \ atm$.

(A) The formula for the $rms$ velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3P}{d}}$.
Given,pressure $P = 1.2 \times 10^{5} \ Nm^{-2}$ and density $d = 4 \ kg \ m^{-3}$.
Substituting the values in the formula:
$v_{rms} = \sqrt{\frac{3 \times 1.2 \times 10^{5}}{4}}$
$v_{rms} = \sqrt{\frac{3.6 \times 10^{5}}{4}}$
$v_{rms} = \sqrt{0.9 \times 10^{5}} = \sqrt{90000} = 300 \ ms^{-1}$.

(A) The de-Broglie wavelength formula is given by $\lambda = \frac{h}{mv}$.
Rearranging the formula to solve for mass $m$,we get $m = \frac{h}{\lambda v}$.
Given values are $h = 6.62 \times 10^{-34} \text{ } Js$,$\lambda = 6.62 \times 10^{-35} \text{ } m$,and $v = 100 \text{ } ms^{-1}$.
Substituting these values into the equation:
$m = \frac{6.62 \times 10^{-34}}{6.62 \times 10^{-35} \times 100}$.
$m = \frac{10^{-34}}{10^{-35} \times 10^2} = \frac{10^{-34}}{10^{-33}} = 10^{-1} = 0.1 \text{ } kg$.
Therefore,$x = 0.1 \text{ } kg$.

(C) The electronic configuration of potassium $(Z=19)$ is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1}$.
The outermost electron is in the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $(n) = 4$.
The azimuthal quantum number $(l)$ for an $s$-orbital is $0$.
The magnetic quantum number $(m_l)$ is $0$.
The spin quantum number $(m_s)$ is $+\frac{1}{2}$ (or $-\frac{1}{2}$).
Thus,the correct set of quantum numbers is $4, 0, 0, \frac{1}{2}$.

(D) Ionisation energy generally increases from left to right in a period due to an increase in effective nuclear charge.
However,nitrogen $(N)$ has a stable half-filled $2p^3$ electronic configuration,which makes it more difficult to remove an electron compared to oxygen $(O)$.
Therefore,the ionisation energy of $N$ is greater than that of $O$.
The correct order is $C < O < N < F$.

(C) The reaction is $2 NH_{3(g)} \longrightarrow N_{2(g)} + 3 H_{2(g)}$.
For any reaction,$\Delta H_r = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants})$.
Since the elements $N_2$ and $H_2$ are in their standard states,their enthalpy of formation is $0$.
$\Delta H_r = [1 \times \Delta H_f^{\circ}(N_2) + 3 \times \Delta H_f^{\circ}(H_2)] - [2 \times \Delta H_f^{\circ}(NH_3)]$.
$\Delta H_r = [0 + 0] - [2 \times (-46 \ kJ \ mol^{-1})]$.
$\Delta H_r = +92 \ kJ$.

(C) The given reaction $H_{2}O_{(l)} \rightleftharpoons H_{2}O_{(g)}$ represents the phase transition of water at its boiling point ($373 \ K$ and $1 \ atm$).
At equilibrium,the Gibbs free energy change is zero,i.e.,$\Delta G = 0$.
Using the thermodynamic relation $\Delta G = \Delta H - T \Delta S$,we substitute $\Delta G = 0$:
$0 = \Delta H - T \Delta S$
Therefore,$\Delta H = T \Delta S$.

(D) The Lucas test is a chemical test used to distinguish between primary,secondary,and tertiary alcohols. It involves the reaction of an alcohol with Lucas reagent ($conc. \ HCl$ and $ZnCl_2$). Tertiary alcohols react immediately to form a cloudy precipitate,secondary alcohols react within $5-10 \ minutes$,and primary alcohols do not react at room temperature.

(B) Dry distillation of a mixture of calcium salts of carboxylic acids produces carbonyl compounds.
$1. (HCOO)_{2}Ca \xrightarrow{\Delta} HCHO + CaCO_{3}$ (Methanal)
$2. (CH_{3}COO)_{2}Ca \xrightarrow{\Delta} CH_{3}COCH_{3} + CaCO_{3}$ (Propanone)
$3. (HCOO)_{2}Ca + (CH_{3}COO)_{2}Ca \xrightarrow{\Delta} 2CH_{3}CHO + 2CaCO_{3}$ (Ethanal)
Propanal $(CH_{3}CH_{2}CHO)$ is not formed in this reaction.

(A) When $1, 1-$dichloroethane $(CH_3CHCl_2)$ is heated with dilute $NaOH$,it undergoes hydrolysis to form an unstable gem-diol intermediate,$1, 1-$ethanediol $(CH_3CH(OH)_2)$.
This unstable intermediate loses a water molecule to form acetaldehyde $(CH_3CHO)$.
The reaction is as follows:
$CH_3CHCl_2 + 2NaOH \rightarrow CH_3CH(OH)_2 + 2NaCl$
$CH_3CH(OH)_2 \rightarrow CH_3CHO + H_2O$

(B) $Step \ 1$: Reduction of acetic acid $(CH_3COOH)$ with $LiAlH_4$ gives ethanol $(CH_3CH_2OH)$ as $X$.
$Step \ 2$: Dehydrogenation of ethanol $(X)$ over $Cu$ at $300^{\circ}C$ gives acetaldehyde $(CH_3CHO)$ as $Y$.
$Step \ 3$: Acetaldehyde $(Y)$ undergoes aldol condensation in the presence of dilute $NaOH$ to form $3-hydroxybutanal$ $(CH_3CH(OH)CH_2CHO)$,which is commonly known as aldol,as $Z$.
Therefore,the correct option is $B$.

(B) Aliphatic amines are more basic than $NH_{3}$ due to the $+I$ effect of alkyl groups.
In an aqueous medium,the basicity is determined by a combination of the $+I$ effect,solvation effect,and steric hindrance.
For methyl-substituted amines in an aqueous solution,the order of basic strength is $(CH_{3})_{2}NH > CH_{3}NH_{2} > (CH_{3})_{3}N > NH_{3}$.
However,looking at the provided options and the standard trend,the correct increasing order is $NH_{3} < (CH_{3})_{3}N < CH_{3}NH_{2} < (CH_{3})_{2}NH$.

(D) Sometimes the blood sugar level of diabetic patients decreases suddenly.
Glucose is a simple sugar that is absorbed directly into the bloodstream.
Therefore,diabetic patients generally carry a packet of glucose,which can increase the blood sugar level almost instantaneously to prevent hypoglycemia.

(A) tripeptide is formed by the combination of $3$ amino acids.
Since there are $20$ different types of naturally occurring amino acids and each position in the tripeptide can be occupied by any of these $20$ amino acids,the total number of possible tripeptides is calculated as $20 \times 20 \times 20 = 20^{3}$.
Therefore,the total number of tripeptides is $8000$.

(D) The oil of wintergreen test is an esterification reaction used to identify salicylic acid.
Salicylic acid reacts with methanol $(CH_{3}OH)$ in the presence of a few drops of concentrated $H_{2}SO_{4}$ (acid catalyst) upon heating to form methyl salicylate.
Methyl salicylate is an ester that has the characteristic smell of oil of wintergreen.
The reaction is: $C_{6}H_{4}(OH)COOH + CH_{3}OH \xrightarrow[Conc. H_{2}SO_{4}]{\Delta} C_{6}H_{4}(OH)COOCH_{3} + H_{2}O$.

(A) The stability of metal oxides is determined by their standard Gibbs free energy of formation $(\Delta G_f^\circ)$.
More negative values of $\Delta G_f^\circ$ indicate higher stability.
Based on the Ellingham diagram,the order of stability is:
$Fe_{2}O_{3} < Cr_{2}O_{3} < Al_{2}O_{3} < MgO$.

(C) The rate law for the reaction $A \rightarrow B$ is given by: $\text{Rate} = k[A]^{n}$,where $n$ is the order of the reaction.
For the first condition: $2 \times 10^{-3} = k(0.05)^{n} \quad \dots(i)$
For the second condition: $1.6 \times 10^{-2} = k(0.1)^{n} \quad \dots(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{1.6 \times 10^{-2}}{2 \times 10^{-3}} = \frac{k(0.1)^{n}}{k(0.05)^{n}}$
$8 = (\frac{0.1}{0.05})^{n}$
$8 = (2)^{n}$
Since $8 = 2^{3}$,we have $2^{3} = 2^{n}$.
Therefore,$n = 3$.

(B) Total time from $1^{st}$ Jan to $1^{st}$ March $2009$ (including both days) $= 31 \text{ (Jan)} + 28 \text{ (Feb)} + 1 \text{ (March)} = 60 \text{ days}$.
Number of half-lives $(n)$ $= \frac{\text{Total time}}{\text{Half-life}} = \frac{60}{15} = 4$.
Amount left $(N)$ $= N_0 \times (\frac{1}{2})^n = 2 \times (\frac{1}{2})^4 = \frac{2}{16} = 0.125 \text{ g}$.

(A) For a zero order reaction,the rate law is given by $r = k[A]^0 = k$.
Integrating the rate equation,we get $[A]_t = [A]_0 - kt$.
At half-life $(t = t_{1/2})$,the concentration of reactant $[A]_t = \frac{[A]_0}{2} = \frac{a}{2}$.
Substituting these values: $\frac{a}{2} = a - kt_{1/2}$.
$kt_{1/2} = a - \frac{a}{2} = \frac{a}{2}$.
Therefore,$t_{1/2} = \frac{a}{2k}$.

(D) The rate law for the reaction is given by: $Rate = k[AB]^n$.
Using the data from the table:
$2.75 \times 10^{-8} = k(0.20)^n$ --- $(i)$
$11.0 \times 10^{-8} = k(0.40)^n$ --- $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{11.0 \times 10^{-8}}{2.75 \times 10^{-8}} = (\frac{0.40}{0.20})^n$
$4 = 2^n$
Since $4 = 2^2$,we have $2^2 = 2^n$.
Therefore,$n = 2$. The reaction is of second order.

(D) In the brown ring complex $[Fe(H_2O)_5NO]SO_4$,the nitrosyl ligand is present as $NO^+$.
The complex ion is $[Fe(H_2O)_5NO]^{2+}$.
Let the oxidation state of $Fe$ be $x$.
$x + 5(0) + 1 = +2$
$x = +1$
Therefore,the oxidation state of $Fe$ is $+1$.

(B) The complex ion is $[Co(NH_3)_5 ONO]^{2+}$.
$1$. The ligand $NH_3$ is named as 'ammine' and there are $5$ of them,so it is 'pentaammine'.
$2$. The ligand $ONO^-$ is an ambidentate ligand attached through oxygen,named as 'nitrito-$O$'.
$3$. The central metal atom is Cobalt $(Co)$.
$4$. Let the oxidation state of $Co$ be $x$. The charge on $NH_3$ is $0$ and on $ONO$ is $-1$. The total charge is $+2$.
$x + 5(0) + (-1) = +2 \implies x = +3$.
$5$. Combining these,the $IUPAC$ name is pentaamminenitrito-$O$-cobalt$(III)$ ion.

(A) The function of $Fe(OH)_{3}$ in the contact process is to remove arsenic impurity.
$Fe(OH)_{3}$ is a positive sol,hence it removes arsenic impurity which is a negative sol.

(C) The oxidation of a secondary alcohol using acidified $K_{2}Cr_{2}O_{7}$ yields a ketone.
Ketones react with phenyl hydrazine to form phenylhydrazones but do not give the silver mirror test (Tollens' test),which is specific to aldehydes.
Among the given options,$(CH_{3})_{2}CHOH$ is a secondary alcohol (propan$-2-$ol),which on oxidation gives acetone $(CH_{3}COCH_{3})$,a ketone.
Therefore,the correct structure of $X$ is $(CH_{3})_{2}CHOH$.

(D) The chromyl chloride test is used for the detection of $Cl^-$ ions.
In this test,$4NaCl + K_2Cr_2O_7 + 3H_2SO_4 \longrightarrow K_2SO_4 + 2Na_2SO_4 + 2CrO_2Cl_2 \uparrow + 3H_2O$.
The $CrO_2Cl_2$ (chromyl chloride) formed appears as reddish-brown vapours.
When these vapours are passed into $NaOH$ solution,it forms sodium chromate: $CrO_2Cl_2 + 4NaOH \longrightarrow 2NaCl + Na_2CrO_4 + 2H_2O$.
This solution,when treated with lead acetate,gives a yellow precipitate of lead chromate: $Na_2CrO_4 + (CH_3COO)_2Pb \longrightarrow 2CH_3COONa + PbCrO_4 \downarrow$.
Chlorine gas is not liberated in this test. Therefore,the statement 'liberation of chlorine' is incorrect.

(C) The reaction for the electrolysis of aqueous $NaCl$ is: $2NaCl + 2H_2O \rightarrow 2NaOH + Cl_2 + H_2$.
Weight of pure $NaCl = 6.5 \ g \times 0.9 = 5.85 \ g$.
Moles of $NaCl = \frac{5.85 \ g}{58.5 \ g/mol} = 0.1 \ mol$.
According to the stoichiometry,$2 \ mol$ of $NaCl$ produces $2 \ mol$ of $NaOH$,so $0.1 \ mol$ of $NaCl$ produces $0.1 \ mol$ of $NaOH$.
For neutralization: $NaOH + CH_3COOH \rightarrow CH_3COONa + H_2O$.
Moles of $CH_3COOH$ required = Moles of $NaOH = 0.1 \ mol$.
Volume of $1 \ M$ acetic acid = $\frac{\text{moles}}{\text{molarity}} = \frac{0.1 \ mol}{1 \ M} = 0.1 \ L = 100 \ cm^{3}$.

(D) The cell reaction is $Fe^{2+} + Zn \longrightarrow Zn^{2+} + Fe$.
Here,$Zn$ is oxidized to $Zn^{2+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).
The formula for the $emf$ of the cell is $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Substituting the given values: $E^{\circ}_{cell} = E^{\circ}_{Fe^{2+}/Fe} - E^{\circ}_{Zn^{2+}/Zn}$.
$E^{\circ}_{cell} = -0.44 \ V - (-0.76 \ V)$.
$E^{\circ}_{cell} = -0.44 \ V + 0.76 \ V = +0.32 \ V$.

(B) The reaction for the evolution of hydrogen at the cathode is: $2H^+ + 2e^- \rightarrow H_2(g)$.
From the stoichiometry,$1 \ mole$ of $H_2$ requires $2 \ moles$ of electrons $(2 \ F)$.
Volume of $H_2$ produced per second $= 1.12 \ cc = 1.12 \ mL$.
At $STP$,$22400 \ mL$ of $H_2$ corresponds to $1 \ mole$.
Therefore,moles of $H_2$ produced per second $= \frac{1.12}{22400} = 5 \times 10^{-5} \ mol$.
Number of moles of electrons required per second $= 2 \times (5 \times 10^{-5}) = 10^{-4} \ mol$.
Since $1 \ mole$ of electrons $= 96500 \ C$,the charge required per second is $Q = 10^{-4} \times 96500 = 9.65 \ C$.
Since $Current (I) = \frac{Q}{t}$ and $t = 1 \ s$,the current is $9.65 \ A$.

(C) Specific conductance (or conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of the solution.
As the solution is diluted,the number of ions present per unit volume $(1 \ cm^3)$ decreases.
Since the number of charge carriers per unit volume decreases,the specific conductance decreases with dilution.

(B) Cinnabar $(HgS)$ is a sulphide ore.
Froth floatation process is specifically used for the concentration of sulphide ores.
Therefore,cinnabar is concentrated by this process.

(D) In the blast furnace used for the extraction of iron,different zones have different temperature ranges.
The slag formation zone,where limestone $(CaCO_3)$ decomposes to $CaO$ and reacts with silica $(SiO_2)$ to form slag $(CaSiO_3)$,occurs at a temperature range of approximately $800-1000^{\circ} C$.
Therefore,the correct option is $D$.

(C) The best method for the conversion of an alcohol into an alkyl chloride is by treating the alcohol with $SOCl_{2}$ in the presence of pyridine.
$ROH + SOCl_{2} \longrightarrow RCl + SO_{2} \uparrow + HCl \uparrow$
In this reaction,the by-products $SO_{2}$ and $HCl$ are gases,which escape from the reaction mixture,leaving behind pure alkyl chloride. This is known as the Darzens process.

(C) The iodine value is a measure of the degree of unsaturation in fats and oils. Higher unsaturation leads to a higher iodine value. $Ghee$ is primarily composed of saturated fatty acids,making it the least unsaturated among the given options. Therefore,it has the least iodine value.

(C) Argon is an inert gas. It is primarily used in high temperature welding and other metallurgical operations that require a non-oxidising atmosphere and the absence of nitrogen to prevent unwanted chemical reactions.

(B) $Proteins$ are the condensation polymers of $\alpha$-amino acids.
$Proteins$ contain peptide bonds,which are represented by the linkage:
$(-CO-NH-)$.

(C) $NH_{3}$ is used in the preparation of Tollen's reagent $([Ag(NH_{3})_{2}]^{+})$.
$NH_{3}$ is also used as a group reagent for the analysis of $III$ group basic radicals (as $NH_{4}OH$).
$NH_{3}$ is not used as a group reagent for the analysis of $IV$ group basic radicals,where $H_{2}S$ in the presence of $NH_{4}OH$ is used.
Nessler's reagent is $K_{2}[HgI_{4}]$ in $KOH$,which is used to test $NH_{4}^{+}$ ions,but $NH_{3}$ is not a component of the reagent itself.

(D) occupies corners,thus number of $A$ atoms per unit cell $= 8 \times \frac{1}{8} = 1$.
$B$ occupies face centres,thus number of $B$ atoms per unit cell $= 6 \times \frac{1}{2} = 3$.
Therefore,the empirical formula of the compound is $AB_{3}$.

(D) According to Raoult's law for an ideal solution,the total vapour pressure $P_{total}$ is given by:
$P_{total} = P_{A}^{\circ} x_{A} + P_{B}^{\circ} x_{B}$
Given: $P_{A}^{\circ} = 70 \ mm \ Hg$,$x_{A} = 0.8$,$P_{total} = 84 \ mm \ Hg$.
Since $x_{A} + x_{B} = 1$,we have $x_{B} = 1 - 0.8 = 0.2$.
Substituting the values:
$84 = (70 \times 0.8) + (P_{B}^{\circ} \times 0.2)$
$84 = 56 + 0.2 P_{B}^{\circ}$
$84 - 56 = 0.2 P_{B}^{\circ}$
$28 = 0.2 P_{B}^{\circ}$
$P_{B}^{\circ} = \frac{28}{0.2} = 140 \ mm \ Hg$.

(A) In countries nearer to the polar region,the roads are sprinkled with $CaCl_{2}$ because $CaCl_{2}$ acts as a freezing point depressant.
It lowers the freezing point of water,which helps in melting the ice on the roads,thereby preventing ice formation and minimizing the wear and tear of the roads.

(A) The molar mass of urea $(NH_2CONH_2)$ is $60 \ g/mol$.
Assuming the density of the solution is $1 \ g/mL$,a $6 \%$ solution means $6 \ g$ of urea in $100 \ mL$ of solution.
Molarity of urea $= \frac{\text{mass of solute}}{\text{molar mass} \times \text{volume of solution in L}} = \frac{6 \ g}{60 \ g/mol \times 0.1 \ L} = 1 \ M$.
Two solutions are isotonic if they have the same molar concentration.
Therefore,a $6 \%$ urea solution is isotonic with a $1 \ M$ solution of glucose.

(A) The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \ \text{BM}$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{15} \ \text{BM}$.
$\sqrt{n(n+2)} = \sqrt{15}$
$n(n+2) = 15$
$n^2 + 2n - 15 = 0$
$(n+5)(n-3) = 0$
Since $n$ cannot be negative,$n = 3$.

(B) Physical adsorption is an exothermic process.
According to Le Chatelier's principle,for an exothermic process,the extent of adsorption decreases with an increase in temperature.
Therefore,the rate of physical adsorption increases with a decrease in temperature.