KCET 2020 Chemistry Question Paper with Answer and Solution

(D) The formula for the refractive index $\mu$ of a prism in terms of the refracting angle $A$ and the angle of minimum deviation $\delta_m$ is given by:
$\mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)}$
Given $\mu = \cot \frac{A}{2}$,we substitute this into the formula:
$\cot \frac{A}{2} = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)}$
Since $\cot \theta = \frac{\cos \theta}{\sin \theta}$,we have:
$\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}}$
$\cos \frac{A}{2} = \sin \left( \frac{A + \delta_m}{2} \right)$
Using the trigonometric identity $\cos \theta = \sin(90^\circ - \theta)$:
$\sin \left( 90^\circ - \frac{A}{2} \right) = \sin \left( \frac{A + \delta_m}{2} \right)$
Comparing the angles:
$90^\circ - \frac{A}{2} = \frac{A + \delta_m}{2}$
$180^\circ - A = A + \delta_m$
$\delta_m = 180^\circ - 2A$

(B) According to Curie's Law,the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature $T$,i.e.,$\chi = \frac{C}{T}$.
The magnetization $I$ is given by $I = \chi H$,where $H$ is the magnetic field intensity. Since $B = \mu_0 H$,we have $H = \frac{B}{\mu_0}$.
Thus,$I = \chi \frac{B}{\mu_0} = \frac{C}{T} \cdot \frac{B}{\mu_0} = K \frac{B}{T}$,where $K$ is a constant.
Given:
$I_1 = 8 \ Am^{-1}$,$B_1 = 0.6 \ T$,$T_1 = 4 \ K$
$I_2 = ?$,$B_2 = 0.2 \ T$,$T_2 = 16 \ K$
Using the ratio $\frac{I_2}{I_1} = \frac{B_2}{B_1} \cdot \frac{T_1}{T_2}$:
$\frac{I_2}{8} = \frac{0.2}{0.6} \cdot \frac{4}{16}$
$\frac{I_2}{8} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$
$I_2 = 8 \cdot \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \ Am^{-1}$.

(D) The formula for the refractive index $\mu$ of a prism in terms of the angle of minimum deviation $\delta_m$ and the refracting angle $A$ is given by:
$\mu = \frac{\sin[(\delta_m + A)/2]}{\sin(A/2)}$
Given that $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)}$,we substitute this into the formula:
$\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin[(\delta_m + A)/2]}{\sin(A/2)}$
Canceling $\sin(A/2)$ from both sides,we get:
$\cos(A/2) = \sin[(\delta_m + A)/2]$
Using the trigonometric identity $\cos(\theta) = \sin(90^\circ - \theta)$,we can write:
$\sin(90^\circ - A/2) = \sin[(\delta_m + A)/2]$
Equating the angles:
$90^\circ - A/2 = (\delta_m + A)/2$
Multiplying by $2$:
$180^\circ - A = \delta_m + A$
Therefore,$\delta_m = 180^\circ - 2A$.

(B) Let the mass of the body be $m$ and the constant acceleration be $a$.
According to Newton's second law of motion,the force $F$ acting on the body is:
$F = ma$
Since $m$ and $a$ are constants,the force $F$ is also constant.
For a body starting from rest $(u = 0)$,the velocity $v$ at time $t$ is given by the equation of motion:
$v = u + at = 0 + at = at$
Thus,$v \propto t$.
Power $P$ delivered to the body is defined as the product of force and velocity:
$P = F \cdot v$
Substituting the expressions for $F$ and $v$:
$P = (ma) \cdot (at) = (ma^2)t$
Since $m$ and $a$ are constants,$ma^2$ is also a constant.
Therefore,$P \propto t$.

(A) The formula for the refractive index $\mu$ of a prism in terms of the refracting angle $A$ and the angle of minimum deviation $\delta_{m}$ is given by:
$\mu = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$
Given that $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)}$,we substitute this into the formula:
$\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$
Canceling $\sin(A/2)$ from both sides,we get:
$\cos(A/2) = \sin((A + \delta_{m})/2)$
Using the trigonometric identity $\cos(\theta) = \sin(90^{\circ} - \theta)$,we can write:
$\sin(90^{\circ} - A/2) = \sin((A + \delta_{m})/2)$
Comparing the angles:
$90^{\circ} - A/2 = (A + \delta_{m})/2$
$180^{\circ} - A = A + \delta_{m}$
$\delta_{m} = 180^{\circ} - 2A$

(D) The reaction is between isobutylmagnesium bromide and methylamine. The Grignard reagent acts as a base and abstracts the acidic proton from the amine to form an alkane.
$ (CH_3)_2CHCH_2MgBr + CH_3NH_2 \rightarrow (CH_3)_2CHCH_3 + Mg(Br)NHCH_3 $
The organic compound $X$ formed is isobutane,which is $2$-methylpropane,$(CH_3)_2CHCH_3$.
The molecular formula of $X$ is $C_4H_{10}$.
The isomers of $C_4H_{10}$ are:
$1$. $n$-butane: $CH_3-CH_2-CH_2-CH_3$
$2$. Isobutane ($2$-methylpropane): $CH_3-CH(CH_3)-CH_3$
Thus,there are $2$ possible isomers for the organic compound $X$.

(A) In $PH_{3}$,the phosphorus atom has one lone pair of electrons. According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion,which causes the bond angle to decrease from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to approximately $93.6^{\circ}$,resulting in a pyramidal geometry.
In $PH_{4}^{+}$,the phosphorus atom has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsions and the presence of four identical bond pair-bond pair interactions,$PH_{4}^{+}$ adopts a perfect tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.

(B) In $NO_{2}^{+}$,the nitrogen atom is $sp$-hybridized. The $s$-character in $sp$-hybridization is $\frac{1}{2} \times 100 = 50 \%$.
In $NO_{3}^{-}$,the nitrogen atom is $sp^{2}$-hybridized. The $s$-character in $sp^{2}$-hybridization is $\frac{1}{3} \times 100 = 33.3 \%$.
Therefore,the percentages are $50 \%$ and $33.3 \%$ respectively.

(D) The formal charge on an atom in a Lewis structure is calculated using the formula:
Formal Charge = (Total number of valence electrons on the free atom) - ($\frac{1}{2}$ $\times$ Total number of shared electrons) - (Total number of unshared valence electrons).
For the central oxygen atom in the ozone $(O_3)$ molecule:
$1$. The central oxygen atom has $6$ valence electrons in its free state.
$2$. It forms one double bond with one oxygen atom and one single bond with another oxygen atom,sharing a total of $6$ electrons ($3$ bonds $\times$ $2$ electrons per bond).
$3$. It has $2$ unshared electrons (one lone pair).
Formal Charge = $6 - \frac{1}{2}(6) - 2 = 6 - 3 - 2 = +1$.

(A) The balanced chemical equation for the reaction is:
$Cr_{2}O_{7}^{2-} + 6I^{-} + 14H^{+} \longrightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O$
From the stoichiometry of the reaction,$1$ mole of $K_{2}Cr_{2}O_{7}$ produces $3$ moles of $I_{2}$.
Therefore,to liberate $6$ moles of $I_{2}$,the required moles of $K_{2}Cr_{2}O_{7}$ is:
$\frac{1 \text{ mol } K_{2}Cr_{2}O_{7}}{3 \text{ mol } I_{2}} \times 6 \text{ mol } I_{2} = 2 \text{ moles of } K_{2}Cr_{2}O_{7}$.

(B) Let $X$ be the number of times an odd number appears in $10$ throws of a die. This follows a binomial distribution with $n = 10$ and $p = \frac{3}{6} = \frac{1}{2}$.
Thus,$q = 1 - p = \frac{1}{2}$.
We need to find the probability of getting an odd number at least once,which is $P(X \geq 1)$.
$P(X \geq 1) = 1 - P(X = 0)$.
Using the binomial probability formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have:
$P(X = 0) = {}^{10}C_{0} \left(\frac{1}{2}\right)^{0} \left(\frac{1}{2}\right)^{10} = 1 \times 1 \times \frac{1}{1024} = \frac{1}{1024}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{1024} = \frac{1023}{1024}$.

(C) Greenhouse gases are gases that form a layer in the atmosphere and prevent reflected sun rays from escaping,thereby contributing to an increase in global temperature.
$CFC$,$CO_{2}$,and $NO_{2}$ are known greenhouse gases.
$O_{2}$ (oxygen) is not a greenhouse gas as it does not absorb infrared radiation.

(D) The strength of a base is inversely proportional to the strength of its conjugate acid.
$1$. $CH_{3}COO^{-}$ has conjugate acid $CH_{3}COOH$ $(pK_{a} \approx 4.75)$.
$2$. $Cl^{-}$ has conjugate acid $HCl$ $(pK_{a} \approx -7)$.
$3$. $OH^{-}$ has conjugate acid $H_{2}O$ $(pK_{a} \approx 15.7)$.
$4$. $CH_{3}O^{-}$ has conjugate acid $CH_{3}OH$ $(pK_{a} \approx 15.5)$.
Comparing the $pK_{a}$ values,$CH_{3}OH$ is the weakest acid among the given options.
Therefore,its conjugate base,$CH_{3}O^{-}$,is the strongest base.

(A) Functional isomers are compounds that have the same molecular formula but possess different functional groups.
In option $A$,both $C_{2}H_{5}OC_{2}H_{5}$ (diethyl ether) and $C_{3}H_{7}OCH_{3}$ (methyl propyl ether) contain the same functional group,which is an ether $(-O-)$. Therefore,they are metamers,not functional isomers.
In option $B$,$CH_{3}CH_{2}OH$ (alcohol) and $CH_{3}OCH_{3}$ (ether) are functional isomers.
In option $C$,$CH_{3}CH_{2}NO_{2}$ (nitroalkane) and $H_{2}NCH_{2}COOH$ (amino acid) represent different functional groups.
In option $D$,$CH_{3}COOH$ (carboxylic acid) and $HCOOCH_{3}$ (ester) are functional isomers.

(B) The reaction of benzene with excess chlorine in the presence of anhydrous $AlCl_3$ (a Lewis acid) in the dark and at cold conditions leads to the electrophilic aromatic substitution of all six hydrogen atoms of the benzene ring by chlorine atoms. This process is known as exhaustive chlorination. The reaction is as follows:
$C_6H_6 + 6Cl_2 \xrightarrow{\text{Anhydrous } AlCl_3, \text{dark, cold}} C_6Cl_6 + 6HCl$
Here,$C_6Cl_6$ is hexachlorobenzene. Therefore,the correct option is $B$.

(D) The conjugate base of a species is formed by the removal of one proton $(H^{+})$ from it.
For $NH_{3}$,the reaction is:
$NH_{3} \longrightarrow H^{+} + NH_{2}^{-}$
Therefore,$NH_{2}^{-}$ is the conjugate base of $NH_{3}$.

(D) The magnetic field $B$ at the centre of a circular coil of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
The magnetic dipole moment $m$ is given by $m = I A = I (\pi r^2)$.
The initial ratio is $x = \frac{B}{m} = \frac{\mu_0 I / 2r}{I \pi r^2} = \frac{\mu_0}{2 \pi r^3}$.
When the current $I$ becomes $I' = 2I$ and the radius $r$ becomes $r' = 2r$,the new magnetic field $B'$ is $B' = \frac{\mu_0 (2I)}{2(2r)} = \frac{\mu_0 I}{2r} = B$.
The new magnetic moment $m'$ is $m' = (2I) \pi (2r)^2 = (2I) \pi (4r^2) = 8 I \pi r^2 = 8m$.
The new ratio $x'$ is $x' = \frac{B'}{m'} = \frac{B}{8m} = \frac{1}{8} \left( \frac{B}{m} \right) = \frac{x}{8}$.

(B) The reaction of a Lewis acid like $BCl_{3}$ with $LiAlH_{4}$ in an ether medium produces diborane $(B_{2}H_{6})$,which is a highly toxic gas.
$4BCl_{3} + 3LiAlH_{4} \rightarrow 2B_{2}H_{6} + 3AlCl_{3} + 3LiCl$
When diborane $(B_{2}H_{6})$ is heated with ammonia $(NH_{3})$,it forms borazine $(B_{3}N_{3}H_{6})$,which is commonly known as inorganic benzene.
$3B_{2}H_{6} + 6NH_{3} \rightarrow 2B_{3}N_{3}H_{6} + 12H_{2}$
Thus,the toxic gas is $B_{2}H_{6}$.

(A) Amphoteric metals react with both acids and bases to liberate $H_{2}$ gas.
$Zn + 2 HCl \text{ (dil.)} \longrightarrow ZnCl_{2} + H_{2}$
$Zn + 2 NaOH + 2 H_{2}O \longrightarrow Na_{2}[Zn(OH)_{4}] + H_{2}$
$Mg$,$Ca$,and $Fe$ are not amphoteric and do not react with aqueous $NaOH$ to produce $H_{2}$.

(B) $NH_4NO_3 \longrightarrow NH_4^+ + NO_3^-$
In $NH_4^+$,let the oxidation number of $N$ be $x$.
$x + 4(+1) = +1 \implies x = -3$.
In $NO_3^-$,let the oxidation number of $N$ be $y$.
$y + 3(-2) = -1 \implies y - 6 = -1 \implies y = +5$.
Thus,the oxidation numbers of nitrogen atoms in $NH_4NO_3$ are $-3$ and $+5$.

(D) Potassium forms various oxides such as $K_{2}O$ (oxide),$K_{2}O_{2}$ (peroxide),and $KO_{2}$ (superoxide).
$K_{2}O_{3}$ (potassium sesquioxide) does not exist because the $O_{3}^{2-}$ ion is not a stable chemical entity.

(B) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$
Moles of $H_2 = \frac{0.4 \text{ g}}{2 \text{ g/mol}} = 0.2 \text{ mol}$
Moles of $Cl_2 = \frac{7.4 \text{ g}}{71 \text{ g/mol}} \approx 0.1042 \text{ mol}$
Since $1 \text{ mol}$ of $H_2$ reacts with $1 \text{ mol}$ of $Cl_2$, $Cl_2$ is the limiting reagent because it is present in a smaller amount.
Moles of $HCl$ formed $= 2 \times \text{moles of } Cl_2 = 2 \times 0.1042 = 0.2084 \text{ mol}$
At $273 \text{ K}$ and $1 \text{ bar}$ ($STP$ conditions), the molar volume of an ideal gas is approximately $22.7 \text{ L/mol}$.
Volume of $HCl = 0.2084 \text{ mol} \times 22.7 \text{ L/mol} \approx 4.73 \text{ L}$.
Given the options, $4.67 \text{ L}$ is the closest value, which is obtained by using the older $STP$ definition $(22.4 \text{ L/mol})$.
Thus, the correct option is $(b)$.

(A) Let the total number of oxide ions be $100$. Then the number of metal ions $M$ is $96$.
Let the number of $M^{2+}$ ions be $x$ and the number of $M^{3+}$ ions be $(96 - x)$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge.
Total negative charge $= 100 \times 2 = 200$.
Total positive charge $= 2x + 3(96 - x) = 200$.
$2x + 288 - 3x = 200$.
$-x = 200 - 288 = -88$.
$x = 88$ (number of $M^{2+}$ ions).
Number of $M^{3+}$ ions $= 96 - 88 = 8$.
Percentage of $M^{3+} = \frac{8}{96} \times 100 = 8.33 \% \approx 8.3 \%$.

(C) Assume the total volume of the gas mixture is $100 \ mL$.
Since the volume percentage is equal to the mole percentage for ideal gases,we have $25 \ mol$ of He and $75 \ mol$ of $CH_4$.
Molar mass of He $= 4 \ g/mol$.
Molar mass of $CH_4 = 16 \ g/mol$.
Mass of He $= 25 \times 4 = 100 \ g$.
Mass of $CH_4 = 75 \times 16 = 1200 \ g$.
Total mass of the mixture $= 100 + 1200 = 1300 \ g$.
Mass percentage of $CH_4 = \frac{\text{Mass of } CH_4}{\text{Total mass}} \times 100 = \frac{1200}{1300} \times 100 \approx 92.3 \%$.
Thus,the percentage by mass of methane is approximately $92 \%$.

(D) $In-situ$ conservation refers to protecting an endangered species in its natural habitat.
$In-situ$ approach includes the protection of a group of ecosystems through a network of protected areas,such as biodiversity hotspots,national parks,sanctuaries,biosphere reserves,and sacred groves and lakes.
Whereas $Ex-situ$ conservation involves the protection of genetic resources of species away from their area of origin or development.
This includes off-site collections such as zoos,botanical gardens,aquariums,gene banks,and $In-vitro$ preservation methods like tissue culture and cryopreservation.
Therefore,a botanical garden is an example of $Ex-situ$ conservation,not $In-situ$ conservation.

(D) The photoelectric effect occurs when light of a suitable frequency strikes the surface of a metal,causing the ejection of electrons.
The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light,provided the frequency is above the threshold frequency.
The kinetic energy of the emitted electrons depends on the frequency of the incident light,not its intensity.
Therefore,the correct statement is that the number of electrons ejected increases with the increase in the intensity of incident light.

(A) We know that the change in entropy is given by $\Delta S = \frac{q_{rev}}{T}$.
Given that the quantity of heat $q_{rev}$ is the same in both cases,we have $\Delta S_{1} = \frac{q_{rev}}{T_{1}}$ and $\Delta S_{2} = \frac{q_{rev}}{T_{2}}$.
Since $T_{1} > T_{2}$,the denominator in the expression for $\Delta S_{1}$ is larger than the denominator in the expression for $\Delta S_{2}$.
Therefore,$\Delta S_{1} < \Delta S_{2}$.

(C) The conversion of propan$-2-$ol to propan$-1-$ol proceeds as follows:
$1$. Propan$-2-$ol reacts with $PCl_{5}$ to form $2-$chloropropane.
$2$. $2-$chloropropane undergoes dehydrohalogenation with alcoholic $KOH$ to form propene.
$3$. Propene undergoes hydroboration-oxidation (reaction with $B_{2}H_{6}$ followed by $H_{2}O_{2}/NaOH$) to form propan$-1-$ol via anti-Markovnikov addition of water.

(D) The Williamson ether synthesis involves the reaction of an alkoxide ion with a primary alkyl halide to form an ether.
$(P)$ $C_6H_5CH_2Br + CH_3ONa \rightarrow C_6H_5CH_2OCH_3 + NaBr$. This is a $S_N2$ reaction between a primary benzylic halide and a strong nucleophile,yielding an ether as the major product.
$(Q)$ $C_6H_5ONa + CH_3Br \rightarrow C_6H_5OCH_3 + NaBr$. This is a $S_N2$ reaction between a phenoxide ion and a primary alkyl halide,yielding an ether as the major product.
$(R)$ $(CH_3)_3CCl + CH_3ONa \rightarrow (CH_3)_2C=CH_2 + CH_3OH + NaCl$. Since $(CH_3)_3CCl$ is a tertiary alkyl halide,the strong base $CH_3ONa$ promotes an $E2$ elimination reaction,giving an alkene as the major product.
$(S)$ $C_6H_5CH=CHCl + CH_3ONa$. Vinyl halides are unreactive towards $S_N2$ reactions due to the partial double bond character of the $C-X$ bond and the instability of the vinylic carbocation,so no ether is formed.
Therefore,both $(P)$ and $(Q)$ give an ether as the major product.

(C) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetaldehyde $(CH_3CHO)$ in the presence of dilute $NaOH$ followed by heating $(\Delta)$ is a cross-aldol condensation reaction.
$1$. The enolate ion formed from acetaldehyde attacks the carbonyl carbon of benzaldehyde to form a $\beta$-hydroxy aldehyde.
$2$. Upon heating,this $\beta$-hydroxy aldehyde undergoes dehydration to form an $\alpha,\beta$-unsaturated aldehyde,which is cinnamaldehyde $(C_6H_5CH=CHCHO)$.

(C) Any carbonyl compound having $\alpha$-hydrogen will undergo aldol condensation.
In $CCl_3CHO$ (trichloro acetaldehyde),the $\alpha$-carbon is bonded to three chlorine atoms and has no $\alpha$-hydrogen atom attached to it.
Therefore,it does not show the aldol condensation reaction.

(B) Hinsberg reagent is $C_{6}H_{5}SO_{2}Cl$ (benzene sulphonyl chloride).
It is used for the detection and differentiation of primary,secondary,and tertiary amines.

(B) Sucrose does not contain a $C_{1}-C_{4}$ glycosidic linkage.
In sucrose,the glycosidic linkage is formed between $C_{1}$ of $\alpha$-glucose and $C_{2}$ of $\beta$-fructose.
Therefore,it is a $C_{1}-C_{2}$ linkage.

(B) Vitamins are classified as fat-soluble or water-soluble.
Fat-soluble vitamins $(A, D, E, K)$ are stored in the body,primarily in the liver and adipose tissue.
Water-soluble vitamins (such as $B$-complex and $C$) are not stored in the body and are excreted through urine.
Therefore,vitamin $B_{6}$ is not stored in adipose tissue.

(C) Hypothyroidism is caused due to the deficiency of thyroxine.
Thyroxine is a hormone secreted by the thyroid gland,which regulates the metabolism of carbohydrates,lipids,and proteins in the body.

(B) The incorrectly matched pair is $XeF_4$: Tetrahedral.
$XeF_4$ has a square planar structure due to $sp^3d^2$ hybridization with two lone pairs of electrons occupying the axial positions.

(C) The boiling point of a compound depends on the strength of intermolecular forces.
$CH_3-CH_2-OH$ (alcohol),$CH_3-CH_2-NH_2$ (amine),and $HCOOH$ (carboxylic acid) all exhibit intermolecular hydrogen bonding,which significantly increases their boiling points.
$CH_3-O-CH_3$ (dimethyl ether) is a polar molecule but lacks hydrogen bonding,relying only on dipole-dipole interactions.
Therefore,$CH_3-O-CH_3$ has the lowest boiling point among the given options.

(B) For the reaction $2 A + 3 B \rightarrow 4 C + D$,the rate expression is given by:
$-\frac{1}{3} \frac{d[B]}{dt} = +\frac{1}{4} \frac{d[C]}{dt}$
Given that the rate of appearance of $C$ is $\frac{d[C]}{dt} = 2.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
The rate of disappearance of $B$ is $-\frac{d[B]}{dt}$.
From the rate expression: $-\frac{d[B]}{dt} = \frac{3}{4} \frac{d[C]}{dt}$
Substituting the given value: $-\frac{d[B]}{dt} = \frac{3}{4} \times (2.8 \times 10^{-3}) \ mol \ L^{-1} \ s^{-1}$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For $60 \%$ completion,$[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$ and $t = 50 \ min$.
$k = \frac{2.303}{50} \log \frac{[A]_0}{0.40[A]_0} = \frac{2.303}{50} \log 2.5$.
For $93.6 \%$ completion,$[A]_t = [A]_0 - 0.936[A]_0 = 0.064[A]_0$.
$k = \frac{2.303}{t'} \log \frac{[A]_0}{0.064[A]_0} = \frac{2.303}{t'} \log \frac{1}{0.064} = \frac{2.303}{t'} \log 15.625$.
Equating the two expressions for $k$:
$\frac{2.303}{50} \log 2.5 = \frac{2.303}{t'} \log 15.625$.
$t' = 50 \times \frac{\log 15.625}{\log 2.5} = 50 \times \frac{\log (2.5)^3}{\log 2.5} = 50 \times 3 = 150 \ min$.

(C) The rate constant $k$ is given by the Arrhenius equation: $k = P Z e^{-E_{a} / R T}$.
To speed up the reaction,the rate constant $k$ must be increased.
From the expression,$k$ is inversely proportional to the exponential term $e^{E_{a} / R T}$.
If the activation energy $E_{a}$ is decreased,the value of the exponent $E_{a} / R T$ decreases.
Consequently,the term $e^{E_{a} / R T}$ decreases,which leads to an increase in the value of $k$.
Therefore,decreasing $E_{a}$ increases the reaction rate.

(B) An $Antagonist$ is a substance that binds to a receptor to block or dampen a biological response,rather than interacting directly with an enzyme's catalytic site in the context of enzyme inhibition mechanisms. $LSD$ is a well-known serotonin antagonist.
Conversely,$Allosteric$ $site$,$Co-enzymes$,and $Enzyme$ $inhibitors$ are all fundamental components or mechanisms involved in drug-enzyme interactions.

(A) $BHA$ (Butylated hydroxyanisole) acts as a food additive which functions as an antioxidant.
It helps in preserving food by preventing the oxidation of fats and oils.
Saccharin is an artificial sweetener,while sugar syrup and salt are commonly used as food preservatives.

(D) Silicon belongs to group $14$ and gallium belongs to group $13$.
When $Si$ is doped with $Ga$,a $p$-type semiconductor is produced.
This is because group $13$ elements have only $3$ valence electrons,while group $14$ elements have $4$.
This creates an electron-deficient bond or an electron vacancy (hole) in the crystal lattice.
These holes can move through the crystal like a positive charge,giving rise to electrical conductivity.

(D) The coordination number of a central metal ion is the total number of sigma bonds formed by ligands with the metal atom.
In $[Fe(C_{2}O_{4})_{3}]^{3-}$,the ligand is oxalate $(C_{2}O_{4}^{2-})$,which is a bidentate ligand. Therefore,the coordination number is $3 \times 2 = 6$.
In $[Co(SCN)_{4}]^{2-}$,the ligand is thiocyanate $(SCN^-)$,which is a monodentate ligand. Therefore,the coordination number is $4 \times 1 = 4$.
Thus,the coordination numbers are $6$ and $4$ respectively.

(D) The complex $[Pt(NH_3)_4][PtCl_4]$ consists of a cationic part $[Pt(NH_3)_4]^{2+}$ and an anionic part $[PtCl_4]^{2-}$.
In the cation $[Pt(NH_3)_4]^{2+}$,the oxidation state of $Pt$ is $x + 4(0) = +2$,so $x = +2$.
In the anion $[PtCl_4]^{2-}$,the oxidation state of $Pt$ is $y + 4(-1) = -2$,so $y = +2$.
The $IUPAC$ name is tetraammineplatinum $(II)$ tetrachloridoplatinate $(II)$.

(D) The complex $[Co(en)_{2} Cl_{2}]^{+}$ exhibits geometrical isomerism and optical isomerism.
$1$. Geometrical isomerism: It exists in two forms,the $cis$-isomer and the $trans$-isomer.
$2$. Optical isomerism: The $cis$-isomer is optically active and exists as a pair of enantiomers ($d$ and $l$ forms),while the $trans$-isomer is optically inactive (achiral) due to the presence of a plane of symmetry.
Therefore,the total number of stereoisomers is $2$ (from $cis$-enantiomers) $+ 1$ (from $trans$-isomer) $= 3$.

(C) substance is paramagnetic if it contains at least one unpaired electron. Let us analyze the electronic configurations of the ions in option $C$:
$Ti$ $(Z=22)$: $[Ar] 4s^{2} 3d^{2}$. $Ti^{2+}$: $[Ar] 3d^{2}$ (contains $2$ unpaired electrons).
$Cu$ $(Z=29)$: $[Ar] 4s^{1} 3d^{10}$. $Cu^{2+}$: $[Ar] 3d^{9}$ (contains $1$ unpaired electron).
$Mn$ $(Z=25)$: $[Ar] 4s^{2} 3d^{5}$. $Mn^{3+}$: $[Ar] 3d^{4}$ (contains $4$ unpaired electrons).
Since all these ions contain unpaired electrons,they are paramagnetic.

(A) In an aqueous medium,$Cu^{2+}$ is more stable than $Cu^{+}$ because the higher negative enthalpy of hydration of $Cu^{2+}_{(aq)}$ more than compensates for the energy required to remove the second electron from $Cu^{+}$.
Therefore,$CuCl_{2}$ (containing $Cu^{2+}$) is more stable than $Cu_{2}Cl_{2}$ (containing $Cu^{+}$) in aqueous solution.

(A) The cell reaction is $2Fe^{3+} + 2I^{-} \longrightarrow 2Fe^{2+} + I_{2}$.
Here,the number of electrons transferred $n = 2$.
The standard cell potential is $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} = 0.76 \ V - 0.55 \ V = 0.21 \ V$.
The relationship between equilibrium constant $K_{c}$ and $E_{\text{cell}}^{\circ}$ is given by $\log K_{c} = \frac{n E_{\text{cell}}^{\circ}}{0.0591} \approx \frac{n E_{\text{cell}}^{\circ}}{0.06}$.
Substituting the values: $\log K_{c} = \frac{2 \times 0.21}{0.06} = \frac{0.42}{0.06} = 7$.
Therefore,$K_{c} = 10^{7} = 1 \times 10^{7}$.

A chemical reaction is possible when $AgNO_{3}$ solution is stirred with copper spoon as copper has lesser positive value of standard electrode reduction potential as compared to $Ag$.
The reaction will be,
$2 AgNO_{3} + Cu \longrightarrow Cu(NO_{3})_{2} + 2 Ag$

(B) The Debye-Huckel-Onsager equation is given by $\Lambda_{m} = \Lambda_{m}^{\circ} - A \sqrt{C}$.
In this equation,the constant $A$ depends on the stoichiometry of the electrolyte,which determines the charge type (e.g.,$1:1, 1:2, 2:2$ electrolytes).
Electrolytes with the same charge type (e.g.,both are $1:1$ electrolytes) will have the same value for the constant $A$.
$NH_{4}Cl$ dissociates into $NH_{4}^{+}$ and $Cl^{-}$ ($1:1$ electrolyte).
$NaBr$ dissociates into $Na^{+}$ and $Br^{-}$ ($1:1$ electrolyte).
Since both are $1:1$ electrolytes,they possess the same value for the constant $A$.

(B) The prolonged exposure of chloroform $(CHCl_{3})$ to air and sunlight leads to its oxidation to form phosgene $(COCl_{2})$,which is a highly poisonous gas.
Phosgene causes damage to the liver and other organs upon inhalation or exposure.
The chemical reaction is:
$2CHCl_{3} + O_{2} \xrightarrow{\text{Sunlight}} 2COCl_{2} + 2HCl$

(A) When sulphide ore is roasted,it is heated in the presence of excess air to form the gas $SO_{2}$.
$S + O_{2} \longrightarrow SO_{2} (X)$
For example: $2ZnS + 3O_{2} \longrightarrow 2ZnO + 2SO_{2}$
Thereafter,$X$ $(SO_{2})$ reacts with $Cl_{2}$ in the presence of activated charcoal to give $Y$ $(SO_{2}Cl_{2})$:
$SO_{2} + Cl_{2} \xrightarrow{\text{activated charcoal}} SO_{2}Cl_{2} (Y)$
Thus,$Y$ is $SO_{2}Cl_{2}$.

(B) In the froth floatation process,collectors like potassium ethyl xanthate are added to the suspension.
These molecules adsorb onto the surface of the mineral particles,making them water-repellent or $hydrophobic$.
This allows the mineral particles to attach to air bubbles and rise to the surface with the froth,effectively separating them from the gangue.

(D) Copper is extracted from copper pyrites $(CuFeS_{2})$ by the auto-reduction process. The steps are as follows:
$2 CuFeS_{2} + O_{2} \longrightarrow Cu_{2}S + 2 FeS + SO_{2}$
$2 FeS + 3 O_{2} \longrightarrow 2 FeO + 2 SO_{2}$
$FeO + SiO_{2} \longrightarrow FeSiO_{3}$
$2 Cu_{2}S + 3 O_{2} \longrightarrow 2 Cu_{2}O + 2 SO_{2}$
$2 Cu_{2}O + Cu_{2}S \longrightarrow 6 Cu + SO_{2}$
In the final step,$Cu_{2}O$ reacts with $Cu_{2}S$ to produce metallic copper,which is known as auto-reduction.

(A) The reactivity of halides towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed in the rate-determining step.
$S_{N}1$ reactivity order is generally $3^{\circ} > 2^{\circ} > 1^{\circ}$.
In $C_{6}H_{5}CH_{2}Cl$,the loss of $Cl^{-}$ generates a benzyl carbocation $(C_{6}H_{5}CH_{2}^{+})$,which is highly stabilized by resonance with the benzene ring.
$CH_{3}CH_{2}Cl$ and $CH_{3}CH_{2}CH_{2}CH_{2}I$ form primary carbocations,which are less stable.
$C_{6}H_{5}Cl$ does not undergo $S_{N}1$ reaction easily due to the partial double bond character of the $C-Cl$ bond and the instability of the phenyl cation.
Therefore,$C_{6}H_{5}CH_{2}Cl$ shows the highest reactivity towards $S_{N}1$ reaction.

(B) $1$. Nitrobenzene reacts with $Br_2/FeBr_3$ to form $m$-bromonitrobenzene $(P)$ because the $-NO_2$ group is meta-directing.
$2$. $m$-Bromonitrobenzene $(P)$ is reduced by $Sn/conc. HCl$ to form $m$-bromoaniline $(Q)$.
$3$. $m$-Bromoaniline $(Q)$ reacts with $NaNO_2/dil. HCl$ at $273 \ K$ to form $m$-bromobenzenediazonium chloride.
$4$. Finally,warming with water replaces the diazonium group with an $-OH$ group to form $m$-bromophenol $(R)$.

(A) Phosphorus pentachloride $(PCl_{5})$ on hydrolysis gives phosphoric acid $(H_{3}PO_{4})$,which is a tribasic acid.
$PCl_{5} + 4H_{2}O \longrightarrow H_{3}PO_{4} + 5HCl$
In $H_{3}PO_{4}$,there are three $P-OH$ groups,making it tribasic.

(B) Polymers are classified based on intermolecular forces into elastomers,fibres,thermoplastics,and thermosetting polymers.
$1$. Elastomers (e.g.,Neoprene) have the weakest intermolecular forces,allowing the polymer chains to be stretched.
$2$. Thermoplastics (e.g.,Polythene,Polystyrene) have intermediate intermolecular forces.
$3$. Fibres (e.g.,Terylene) have the strongest intermolecular forces,such as hydrogen bonding or dipole-dipole interactions,which result in high tensile strength and crystalline nature.
Therefore,Terylene,being a fibre,has the strongest intermolecular forces among the given options.

(B) Condensation polymerisation occurs when monomers containing two or more functional groups (such as $-NH_2$ and $-COOH$) react,resulting in the loss of small molecules like $H_2O$,$HCl$,etc.
Glycine $(NH_2CH_2COOH)$ contains both an amino group $(-NH_2)$ and a carboxylic acid group $(-COOH)$. Therefore,it can undergo condensation polymerisation to form polyglycine.
Styrene,isoprene,and propene contain carbon-carbon double bonds and undergo addition polymerisation.

(A) The salt $(A)$ reacts with $BaCl_{2}$ to form a white precipitate. Sulfite salts $(SO_{3}^{2-})$ form $BaSO_{3}$ (white precipitate) with $BaCl_{2}$.
$BaSO_{3(s)} + 2HCl_{(aq)} \longrightarrow BaCl_{2(aq)} + H_{2}O_{(l)} + SO_{2(g)}$.
The gas $(B)$ is $SO_{2}$,which is a reducing agent and decolourises acidified $KMnO_{4}$ solution by reducing $Mn^{7+}$ to $Mn^{2+}$.
Therefore,$(A)$ is $Na_{2}SO_{3}$ and $(B)$ is $SO_{2}$.

(D) During the electrolysis of an aqueous solution of $NaF$ using inert electrodes,water molecules are preferentially oxidized at the anode rather than fluoride ions because the oxidation potential of water is lower than that of fluoride ions.
At the cathode: $2H_{2}O + 2e^{-} \longrightarrow H_{2} + 2OH^{-}$.
At the anode: $2H_{2}O \longrightarrow O_{2} + 4H^{+} + 4e^{-}$.
Even if $F_{2}$ were produced,it would react with water to form $O_{2}$ as follows: $2F_{2} + 2H_{2}O \longrightarrow 4HF + O_{2}$.
Therefore,the final product obtained at the anode is $O_{2}$.

(B) For an $FCC$ unit cell,the relationship between the edge length $a$ and the metallic radius $r$ is given by $4r = \sqrt{2}a$.
Given $r = \sqrt{2} \ \mathring{A}$.
Substituting the value of $r$: $a = \frac{4r}{\sqrt{2}} = \frac{4 \times \sqrt{2}}{\sqrt{2}} = 4 \ \mathring{A}$.
The volume of the unit cell is $V = a^{3} = (4 \ \mathring{A})^{3} = 64 \ \mathring{A}^{3}$.
Since $1 \ \mathring{A} = 10^{-10} \ m$,then $1 \ \mathring{A}^{3} = 10^{-30} \ m^{3}$.
Therefore,$V = 64 \times 10^{-30} \ m^{3} = 6.4 \times 10^{-29} \ m^{3}$.

(A) Two solutions are isotonic if their osmotic pressures are equal.
Osmotic pressure $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is molarity,$R$ is the gas constant,and $T$ is temperature.
For isotonic solutions at the same temperature,the effective concentration $(i \times C)$ must be equal.
$(A)$ For $0.01 \ M \ BaCl_2$: $i = 3$,$C = 0.01$,$i \times C = 0.03$. For $0.015 \ M \ NaCl$: $i = 2$,$C = 0.015$,$i \times C = 0.03$. Since $0.03 = 0.03$,this pair is isotonic.
$(B)$ For $0.001 \ M \ Al_2(SO_4)_3$: $i = 5$,$C = 0.001$,$i \times C = 0.005$. For $0.001 \ M \ BaCl_2$: $i = 3$,$C = 0.001$,$i \times C = 0.003$. Not isotonic.
$(C)$ For $0.001 \ M \ CaCl_2$: $i = 3$,$C = 0.001$,$i \times C = 0.003$. For $0.001 \ M \ Al_2(SO_4)_3$: $i = 5$,$C = 0.001$,$i \times C = 0.005$. Not isotonic.
$(D)$ For $0.01 \ M \ BaCl_2$: $i \times C = 0.03$. For $0.001 \ M \ CaCl_2$: $i \times C = 0.003$. Not isotonic.
Thus,the correct option is $(A)$.

(D) Given:
Mass of solute $(w_B)$ = $2.5 \ g$
Mass of solvent $(w_A)$ = $100 \ g$
Elevation in boiling point $(\Delta T_b)$ = $0.3 \ K$
$K_b = 0.52 \ K \ kg \ mol^{-1}$
Degree of association $(\alpha)$ = $0.8$
For dimerization: $2X \rightleftharpoons X_2$
Van't Hoff factor $(i)$ = $1 - \alpha + \frac{\alpha}{2} = 1 - 0.8 + \frac{0.8}{2} = 0.2 + 0.4 = 0.6$
Formula for elevation in boiling point:
$\Delta T_b = i \times K_b \times m$
$\Delta T_b = i \times K_b \times \left( \frac{w_B \times 1000}{M_B \times w_A} \right)$
$0.3 = 0.6 \times 0.52 \times \left( \frac{2.5 \times 1000}{M_B \times 100} \right)$
$0.3 = 0.6 \times 0.52 \times \left( \frac{25}{M_B} \right)$
$M_B = \frac{0.6 \times 0.52 \times 25}{0.3}$
$M_B = 2 \times 0.52 \times 25 = 26 \ g \ mol^{-1}$

(D) The $6$th period corresponds to the filling of the $6s, 4f, 5d,$ and $6p$ orbitals.
The last element of the $6$th period is the noble gas Radon $(Rn)$,which has an atomic number of $86$.
The electronic configuration of Radon is $[Xe] 4f^{14} 5d^{10} 6s^2 6p^6$.
Therefore,the outermost electronic configuration is $4f^{14}, 5d^{10}, 6s^2, 6p^6$.

(D) The reaction between $AgNO_{3}$ and $KI$ is: $AgNO_{3} + KI \longrightarrow AgI(s) + KNO_{3}$.
Since equal volumes are mixed,the number of moles of $KI$ $(0.2 \ M)$ is greater than the number of moles of $AgNO_{3}$ $(0.1 \ M)$.
Thus,$KI$ is in excess.
The $AgI$ particles preferentially adsorb the common ion present in excess from the dispersion medium.
In this case,$I^{-}$ ions are in excess and are adsorbed on the surface of $AgI$ particles,resulting in a negatively charged sol.

(A) Adsorption is a spontaneous process,so $\Delta G < 0$.
It is an exothermic process,which means heat is released,so $\Delta H < 0$.
As gas molecules get adsorbed on the solid surface,their freedom of movement decreases,leading to a decrease in entropy,so $\Delta S < 0$.
According to the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T \Delta S$. At low temperatures,the negative $\Delta H$ term dominates,ensuring $\Delta G < 0$.