KCET 2022 Chemistry Question Paper with Answer and Solution

(C) When a charge $q$ is released in a uniform electric field $E$,the force acting on it is $F = qE.$
Since $F = ma$,the acceleration of the particle is $a = \frac{qE}{m}.$
Since the particle starts from rest $(u = 0)$,its velocity $v$ after time $t$ is given by $v = u + at = 0 + \left(\frac{qE}{m}\right)t = \frac{qEt}{m}.$
The kinetic energy $(KE)$ of the particle is given by $KE = \frac{1}{2}mv^2.$
Substituting the value of $v$,we get $KE = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{1}{2}m\left(\frac{q^2E^2t^2}{m^2}\right) = \frac{q^2E^2t^2}{2m}.$

(D) According to the lens maker's formula,the focal length $f$ of a lens is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
This implies $f \propto \frac{1}{\mu - 1}$.
According to Cauchy's dispersion formula,the refractive index $\mu$ is related to the wavelength $\lambda$ as $\mu = A + \frac{B}{\lambda^2}$. Thus,$\mu$ is inversely proportional to $\lambda$ $(\mu \propto \frac{1}{\lambda})$.
Combining these,we get $f \propto \lambda$.
Since the wavelength of red light $({\lambda_r})$ is greater than the wavelength of blue,green,or yellow light,the focal length $f$ will be maximum for red light.

(D) $ (d) $ Osmotic pressure method is especially suitable for the determination of molecular masses of macromolecules such as proteins and polymers.
For these substances,the values of other colligative properties,such as elevation in boiling point or depression in freezing point,are too small to be measured accurately.
On the other hand,the osmotic pressure of such substances is large enough to be measured precisely even at low concentrations.

(D) An electron-deficient hydride is one that does not have sufficient electrons to form normal covalent bonds between all the bonded atoms.
In $B_2H_6$ (diborane),there are only $12$ valence electrons available for bonding,which is insufficient to form the expected number of two-center two-electron bonds.
Therefore,$B_2H_6$ is an electron-deficient hydride.

(B) In an $LC$ circuit,the charge on the capacitor at any time $t$ is given by $q(t) = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.
The energy stored in the electric field is $U_E = \frac{q^2}{2C}$ and the total energy is $U_{total} = \frac{q_0^2}{2C}$.
We are given that the energy is shared equally between the electric and magnetic fields,so $U_E = \frac{1}{2} U_{total}$.
$\frac{q^2}{2C} = \frac{1}{2} \left( \frac{q_0^2}{2C} \right) \implies q^2 = \frac{q_0^2}{2} \implies q = \frac{q_0}{\sqrt{2}}$.
Substituting $q(t) = q_0 \cos(\omega t)$,we get $q_0 \cos(\omega t) = \frac{q_0}{\sqrt{2}}$,which means $\cos(\omega t) = \frac{1}{\sqrt{2}}$.
Thus,$\omega t = \frac{\pi}{4}$.
Substituting $\omega = \frac{1}{\sqrt{LC}}$,we get $t = \frac{\pi}{4\omega} = \frac{\pi}{4} \sqrt{LC}$.

(C) When a charge $q$ is released in a uniform electric field $E$,the force acting on it is $F = qE$.
According to Newton's second law,the acceleration $a$ is given by $a = \frac{F}{m} = \frac{qE}{m}$.
Since the electric field is uniform,the acceleration is constant.
Using the equation of motion $v = u + at$,where initial velocity $u = 0$,the velocity after time $t$ is $v = \left(\frac{qE}{m}\right)t$.
The kinetic energy $KE$ is given by $KE = \frac{1}{2}mv^2$.
Substituting the value of $v$,we get $KE = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{1}{2}m \cdot \frac{q^2E^2t^2}{m^2} = \frac{q^2E^2t^2}{2m}$.

(D) The total energy in an $LC$ circuit is constant and is given by $U = U_E + U_B$,where $U_E$ is the electric energy and $U_B$ is the magnetic energy.
At $t = 0$,the charge on the capacitor is $q_0$,so the total energy is $U = \frac{q_0^2}{2C}$.
We want to find the time $t$ when $U_E = U_B = \frac{U}{2}$.
Since $U_E = \frac{q^2}{2C}$,we have $\frac{q^2}{2C} = \frac{1}{2} \frac{q_0^2}{2C}$,which implies $q^2 = \frac{q_0^2}{2}$,or $q = \frac{q_0}{\sqrt{2}}$.
The charge on the capacitor at time $t$ is given by $q(t) = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.
Setting $q(t) = \frac{q_0}{\sqrt{2}}$,we get $\cos(\omega t) = \frac{1}{\sqrt{2}}$.
This implies $\omega t = \frac{\pi}{4}$.
Substituting $\omega = \frac{1}{\sqrt{LC}}$,we get $t = \frac{\pi}{4} \sqrt{LC}$.

(D) An electron-deficient hydride is one that does not have sufficient electrons to form the conventional electron-pair bonds between all its bonded atoms.
$B_2H_6$ (diborane) is a classic example of an electron-deficient hydride because it has fewer electrons than required to form standard two-center two-electron bonds for all its atoms.
$NaH$ and $CaH_2$ are saline (ionic) hydrides,and $CH_4$ is an electron-precise hydride (octet is complete).

(B) Given: $R = 300 \Omega, L = 0.9 \text{ H}, C = 2.0 \mu\text{F} = 2.0 \times 10^{-6} \text{ F}, \omega = 1000 \text{ rad/s}$.
The impedance $Z$ of a series $LCR$ circuit is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
First,calculate the inductive reactance $X_L$:
$X_L = \omega L = 1000 \times 0.9 = 900 \Omega$
Next,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 2.0 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega$
Now,substitute these values into the impedance formula:
$Z = \sqrt{300^2 + (900 - 500)^2}$
$Z = \sqrt{90000 + (400)^2}$
$Z = \sqrt{90000 + 160000}$
$Z = \sqrt{250000} = 500 \Omega$
Therefore,the impedance of the circuit is $500 \Omega$.

(B) According to Molecular Orbital Theory $(MOT)$:
For $O_2$ ($16$ electrons),the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. The bond order is $2$,consisting of $1 \sigma$ and $1 \pi$ bond.
For $C_2$ ($12$ electrons),the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. The bond order is $2$,consisting of $2 \pi$ bonds (no $\sigma$ bond).
Thus,option $(B)$ is correct.

(B) The reaction is: $2HI \rightleftharpoons H_2 + I_2$
Initial moles: $1 \ mol$ of $HI$,$0 \ mol$ of $H_2$,$0 \ mol$ of $I_2$.
At equilibrium,$0.5 \ mol$ of $HI$ is dissociated. Therefore,remaining $HI = 1 - 0.5 = 0.5 \ mol$.
According to stoichiometry,$2 \ mol$ of $HI$ produces $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.
So,$0.5 \ mol$ of $HI$ produces $0.25 \ mol$ of $H_2$ and $0.25 \ mol$ of $I_2$.
Equilibrium concentrations in $2 \ L$ container:
$[HI] = 0.5 / 2 = 0.25 \ M$
$[H_2] = 0.25 / 2 = 0.125 \ M$
$[I_2] = 0.25 / 2 = 0.125 \ M$
$K_C = \frac{[H_2][I_2]}{[HI]^2} = \frac{0.125 \times 0.125}{(0.25)^2} = \frac{0.015625}{0.0625} = 0.25$

(A) The relation is defined as $3 a + 2 b = 27$,where $a, b \in N$ (natural numbers).
Since $b = \frac{27 - 3 a}{2}$,for $b$ to be a natural number,$(27 - 3 a)$ must be even and positive.
If $a = 1$,$b = \frac{27 - 3}{2} = 12$.
If $a = 3$,$b = \frac{27 - 9}{2} = 9$.
If $a = 5$,$b = \frac{27 - 15}{2} = 6$.
If $a = 7$,$b = \frac{27 - 21}{2} = 3$.
If $a = 9$,$b = \frac{27 - 27}{2} = 0$,which is not in $N$.
Thus,the relation $R = \{(1, 12), (3, 9), (5, 6), (7, 3)\}$.
Therefore,option $A$ is correct.

(C) In an $LC$ circuit,the total energy $E$ is constant and oscillates between the electric field of the capacitor and the magnetic field of the inductor.
The energy stored in the capacitor is $U_E = \frac{q^2}{2C}$ and the energy stored in the inductor is $U_B = \frac{1}{2}LI^2$.
At $t=0$,the charge is $q_0$,so the total energy is $E = \frac{q_0^2}{2C}$.
We want the time $t$ when $U_E = U_B = \frac{E}{2}$.
Since $U_E = \frac{q^2}{2C} = \frac{1}{2} \left( \frac{q_0^2}{2C} \right)$,we have $q^2 = \frac{q_0^2}{2}$,which implies $q = \frac{q_0}{\sqrt{2}}$.
Given $q = q_0 \cos(\omega t)$,we set $q_0 \cos(\omega t) = \frac{q_0}{\sqrt{2}}$,so $\cos(\omega t) = \frac{1}{\sqrt{2}}$.
This gives $\omega t = \frac{\pi}{4}$.
Since $\omega = \frac{1}{\sqrt{LC}}$,we have $t = \frac{\pi}{4\omega} = \frac{\pi}{4} \sqrt{LC}$.

(B) Given,reading of $1$ main scale division $(MSD)$ = $0.5 \text{ mm}$.
According to the given situation,$50$ vernier scale divisions $(VSD)$ = $49$ main scale divisions $(MSD)$.
Therefore,$1$ vernier scale division $(VSD)$ = $\frac{49}{50} \text{ MSD}$.
The least count of the microscope is defined as the difference between $1 \text{ MSD}$ and $1 \text{ VSD}$.
$\text{Least count} = 1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ MSD} - \frac{49}{50} \text{ MSD} = \frac{1}{50} \text{ MSD}$.
Substituting the value of $1 \text{ MSD} = 0.5 \text{ mm}$:
$\text{Least count} = \frac{1}{50} \times 0.5 \text{ mm} = 0.01 \text{ mm}$.

(C) $1$. Identify the principal functional group. The carboxylic acid group $(-COOH)$ has higher priority than the ketone group $(>C=O)$.
$2$. Number the carbon chain starting from the carbon of the carboxylic acid group as $C-1$.
$3$. The chain is $CH_3(5)-CO(4)-CH_2(3)-CH_2(2)-COOH(1)$.
$4$. The ketone group is at position $4$,so it is named as an oxo substituent.
$5$. The parent chain has $5$ carbons,so the root is pentane. Adding the suffix for the acid gives pentanoic acid.
$6$. Combining these,the $IUPAC$ name is $4-$oxopentanoic acid.

(A) $1$. The reaction of propene with $HBr$ in the presence of benzoyl peroxide follows the anti-Markovnikov addition (peroxide effect or Kharasch effect). In this reaction,the $Br^-$ adds to the less substituted carbon atom of the double bond,resulting in $1$-bromopropane $(CH_3-CH_2-CH_2-Br)$ as the major product $(A)$.
$2$. The reaction of propene with $HI$ follows the Markovnikov addition rule. In this reaction,the $H^+$ adds to the carbon with more hydrogen atoms,and the $I^-$ adds to the more substituted carbon atom,resulting in $2$-iodopropane $(CH_3-CH(I)-CH_3)$ as the major product $(B)$.

(C) An electron-deficient hydride is one that does not have sufficient electrons to form the required number of covalent bonds for its structure.
In $B_2H_6$ (diborane),there are $8$ covalent bonds,which would normally require $16$ electrons.
However,$B_2H_6$ has only $12$ valence electrons available for bonding.
Therefore,it is an electron-deficient hydride.

(A) For an aqueous solution,the $pH$ scale ranges from $0$ to $14$. Acidic solutions have $pH < 7$,neutral solutions have $pH = 7$,and basic solutions have $pH > 7$.
Among the given options,$H_2SO_4$ and $HCl$ are strong acids,while $NaOH$ is a strong base. Therefore,the basic solutions will have a higher $pH$ than the acidic solutions.
Comparing the two basic solutions ($1 \ M \ NaOH$ and $0.1 \ M \ NaOH$):
For $1 \ M \ NaOH$: $[OH^-] = 1 \ M$. Thus,$pOH = -\log(1) = 0$. Since $pH + pOH = 14$,$pH = 14 - 0 = 14$.
For $0.1 \ M \ NaOH$: $[OH^-] = 0.1 \ M$. Thus,$pOH = -\log(0.1) = 1$. Since $pH + pOH = 14$,$pH = 14 - 1 = 13$.
Comparing the values,$1 \ M \ NaOH$ has the highest $pH$.

(B) The property of low solubility of carbon dioxide in water makes it biologically and geo-chemically important.
Carbon dioxide on reaction with water forms carbonic acid,which dissociates to give $HCO_3^{-}$ ions.
The $H_2CO_3 / HCO_3^{-}$ buffer system helps to maintain the $pH$ of blood between $7.26$ and $7.42$.

(A) In $NH_4NO_3$,the nitrogen element exhibits two different oxidation states.
$NH_4NO_3$ is an ionic compound consisting of $NH_4^+$ and $NO_3^-$ ions.
For $NH_4^+$: Let the oxidation state of $N$ be $x$. Then $x + 4(+1) = +1$,which gives $x = -3$.
For $NO_3^-$: Let the oxidation state of $N$ be $y$. Then $y + 3(-2) = -1$,which gives $y = +5$.
Thus,nitrogen exists in $-3$ and $+5$ oxidation states in $NH_4NO_3$.

(C) $SnO_2$ (Tin$(IV)$ oxide) is amphoteric in nature,meaning it can react with both acids and bases.
$CO_2$ is an acidic oxide,and $Ag_2O$ is a basic oxide.

(B) The given circuit consists of two $NOT$ gates connected to the inputs $A$ and $B$,followed by a $NOR$ gate.
Let the inputs to the $NOR$ gate be $\bar{A}$ and $\bar{B}$.
The output $Y$ of the $NOR$ gate is given by:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's Law,$\overline{x + y} = \bar{x} \cdot \bar{y}$,we get:
$Y = \overline{\bar{A}} \cdot \overline{\bar{B}}$
Since $\overline{\bar{A}} = A$ and $\overline{\bar{B}} = B$,the expression simplifies to:
$Y = A \cdot B$
This is the Boolean expression for an $AND$ gate.
Therefore,the combination represents an $AND$ gate.

(B) Given,mass of $CO = 2.8 \ g$.
Temperature $T = 27 + 273 = 300 \ K$.
Pressure $p = 0.821 \ atm$.
Gas constant $R = 0.08210 \ L \ atm \ K^{-1} \ mol^{-1}$.
Molar mass of $CO = 12 + 16 = 28 \ g \ mol^{-1}$.
Number of moles $n = \frac{2.8 \ g}{28 \ g \ mol^{-1}} = 0.1 \ mol$.
Using the ideal gas equation $pV = nRT$:
$V = \frac{nRT}{p} = \frac{0.1 \times 0.08210 \times 300}{0.821} = \frac{2.463}{0.821} = 3 \ L$.

(A) $Cuscuta$ is an example of ectoparasitism.
$Cuscuta$ is a parasitic plant that lacks chlorophyll and leaves. It derives its nutrition from the host plant by growing on its surface,which classifies it as an ectoparasite.

(A) Elements $X, Y$ and $Z$ with atomic numbers $19, 37$ and $55$ are $K, Rb$ and $Cs$ respectively.
These elements belong to group-$I$ of the periodic table.
On moving down the group,the atomic size increases,which leads to a decrease in the ionisation potential.
Therefore,the order of ionisation potential is $X > Y > Z$.
Thus,$Y$ has an ionisation potential between those of $X$ and $Z$.

(D) Given,wavelength of photon,$\lambda = 2.2 \times 10^{-11} \ m$.
Planck constant,$h = 6.6 \times 10^{-34} \ J \ s$.
According to the de Broglie relation,$\lambda = \frac{h}{p}$.
Therefore,the momentum $p = \frac{h}{\lambda}$.
Substituting the values: $p = \frac{6.6 \times 10^{-34} \ J \ s}{2.2 \times 10^{-11} \ m} = 3 \times 10^{-23} \ kg \ m \ s^{-1}$.

(D) Given: $n = 2 \ mol$,$V_1 = 1 \ L$,$V_2 = 10 \ L$,$T = 300 \ K$,$R = 0.0083 \ kJ \ K^{-1} \ mol^{-1}$.
For a reversible isothermal expansion,the work done is given by the formula:
$W = -2.303 \ nRT \log \frac{V_2}{V_1}$.
Substituting the values:
$W = -2.303 \times 2 \times 0.0083 \times 300 \times \log \frac{10}{1}$.
$W = -2.303 \times 2 \times 0.0083 \times 300 \times 1$.
$W = -11.47 \ kJ \approx -11.5 \ kJ$.
The magnitude of work done by the system is $11.5 \ kJ$.

(D) The reaction of ethanol with excess concentrated $H_2SO_4$ at $443 \ K$ is a dehydration reaction.
In this process,a molecule of water is removed from ethanol to form ethene.
The chemical equation is:
$CH_3CH_2OH \xrightarrow[443 \ K]{\text{Conc. } H_2SO_4} CH_2=CH_2 + H_2O$

(D) In Kolbe's reaction,phenol is first converted to sodium phenolate by reaction with $NaOH$.
Then,sodium phenolate reacts with $CO_2$ under pressure,followed by acidification to form salicylic acid.
The reaction is as follows:
$C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$
$C_6H_5ONa + CO_2 \xrightarrow{H^+} \text{Salicylic acid}$
Thus,the reacting substances for the carboxylation step are sodium phenolate and $CO_2$.

(D) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond between the methyl group and the oxygen atom.
This occurs because the $C_6H_5-O$ bond has partial double bond character due to resonance,making it stronger and harder to break.
The reaction is:
$C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$
Thus,the products formed are phenol and iodomethane.

(A) The reaction in option $(A)$ involves Clemmensen reduction conditions $(Zn-Hg/HCl)$,which are used to reduce carbonyl groups (aldehydes or ketones) to methylene groups $(-CH_2-)$. Carboxylic acids are generally inert to these conditions,so benzaldehyde cannot be prepared from benzoic acid using this method.
Option $(B)$ is the Etard reaction,which produces benzaldehyde from toluene.
Option $(C)$ is the Rosenmund reduction,which produces benzaldehyde from benzoyl chloride.
Option $(D)$ is the Gattermann-Koch reaction,which produces benzaldehyde from benzene.

(A) Aldehydes react with alcohols in the presence of dry $HCl$ gas to form hemiacetals,which further react with another molecule of alcohol to form acetals.
The reaction is as follows:
$R-CHO + R'-OH \rightleftharpoons R-CH(OH)(OR')$ (Hemiacetal)
$R-CH(OH)(OR') + R'-OH \rightleftharpoons R-CH(OR')_2 + H_2O$ (Acetal)

(C) $\text{Pentan-2-one}$ contains a $CH_3CO-$ group,which gives a positive iodoform test when reacted with $I_2$ and $NaOH$,resulting in yellow precipitates of iodoform $(CHI_3)$.
$\text{Pentan-3-one}$ does not contain a $CH_3CO-$ group and therefore does not give the iodoform test.
The reaction for $\text{pentan-2-one}$ is:
$CH_3CH_2CH_2COCH_3 + 3NaOI \rightarrow CH_3CH_2CH_2COONa + CHI_3 \downarrow + 2NaOH$

(B) The carbylamine test is a chemical test for the detection of primary amines. In this reaction,a primary amine is heated with chloroform $(CHCl_3)$ and an alcoholic solution of potassium hydroxide $(KOH)$. The reaction produces an isocyanide (also known as a carbylamine),which has a characteristic foul smell. The general reaction is: $R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$. Among the given options,the general product is an isocyanide $(R-NC)$.

(C) secondary amine is a compound in which $2$ of the hydrogen atoms of $NH_3$ have been replaced by organic groups. The general structure is $R_2NH$,where $R$ represents an alkyl or aryl group. For example,dimethylamine $(CH_3)_2NH$ is a secondary amine.

(C) The $2-$deoxy$-D-$glucose $(2-DG)$ was developed by the Institute of Nuclear Medicine and Allied Sciences $(INMAS)$,a laboratory of the Defence Research and Development Organisation $(DRDO)$,in collaboration with Dr. Reddy's Laboratories. The emergency use of this drug as an additive therapy for $COVID-19$ patients was approved by the Drug Controller General of India $(DCGI)$.

(C) nucleotide is composed of three components: a nitrogenous base (purine or pyrimidine),a pentose sugar,and ortho-phosphoric acid.
These three components are linked together to form the monomeric unit of nucleic acids.

(A) The Hell-Volhard Zelinsky $(HVZ)$ reaction is specific to carboxylic acids that contain at least one $\alpha$-hydrogen atom.
Ethanoic acid $(CH_3COOH)$ contains an $\alpha$-carbon atom attached to three $\alpha$-hydrogen atoms,allowing it to undergo the $HVZ$ reaction.
Methanoic acid $(HCOOH)$ does not have an $\alpha$-carbon atom,and consequently,it lacks $\alpha$-hydrogen atoms,which prevents it from undergoing the $HVZ$ reaction.

(B) The given rate law is $\text{rate} = k[CH_3COOC_2H_5][NaOH]$.
Since the sum of the powers of the concentration terms is $1 + 1 = 2$,the reaction is of second order.
The unit of the rate constant $k$ for a reaction of order $n$ is given by $(mol \ L^{-1})^{1-n} \ s^{-1}$.
For $n = 2$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

(C) For a reaction of $n$th order,the integrated rate law is given by $k = \frac{1}{(n-1)t} [\frac{1}{(A_t)^{n-1}} - \frac{1}{(A_0)^{n-1}}]$.
At half-life,$t = t_{1/2}$ and $A_t = A_0 / 2$.
Substituting these values,we get $t_{1/2} \propto \frac{1}{(A_0)^{n-1}}$,which can also be written as $t_{1/2} \propto (A_0)^{1-n}$.
Therefore,the half-life period is directly proportional to $a^{1-n}$ where $a$ is the initial concentration.

(C) The half-life period $(t_{1/2})$ of an $n$th order reaction is related to the initial concentration $(a)$ as:
$t_{1/2} \propto \frac{1}{a^{n-1}}$
Given that $t_{1/2} \propto \frac{1}{a^5}$,we can compare the exponents:
$n - 1 = 5$
$n = 6$
Therefore,the order of the reaction is $6$.

(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 45 \ min$,so $k = \frac{0.693}{45} \ min^{-1}$.
For $99.9 \%$ completion,the amount remaining $(a-x)$ is $100 - 99.9 = 0.1$.
The time $t$ required is $t = \frac{2.303}{k} \log \frac{a}{a-x}$.
Substituting the values: $t = \frac{2.303}{0.693 / 45} \log \frac{100}{0.1} = \frac{2.303 \times 45}{0.693} \log 1000$.
Since $\log 1000 = 3$,we have $t = \frac{2.303 \times 45 \times 3}{0.693} \approx 448.5 \ min$.
Converting to hours: $t = \frac{448.5}{60} \approx 7.475 \ hrs \approx 7.5 \ hours$.

(D) The chemical formula of $cis-$platin is $[Pt(NH_3)_2Cl_2]$.
In this coordination complex,the oxidation state of $Pt$ is calculated as $x + 2(0) + 2(-1) = 0$,which gives $x = +2$.
According to $IUPAC$ nomenclature rules for coordination compounds,ligands are named in alphabetical order: ammine $(NH_3)$ comes before chlorido $(Cl^-)$.
Thus,the name is diamminedichloridoplatinum $(II)$.

(B) When a transition metal is in its highest oxidation state,it cannot be further oxidized.
Instead,it can easily gain electrons to reach a lower oxidation state.
Therefore,it acts as an oxidising agent.

(D) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{24} \ BM$,we have $\sqrt{n(n+2)} = \sqrt{24}$.
Squaring both sides,$n(n+2) = 24$,which simplifies to $n^2 + 2n - 24 = 0$.
Solving the quadratic equation: $(n+6)(n-4) = 0$.
Since the number of electrons cannot be negative,$n = 4$.
The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
For $Fe^{2+}$,the configuration is $[Ar] 3d^6$,which has $4$ unpaired electrons.
Thus,$x = 2$.

(D) The chemical formula of hexamineplatinum$(IV)$ chloride is $[Pt(NH_3)_6]Cl_4$.
Upon ionization in an aqueous solution,it dissociates as follows:
$[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$.
This results in one complex cation $[Pt(NH_3)_6]^{4+}$ and four chloride anions $4Cl^-$.
Total number of ions = $1 + 4 = 5$ ions.

(B) $[CoCl_6]^{4-}$ is an octahedral complex,so its $CFSE$ value is $\Delta_0 = 18000 \ cm^{-1}$.
$[CoCl_4]^{2-}$ is a tetrahedral complex,so its $CFSE$ value is $\Delta_t$.
The relationship between tetrahedral and octahedral splitting is given by $\Delta_t = \frac{4}{9} \Delta_0$.
Substituting the value: $\Delta_t = \frac{4}{9} \times 18000 \ cm^{-1} = 8000 \ cm^{-1}$.

(A) $Cu^{2+}$ is a strong oxidizing agent and $I^{-}$ is a strong reducing agent. $Cu^{2+}$ oxidizes $I^{-}$ to $I_2$ as follows:
$2Cu^{2+} + 4I^{-}$ $\longrightarrow 2CuI_2$ $\longrightarrow 2CuI + I_2$.
Thus,$CuI_2$ is unstable and decomposes to form $CuI$ and $I_2$.

(B) Among the lanthanoids (atomic numbers $57-71$),$Promethium$ ($Pm$,atomic number $61$) is the only element that is radioactive.

(C) For a cell reaction to be spontaneous,the Gibbs free energy change $(\Delta G)$ must be negative.
Since $\Delta G = -nFE_{\text{cell}}$,a negative $\Delta G$ implies that the cell potential $(E_{\text{cell}})$ must be positive.

(D) In fuel cells,$Pt$ stands for platinum and $Pd$ stands for palladium. These metals are used as catalysts to increase the rate of the electrochemical reactions occurring at the electrodes.

(C) We know that,molar conductivity,$\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
This implies that $\Lambda_{m}$ is inversely proportional to the molarity $(M)$.
The solution with the lowest concentration will have the maximum molar conductivity.
Comparing the given concentrations,$0.001 \ M$ is the lowest concentration.
Therefore,the solution with $0.001 \ M$ concentration has the maximum molar conductivity.

(D) Given,specific conductance,$\kappa = 6.3 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$.
$HNO_3$ concentration,$c = 0.1 \ M$.
Molar conductance formula is $\Lambda_m = \frac{\kappa \times 1000}{c}$.
Substituting the values: $\Lambda_m = \frac{6.3 \times 10^{-2} \times 1000}{0.1} = \frac{63}{0.1} = 630 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(A) The rate of $S_N 1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
The stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ}$.
In pair $(i)$,$2$-chloro$-2-$methylpropane forms a $3^{\circ}$ carbocation,which is more stable than the $2^{\circ}$ carbocation formed by $2$-chloropentane.
In pair (ii),$2$-chloropentane forms a $2^{\circ}$ carbocation,which is more stable than the $1^{\circ}$ carbocation formed by $1$-chloropentane.
Therefore,both $2$-chloro$-2-$methylpropane and $2$-chloropentane undergo faster $S_N 1$ reactions compared to their respective counterparts in the given pairs.

(B) $R-X + AgCN \longrightarrow R-NC + AgX$
$R-X + KCN \longrightarrow R-CN + KX$
$KCN$ is predominantly ionic in nature,providing $CN^-$ ions in solution. The carbon atom is more nucleophilic,leading to the formation of alkyl cyanide $(R-CN)$.
$AgCN$ is predominantly covalent in nature. The nitrogen atom has a lone pair available for bonding,while the carbon atom is involved in the covalent bond with silver. This leads to the formation of alkyl isocyanide $(R-NC)$.

(B) The molecular formula $C_7H_8O$ corresponds to cresols or benzyl alcohol.
Since the compound dissolves in $NaOH$,it must be acidic in nature,which indicates it is a phenol (cresol).
It gives a characteristic violet color with neutral $FeCl_3$,which is a test for phenolic groups.
Treatment with bromine water leads to a tribromo derivative,indicating the presence of three reactive positions (ortho and para) relative to the $-OH$ group.
All three isomers of cresol ($o$-,$m$-,and $p$-) satisfy these conditions. However,in many textbook contexts,$m$-cresol is often cited for such reactions.
Specifically,the reaction with $Br_2$ water on $m$-cresol yields $2,4,6$-tribromo-$m$-cresol.

(B) Solvolysis proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate. The rate of solvolysis depends on the stability of the carbocation formed.
$(a)$ $C_6H_5Cl$: The carbocation formed would be a phenyl cation,which is highly unstable.
$(b)$ $C_6H_5CH_2Cl$: The carbocation formed is $C_6H_5CH_2^+$,which is resonance-stabilized by the benzene ring (benzylic carbocation). This is the most stable among the options.
$(c)$ $C_6H_5CH_2CH_2Cl$: The carbocation formed is a primary alkyl carbocation,which is less stable than the benzylic carbocation.
$(d)$ $CH_2=CHCl$: The carbocation formed would be a vinyl cation,which is highly unstable due to the $sp$ hybridization of the positively charged carbon.
Therefore,$C_6H_5CH_2Cl$ readily undergoes solvolysis.

(A) The thermal decomposition of ammonium dichromate $(NH_4)_2Cr_2O_7$ produces nitrogen gas $(N_2)$:
$(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 \uparrow + Cr_2O_3 + 4H_2O$
Similarly,ammonium nitrite $(NH_4NO_2)$ on heating also liberates nitrogen gas $(N_2)$:
$NH_4NO_2 \stackrel{\Delta}{\longrightarrow} N_2 \uparrow + 2H_2O$
Thus,the same gas is obtained by heating $NH_4NO_2$.

(B) The strong reducing property of hypophosphorus acid $(H_3PO_2)$ is due to the presence of two $P-H$ bonds. The hydrogen atoms directly bonded to the phosphorus atom are responsible for its reducing character. The structure of $H_3PO_2$ is as follows:
$O=P(H)(H)(OH)$

(C) Down the group,electron gain enthalpies become less negative with the increase in the size of the halogen. However,the electron gain enthalpy of $F$ is less negative than $Cl$. This is due to the small size of the $F$-atom. As a result,there are strong inter-electronic repulsions in the relatively small $2p$-orbitals of $F$,and thus incoming electrons experience much less attraction. Thus,the correct order is $Cl > F > Br > I$. Therefore,the option $F > Cl > Br > I$ for electron gain enthalpy is incorrect.

(D) Among the noble gases,$He$ (Helium) has the least tendency to form compounds.
This is because $He$ has the highest first ionization enthalpy among all elements in the periodic table,making it extremely stable and chemically inert.

(B) The correct pair is $Bakelite$ and $Novolac$.
$Bakelite$ is formed by the polymerization of $phenol$ and $formaldehyde$ monomers.
$Teflon$ is the polymer of $tetrafluoroethene$.
$Polyester$ is the polymer of $ethylene glycol$ and $terephthalic acid$.
$Nylon 6$ is the polymer of $caprolactum$.

(C) Statement $(C)$ is incorrect.
Amorphous solids are isotropic in nature,meaning their physical properties (such as electrical conductivity or refractive index) are the same in all directions.
Crystalline solids,on the other hand,are anisotropic.

(D) For a $BCC$ unit cell,the number of atoms per unit cell is $2$.
The packing efficiency is calculated as:
$\text{Packing Efficiency} = \frac{Z \times \frac{4}{3} \pi r^3}{a^3} \times 100$.
For $BCC$,the relation between edge length $a$ and radius $r$ is $a = \frac{4r}{\sqrt{3}}$.
Substituting this,$\text{Packing Efficiency} = \frac{2 \times \frac{4}{3} \pi r^3}{(\frac{4}{\sqrt{3}} r)^3} \times 100 = 68 \%$.
Therefore,the vacant space (void space) is $100 \% - 68 \% = 32 \%$.

(C) cube has $8$ corners,and each corner atom contributes $\frac{1}{8}$ to the unit cell. So,atoms from corners $= 8 \times \frac{1}{8} = 1$.
$A$ cube has $4$ body diagonals. Each body diagonal contains $2$ atoms. Since these atoms are inside the body,they contribute fully $(1)$ to the unit cell.
So,atoms from body diagonals $= 4 \times 2 = 8$.
Total number of atoms $= 1 + 8 = 9$.

(B) Dislocation defect (Frenkel defect) occurs in ionic crystals where there is a large difference in the size of the ions,allowing the smaller ion (usually the cation) to occupy an interstitial site. In alkali halides,the sizes of the cations and anions are almost equal,which prevents the smaller ion from fitting into the interstitial spaces. Therefore,they do not show dislocation defect.

(C) Viscosity is directly proportional to the extent of intermolecular hydrogen bonding.
Among the given options,$Glycerol$ $(CH_2OH-CHOH-CH_2OH)$ has the highest viscosity because it contains $3$ $-OH$ groups,allowing for extensive hydrogen bonding.
In comparison,$Ethylene \ glycol$ has $2$ $-OH$ groups,while $Ethanol$ and $Methanol$ each have only $1$ $-OH$ group.

(B) According to Henry's Law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas $(p)$ above the liquid surface.
Additionally,the dissolution of a gas in a liquid is an exothermic process. According to Le Chatelier's principle,decreasing the temperature $(T)$ favors the forward reaction,thereby increasing the solubility of the gas.

(A) Given: Mass of glucose $(w) = 1.8 \ g$
Mass of solvent $(W) = 100 \ g$
Elevation in boiling point $(\Delta T_b) = 0.1^{\circ} C$
Molar mass of glucose $(M) = 180 \ g / mol$
The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is molality.
Molality $(m) = \frac{w \times 1000}{M \times W} = \frac{1.8 \times 1000}{180 \times 100} = 0.1 \ mol / kg$
Now,$K_b = \frac{\Delta T_b}{m} = \frac{0.1}{0.1} = 1 \ K \ kg / mol$.

(B) Given: Molar mass of glucose $(M_B) = 180 \ g \ mol^{-1}$.
Mass of glucose $(W_B) = 3 \ g$.
Mass of water $(W_A) = 60 \ g$.
Temperature $(T) = 15^{\circ} C = 273 + 15 = 288 \ K$.
Assuming the density of the solution is approximately $1 \ g \ mL^{-1}$,the volume of the solution $(V)$ is approximately $60 \ mL = 0.06 \ L$.
Number of moles of glucose $(n_B) = \frac{W_B}{M_B} = \frac{3}{180} = 0.01667 \ mol$.
Molarity $(C) = \frac{n_B}{V(L)} = \frac{0.01667}{0.06} = 0.2778 \ mol \ L^{-1}$.
Using the formula for osmotic pressure: $\pi = CRT$.
$\pi = 0.2778 \times 0.0821 \times 288 = 6.568 \ atm \approx 6.57 \ atm$.

(C) Among the given colligative properties,osmotic pressure can provide the molar mass of proteins,polymers,and colloids with greater precision.
This is because,for these substances,the values of other colligative properties are too small to be measured accurately.
Secondly,this method uses molarity at room temperature,which is more convenient than molality for these large molecules.

(D) Given,amount of ethyl alcohol $= 414 \ g$.
Amount of water $= 18 \ g$.
Molar mass of water $(H_2O) = 18 \ g/mol$.
Molar mass of ethyl alcohol $(C_2H_5OH) = 46 \ g/mol$.
Moles of $C_2H_5OH = \frac{414}{46} = 9 \ mol$.
Moles of $H_2O = \frac{18}{18} = 1 \ mol$.
Mole fraction of $H_2O = \frac{\text{Moles of } H_2O}{\text{Moles of } H_2O + \text{Moles of } C_2H_5OH} = \frac{1}{1+9} = \frac{1}{10} = 0.1$.

(A) Palladium has a unique property of adsorbing a very large volume of hydrogen gas on its surface. This phenomenon is known as occlusion. Among the given options,the colloidal solution of palladium provides a very high surface area,allowing it to adsorb a significantly larger volume of hydrogen gas compared to the others.

(D) Among the given options,colloidal sulphur is commonly used in the treatment of skin diseases,often in the form of lotions or ointments to treat conditions like acne or scabies.