MHT CET 2007 Chemistry Question Paper with Answer and Solution

(A) Silicon carbide $(SiC)$,also known as carborundum,is a semiconductor containing silicon and carbon.
It occurs in nature as the extremely rare mineral moissanite.

(C) The given equation $xy + 2x + 2y + 4 = 0$ can be factored as $(x + 2)(y + 2) = 0$,which represents two lines: $x = -2$ and $y = -2$.
The third line is $x + y + 2 = 0$.
To find the vertices of the triangle,we find the intersection points of these lines:
$1$. Intersection of $x = -2$ and $y = -2$ is $C(-2, -2)$.
$2$. Intersection of $x = -2$ and $x + y + 2 = 0$ is $A(-2, 0)$.
$3$. Intersection of $y = -2$ and $x + y + 2 = 0$ is $B(0, -2)$.
Since the lines $x = -2$ and $y = -2$ are perpendicular,the triangle is a right-angled triangle with the right angle at $C(-2, -2)$.
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $AB$.
The midpoint of $AB$ is $(\frac{-2 + 0}{2}, \frac{0 - 2}{2}) = (-1, -1)$.

(A) Given $y = \log_{\cos x} \sin x$.
Using the change of base formula,$\log_a b = \frac{\log b}{\log a}$,we can write:
$y = \frac{\log \sin x}{\log \cos x}$.
Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$\frac{dy}{dx} = \frac{(\log \cos x) \frac{d}{dx}(\log \sin x) - (\log \sin x) \frac{d}{dx}(\log \cos x)}{(\log \cos x)^2}$.
Calculating the derivatives:
$\frac{d}{dx}(\log \sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
$\frac{d}{dx}(\log \cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
Substituting these back:
$\frac{dy}{dx} = \frac{(\log \cos x)(\cot x) - (\log \sin x)(-\tan x)}{(\log \cos x)^2}$.
$\frac{dy}{dx} = \frac{\cot x \log \cos x + \tan x \log \sin x}{(\log \cos x)^2}$.
Thus,the correct option is $A$.

(A) Given the function $f(x) = \log(1 + x) - \frac{2x}{2 + x}$.
To find the interval where the function is increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{d}{dx} [\log(1 + x)] - \frac{d}{dx} \left[ \frac{2x}{2 + x} \right]$
Using the quotient rule for the second term:
$f'(x) = \frac{1}{1 + x} - \frac{(2 + x)(2) - (2x)(1)}{(2 + x)^2}$
$f'(x) = \frac{1}{1 + x} - \frac{4 + 2x - 2x}{(2 + x)^2}$
$f'(x) = \frac{1}{1 + x} - \frac{4}{(2 + x)^2}$
Simplify the expression:
$f'(x) = \frac{(2 + x)^2 - 4(1 + x)}{(1 + x)(2 + x)^2}$
$f'(x) = \frac{4 + 4x + x^2 - 4 - 4x}{(1 + x)(2 + x)^2}$
$f'(x) = \frac{x^2}{(1 + x)(2 + x)^2}$
For the function to be increasing,we require $f'(x) > 0$.
Since $x^2 \ge 0$ and $(2 + x)^2 > 0$ for all $x > -1$ (domain of $\log(1+x)$),the sign of $f'(x)$ depends on $(1 + x)$.
$f'(x) > 0$ when $1 + x > 0$,which means $x > -1$.
However,looking at the options provided,$f'(x) > 0$ for all $x > 0$ is clearly satisfied.
Thus,$f(x)$ is increasing on $(0, \infty)$.

(B) We know that $e^{\log(\sin x)} = \sin x$.
Therefore,the integral becomes $I = \int \cos^3 x \sin x \, dx$.
Let $t = \cos x$. Then $dt = -\sin x \, dx$,which implies $\sin x \, dx = -dt$.
Substituting these into the integral:
$I = \int t^3 (-dt) = -\int t^3 \, dt$.
Integrating with respect to $t$,we get $I = -\frac{t^4}{4} + c$.
Substituting back $t = \cos x$,we get $I = -\frac{\cos^4 x}{4} + c$.

(B) In $NH_4Cl$,the ammonium ion $(NH_4^+)$ is held together by covalent bonds between nitrogen and hydrogen atoms. The interaction between the ammonium ion $(NH_4^+)$ and the chloride ion $(Cl^-)$ is ionic in nature. Therefore,$NH_4Cl$ contains both covalent and ionic bonds.

(B) For the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,we rewrite it as $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$.
Here,$a^2 = \frac{144}{25}$ and $b_H^2 = \frac{81}{25}$.
The eccentricity $e_H$ is given by $e_H^2 = 1 + \frac{b_H^2}{a^2} = 1 + \frac{81/25}{144/25} = 1 + \frac{81}{144} = \frac{225}{144}$.
So,$e_H = \frac{15}{12} = \frac{5}{4}$.
The foci are $(\pm ae_H, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm ae_E, 0)$ where $a^2 = 16$.
Since the foci coincide,$ae_E = 3$,so $4e_E = 3$,which means $e_E = \frac{3}{4}$.
Using $b^2 = a^2(1 - e_E^2)$,we get $b^2 = 16(1 - (\frac{3}{4})^2) = 16(1 - \frac{9}{16}) = 16(\frac{7}{16}) = 7$.

(D) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$.
Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.
The eccentricity $e_h$ of the hyperbola is given by $e_h^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{81/25}{144/25} = 1 + \frac{81}{144} = \frac{225}{144}$.
So,$e_h = \frac{15}{12} = \frac{5}{4}$.
The foci of the hyperbola are $(\pm a_h e_h, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm ae, 0) = (\pm \sqrt{16 - b^2}, 0)$.
Since the foci coincide,$\sqrt{16 - b^2} = 3$.
Squaring both sides,$16 - b^2 = 9$.
Therefore,$b^2 = 16 - 9 = 7$.

(B) Every object with a temperature above absolute zero emits electromagnetic radiation. The wavelength of this radiation depends on the temperature of the object.
The human body,having a temperature of approximately $37^{\circ}C$ $(310 \ K)$,emits electromagnetic radiation primarily in the infrared region of the spectrum.
This occurs because the thermal vibrations of atoms and molecules in the body correspond to frequencies that fall within the infrared range. Therefore,the radiation emitted by the human body is in the infrared region.

(A) Citrus fruits like lemon contain $Citric \ acid$.
This acid is responsible for the sour taste of lemon.

(C) Ammonium chloride $(NH_{4}Cl)$ contains both covalent and ionic bonds.
In the ammonium ion $(NH_{4}^{+})$,the nitrogen atom is bonded to four hydrogen atoms by covalent bonds.
The ammonium ion $(NH_{4}^{+})$ and the chloride ion $(Cl^{-})$ are held together by an ionic bond.

(A) In $SiO_{4}^{4-}$,the central silicon atom $(Si)$ has $4$ valence electrons.
Each of the $4$ oxygen atoms contributes $1$ electron to form $4$ single bonds with $Si$,and carries a negative charge.
The total number of electron pairs around the central $Si$ atom is $4$ (all bond pairs).
According to $VSEPR$ theory,$4$ bond pairs correspond to $sp^3$ hybridization.
Therefore,the geometry of the $SiO_{4}^{4-}$ anion is tetrahedral.

(A) In $NH_3$,the nitrogen atom undergoes $sp^3$ hybridization.
It has $3$ bond pairs and $1$ lone pair of electrons.
According to $VSEPR$ theory,the presence of a lone pair causes repulsion,which distorts the expected tetrahedral geometry into a pyramidal shape.

(C) The bond energy (or bond dissociation enthalpy) is defined as the amount of energy required to break one mole of similar bonds in the gaseous state to separate the bonded atoms.

(A) $1$. Identify the longest carbon chain containing the functional group. The longest chain has $5$ carbon atoms,so the parent alkane is pentane.
$2$. The functional group is an ether group $(-OC_2H_5)$,which is named as an alkoxy group (ethoxy).
$3$. Number the chain from the end that gives the lowest locant to the substituent. Numbering from left to right,the ethoxy group is at position $2$.
$4$. Therefore,the $IUPAC$ name is $2-$ethoxy pentane.

(C) The synthesis of $but-1-ene$ $(CH_2=CH-CH_2-CH_3)$ from $CH_3MgI$ (a Grignard reagent) involves a nucleophilic substitution reaction with an alkyl halide containing a terminal double bond.
Specifically,$CH_3MgI$ reacts with allyl chloride $(CH_2=CH-CH_2Cl)$ to form $but-1-ene$ and magnesium salts $(Mg(Cl)I)$:
$CH_2=CH-CH_2Cl + CH_3MgI \rightarrow CH_2=CH-CH_2-CH_3 + Mg(Cl)I$
Thus,the correct reagent is allyl chloride.

(C) Ammonia solution contains $NH_4OH$ which is a weak base and ionizes as $NH_4OH \rightleftharpoons NH_4^{+} + OH^{-}$.
When $NH_4Cl$ is added,it provides $NH_4^{+}$ ions,which is a common ion.
Due to the common ion effect,the equilibrium shifts to the left,decreasing the concentration of $OH^{-}$ ions.
Since $pH = 14 - pOH$ and $pOH = -\log[OH^{-}]$,a decrease in $[OH^{-}]$ leads to an increase in $pOH$,which consequently results in a decrease in the $pH$ value.

(C) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
The total concentration of $H^+$ ions is given by:
$[H^+]_{total} = [H^+]_{HCl} + [H^+]_{water} = 10^{-8} \ M + 10^{-7} \ M$.
$[H^+]_{total} = (0.1 \times 10^{-7} + 1 \times 10^{-7}) \ M = 1.1 \times 10^{-7} \ M$.
Now,calculate the $pH$:
$pH = -\log[H^+]_{total} = -\log(1.1 \times 10^{-7})$.
$pH = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$.
Thus,the $pH$ lies between $6$ and $7$.

(C) $Na_{2}CO_{3}$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_{2}CO_{3})$.
When dissolved in water,it dissociates into $2Na^{+}$ and $CO_{3}^{2-}$ ions.
The $Na^{+}$ ion does not undergo hydrolysis because it is the conjugate acid of a strong base.
The $CO_{3}^{2-}$ ion,being the conjugate base of a weak acid $(HCO_{3}^{-})$,undergoes anionic hydrolysis:
$CO_{3}^{2-} + H_{2}O \rightleftharpoons HCO_{3}^{-} + OH^{-}$
Thus,the hydrolysis involves the reaction between $CO_{3}^{2-}$ and water.

(C) The solubility product constant $(K_{sp})$ of $AgCl$ is calculated as:
$K_{sp} = (\text{solubility})^2 = (1 \times 10^{-5})^2 = 1 \times 10^{-10}$.
Let the solubility of $AgCl$ in $0.1 \ M \ NaCl$ be $x \ mol/L$.
In the presence of $0.1 \ M \ NaCl$,the concentration of chloride ions is $[Cl^-] = (x + 0.1) \ M$.
Since $x$ is very small compared to $0.1$,we can approximate $[Cl^-] \approx 0.1 \ M$.
The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.
Substituting the values: $1 \times 10^{-10} = x \times 0.1$.
Solving for $x$: $x = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \ mol/L$.

(C) An amphoteric oxide is one that can react with both acids and bases to form salt and water.
$Al_{2}O_{3}$ is a well-known amphoteric oxide.
It reacts with acids: $Al_{2}O_{3} + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}O$.
It reacts with bases: $Al_{2}O_{3} + 2NaOH + 3H_{2}O \rightarrow 2Na[Al(OH)_{4}]$.
$SO_{3}$ and $P_{4}O_{10}$ are acidic oxides,while $MgO$ is a basic oxide.

(A) Carborundum is the common name for silicon carbide $(SiC)$.
It is produced by heating silica $(SiO_{2})$ with excess carbon in an electric furnace at high temperatures.
The chemical reaction is:
$SiO_{2} + 3C \longrightarrow SiC + 2CO$

(A) Monosilane $(SiH_{4})$ is highly reactive and spontaneously ignites in air to produce silica $(SiO_{2})$ and water. The combustion reaction is: $SiH_{4} + 2O_{2} \longrightarrow SiO_{2} + 2H_{2}O$. The white smoke produced consists of finely divided particles of $SiO_{2}$,which form characteristic vortex rings.

(C) The high electronegativity of fluorine atoms leads to intermolecular hydrogen bonding in $(HF)_{n}$.
Hydrogen bonding facilitates the association of $HF$ molecules,which is why $HF$ exists in the liquid state.
$H-F \dots H-F \dots H-F$
Here,the bond between $F \dots H$ is the hydrogen bond.

(D) Sublimation is a process in which a solid substance converts directly into vapours without passing through the liquid state.
$I_{2}$ (Iodine) is a solid at room temperature and undergoes sublimation upon heating.
$F_{2}$ and $Cl_{2}$ are gases,while $Br_{2}$ is a liquid at room temperature.
Therefore,$I_{2}$ can be purified by sublimation.

(C) State functions are properties that depend only on the initial and final states of the system,such as internal energy $(U)$,enthalpy $(H)$,and entropy $(S)$.
Path functions are properties that depend on the path taken to reach the state,such as work $(w)$ and heat $(q)$.

(B) Hess's law states that the total enthalpy change for a reaction is the same,whether it occurs in one step or several steps.
This is a direct consequence of the law of conservation of energy,which is also known as the first law of thermodynamics.
Therefore,both $B$ and $C$ are technically correct,but since the law is fundamentally a statement of the conservation of energy,$B$ is the most appropriate answer.

(C) For an ideal gas,the heat of reaction at constant pressure $(q_{p})$ and constant volume $(q_{v})$ are related by the enthalpy change equation.
Since $\Delta H = \Delta E + \Delta n RT$ and $\Delta H = q_{p}$ (at constant pressure) and $\Delta E = q_{v}$ (at constant volume),
Therefore,$q_{p} = q_{v} + \Delta n RT$.

(A) The given reaction is $2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_{2}O_{(g)}$ with $\Delta H^{\circ} = -573.2 \ kJ$.
This is the heat of formation for $2 \text{ moles}$ of water.
The decomposition reaction is the reverse of the formation reaction: $H_{2}O_{(g)} \longrightarrow H_{2(g)} + \frac{1}{2} O_{2(g)}$.
For this reaction,the enthalpy change is $\Delta H = -(\frac{-573.2 \ kJ}{2}) = +286.6 \ kJ/mol$.

(A) The given equation is $y = a \cos x + b \sin x + c e^{-x}$.
Here,$a$,$b$,and $c$ are three arbitrary constants (parameters).
The order of a differential equation is equal to the number of independent arbitrary constants present in its general solution.
Since there are $3$ arbitrary constants,the order of the differential equation is $3$.

(C) Given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$.
Here,$a^{2}=16$,so $a=4$.
The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{16}}=\frac{\sqrt{16-b^{2}}}{4}$.
Thus,the foci of the ellipse are $(\pm ae, 0) = (\pm \sqrt{16-b^{2}}, 0)$.
Given equation of the hyperbola is $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$,which can be written as $\frac{x^{2}}{(12/5)^{2}}-\frac{y^{2}}{(9/5)^{2}}=1$.
Here,$a^{2}=\frac{144}{25}$ and $b^{2}=\frac{81}{25}$.
The eccentricity $e$ is $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4}$.
The foci of the hyperbola are $(\pm ae, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
Since the foci coincide,we have $\sqrt{16-b^{2}}=3$.
Squaring both sides,$16-b^{2}=9$,which gives $b^{2}=7$.

(D) The equation of any tangent to the parabola $y^{2}=4x$ is given by $y=mx+\frac{1}{m}$.
This line also touches the parabola $x^{2}=-32y$. Substituting $y=mx+\frac{1}{m}$ into the second equation:
$x^{2}=-32(mx+\frac{1}{m})$
$x^{2}+32mx+\frac{32}{m}=0$
Since the line is a tangent,the discriminant of this quadratic equation must be zero:
$D = (32m)^{2}-4(1)(\frac{32}{m}) = 0$
$1024m^{2}-\frac{128}{m}=0$
$1024m^{3}=128$
$m^{3}=\frac{128}{1024}=\frac{1}{8}$
$m=\frac{1}{2}$
Substituting $m=\frac{1}{2}$ into the tangent equation $y=mx+\frac{1}{m}$:
$y=\frac{1}{2}x+\frac{1}{1/2}$
$y=\frac{1}{2}x+2$
$2y=x+4$
$x-2y+4=0$

(D) The capacitance of a parallel plate capacitor with a dielectric medium is given by $C = \frac{K \varepsilon_{0} A}{d}$.
Here,$K$ is the dielectric constant,$\varepsilon_{0}$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between them.
Given that with oil $(K = 2)$,the capacitance is $C = \frac{2 \varepsilon_{0} A}{d}$.
When the oil is removed,the medium becomes air (or vacuum) with $K = 1$. Let the new capacitance be $C_{0}$.
Then,$C_{0} = \frac{\varepsilon_{0} A}{d}$.
Comparing the two equations,we get $C = 2 C_{0}$,which implies $C_{0} = \frac{C}{2}$.

(A) The hydrolysis of $2,2-$dichloro propane $(CH_3-CCl_2-CH_3)$ involves the replacement of two chlorine atoms with two hydroxyl groups to form $2,2-$propane diol $(CH_3-C(OH)_2-CH_3)$.
Since two hydroxyl groups are attached to the same carbon atom,the resulting gem-diol is unstable.
It readily loses a water molecule $(H_2O)$ to form a stable carbonyl compound,which is acetone $(CH_3-CO-CH_3)$.

(C) The term 'anhydride' in this context refers to the compound formed by the removal of a water molecule $(H_2O)$ from a diol.
$2,3-$butanediol has the structure $CH_3-CH(OH)-CH(OH)-CH_3$.
Upon dehydration,it loses a water molecule to form an epoxide (oxirane derivative).
The reaction is: $CH_3-CH(OH)-CH(OH)-CH_3 \xrightarrow{-H_2O} CH_3-CH(O)CH-CH_3$.
The product formed is $2,3-$epoxybutane.
Therefore,$2,3-$epoxybutane is the anhydride of $2,3-$butanediol.

(B) Fehling's solution is a mild oxidizing agent used to distinguish aliphatic aldehydes from aromatic aldehydes and ketones.
Aliphatic aldehydes like $CH_{3}CHO$ and reducing sugars like $C_{6}H_{12}O_{6}$ react with Fehling's solution to give a red precipitate of $Cu_{2}O$.
Formic acid $(HCOOH)$ also gives a positive test with Fehling's solution due to the presence of an aldehydic group.
Aromatic aldehydes like $C_{6}H_{5}CHO$ do not react with Fehling's solution because the resonance stabilization of the benzene ring makes the carbonyl carbon less susceptible to oxidation.

(A) The aldol condensation of two molecules of acetaldehyde $(CH_{3}CHO)$ in the presence of a dilute base $(OH^-)$ yields $3-$hydroxybutanal,commonly known as aldol.
The reaction is as follows:
$2CH_{3}CHO \xrightarrow{OH^-} CH_{3}CH(OH)CH_{2}CHO$ (aldol)

(A) Glucose gives a silver mirror test (Tollens' test) with ammoniacal silver nitrate due to the presence of the $-CHO$ (aldehyde) group in its open-chain structure.
$CH_2OH(CHOH)_4CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_2OH(CHOH)_4COO^- + 2Ag(s) + 4NH_3 + 2H_2O$
The formation of metallic silver $(Ag)$ results in the silver mirror.

(C) The reaction of a primary amine with an excess of acetyl chloride $(CH_{3}COCl)$ in the presence of a base leads to diacylation.
The reaction is:
$(CH_{3})_{2}CHNH_{2} + 2CH_{3}COCl \rightarrow (CH_{3})_{2}CHN(COCH_{3})_{2} + 2HCl$.
Here,the primary amine $(CH_{3})_{2}CHNH_{2}$ reacts with two equivalents of acetyl chloride to form $N,N$-diacetylisopropylamine,which is $((CH_{3})_{2}CHN(COCH_{3})_{2})$.

(C) The reaction of a primary amine with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the carbylamine reaction.
The chemical equation is: $C_2H_5NH_2 + CHCl_3 + 3KOH(alc.) \longrightarrow C_2H_5NC + 3KCl + 3H_2O$.
The product formed is ethyl isocyanide $(C_2H_5NC)$.

(C) Sucrose on hydrolysis with dilute acid or the enzyme invertase yields an equimolar mixture of glucose and fructose.
$C_{12}H_{22}O_{11} + H_2O$ $\xrightarrow{\text{H}^+ \text{ or Invertase}} C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$

(D) zwitter ion is a dipolar ion that contains both a positive and a negative charge,typically formed by amino acids.
Glycine $(NH_2CH_2COOH)$ is the simplest amino acid.
In an aqueous solution,the acidic carboxyl group $(-COOH)$ loses a proton to the basic amino group $(-NH_2)$,resulting in the formation of a zwitter ion:
$NH_3^+CH_2COO^-$

(C) When proteins or amino acids are heated with a dilute solution of $Ninhydrin$ ($triketo-hydrindin$ hydrate),a characteristic blue or violet colour is produced.
This reaction is commonly used for the detection of proteins and amino acids.
$\text{Protein} + \text{Ninhydrin} \rightarrow \text{Blue colour}$

(C) The process of converting a freshly prepared precipitate into a colloidal sol by adding a suitable electrolyte is known as peptization.
In this process,the precipitate particles adsorb ions from the electrolyte,which causes them to break down and disperse into the colloidal state.

(B) Fats and oils are triesters of glycerol with long-chain fatty acids,commonly known as triglycerides.
They are formed by the esterification of glycerol $(CH_2OH-CHOH-CH_2OH)$ with three molecules of higher carboxylic acids (fatty acids).
Therefore,fats are esters of glycerol.

(A) The reaction between a carboxylic acid and an alcohol in the presence of an acid catalyst produces an ester and water. This process is known as esterification.
$RCOOH + R^{\prime}OH \xrightarrow{H^+} RCOOR^{\prime} + H_2O$
Here,$RCOOH$ is the carboxylic acid,$R^{\prime}OH$ is the alcohol,and $RCOOR^{\prime}$ is the ester.

(A) Corrosive sublimate $(HgCl_{2})$ acts as a mild oxidizing agent. It reacts with formic acid $(HCOOH)$ to form a white precipitate of calomel $(Hg_{2}Cl_{2})$,which may appear greyish-black due to the presence of finely divided mercury. Acetic acid $(CH_{3}COOH)$ does not react with $HgCl_{2}$.
$2 HCOOH + 2 HgCl_{2} \longrightarrow Hg_{2}Cl_{2} + 2 CO_{2} + 2 HCl$
$CH_{3}COOH + HgCl_{2} \longrightarrow \text{No reaction}$
Thus,it is used to distinguish between formic acid and acetic acid.

(C) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{total time}}{\text{half-life period}} = \frac{90 \ min}{30 \ min} = 3$.
The number of remaining atoms $(N)$ is given by the formula: $N = N_0 \times (\frac{1}{2})^n$.
Substituting the values: $N = 600 \times (\frac{1}{2})^3$.
$N = 600 \times \frac{1}{8} = 75 \ atoms$.

(A) Let the reaction be ${}_{90}Th^{232} \rightarrow {}_{82}Pb^{208} + n_{\alpha} ({}_{2}He^{4}) + n_{\beta} ({}_{-1}e^{0})$.
Mass balance: $232 = 208 + 4n_{\alpha}$ $\Rightarrow 4n_{\alpha} = 24$ $\Rightarrow n_{\alpha} = 6$.
Atomic number balance: $90 = 82 + 2n_{\alpha} - n_{\beta}$ $\Rightarrow 90 = 82 + 2(6) - n_{\beta}$ $\Rightarrow 90 = 94 - n_{\beta}$ $\Rightarrow n_{\beta} = 4$.
Therefore,$6 \alpha$ and $4 \beta$ particles are emitted.

(C) $40 \%$ aqueous solution of formaldehyde $(methanal)$ is known as formalin.
Formalin is commonly used as a disinfectant and as a preservative for biological specimens.

(C) $Uranium$ and $plutonium$ are commonly used as fuel in an atomic pile (nuclear reactor).

(D) The reducing power of a substance is inversely proportional to its standard reduction potential $(E^{\circ}_{red})$.
Substances with more negative $E^{\circ}_{red}$ values are stronger reducing agents.
Given values are:
$A: +0.68 \ V$
$C: -0.50 \ V$
$B: -2.54 \ V$
Comparing the values: $-2.54 < -0.50 < +0.68$.
Therefore,the order of reducing power is $B > C > A$.

(D) The reduction reaction for a hydrogen electrode is: $2H^{+} + 2e^{-} \longrightarrow H_{2}(g)$.
According to the Nernst equation at $298 \ K$:
$E = E^{\circ} - \frac{0.0591}{n} \log \frac{P_{H_{2}}}{[H^{+}]^{2}}$.
Given $E^{\circ} = 0 \ V$,$n = 2$,and $P_{H_{2}} = 1 \ atm$:
$E = 0 - \frac{0.0591}{2} \log \frac{1}{[H^{+}]^{2}}$.
$E = -\frac{0.0591}{2} \times (-2 \log [H^{+}])$.
Since $pH = -\log [H^{+}]$,we get:
$E = -0.0591 \ pH$.

(C) The reduction reaction of silver ions is: $Ag^{+} + e^{-} \rightarrow Ag$.
From the reaction,$1 \ \text{mole}$ of electrons $(1 \ F)$ deposits $1 \ \text{mole}$ of silver $(Ag)$.
The atomic mass of $Ag$ is $108 \ g/mol$.
Therefore,$1 \ F$ of electricity deposits $108 \ g$ of $Ag$.
For $2 \ F$ of electricity,the amount of $Ag$ deposited is:
$2 \times 108 \ g = 216 \ g$.

(A) Phenol is more acidic than alcohol because the phenoxide ion formed after the loss of a proton $(H^+)$ is stabilised by resonance. The negative charge on the oxygen atom is delocalised over the benzene ring,which increases the stability of the phenoxide ion compared to the alkoxide ion formed from alcohols.

(C) Williamson's synthesis is a reaction where an alkoxide ion acts as a nucleophile and attacks an alkyl halide to form an ether.
The general reaction is: $R-ONa + R'-X \longrightarrow R-O-R' + NaX$.
This reaction proceeds via an $S_{N}2$ mechanism,where the nucleophile attacks the alkyl halide from the backside,leading to the displacement of the halide ion in a single step.

(A) The reaction of $1$-chlorobutane $(CH_3-CH_2-CH_2-CH_2-Cl)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.
In this reaction,an $H$ atom is removed from the $\beta$-carbon and the $Cl$ atom is removed from the $\alpha$-carbon,resulting in the formation of an alkene.
The reaction proceeds as follows:
$CH_3-CH_2-CH_2-CH_2-Cl + \text{alc. } KOH \xrightarrow{\Delta} CH_3-CH_2-CH=CH_2 + KCl + H_2O$
The product formed is but-$1$-ene.

(C) Ruby is a mineral of aluminium,i.e.,$Al_{2}O_{3}$. It does not contain silicon.

(B) Glass contains silica $(SiO_{2})$,which reacts with hydrofluoric acid $(HF)$.
$SiO_{2} + 4HF \longrightarrow SiF_{4} + 2H_{2}O$
$SiF_{4} + 2HF \longrightarrow H_{2}SiF_{6}$ (fluorosilicic acid)
$HF$ is commonly used for the etching of glass.

(D) Terylene (Dacron) is a polyester fibre which is prepared by the condensation polymerisation of ethylene glycol and terephthalic acid with the elimination of water molecules.
The chemical reaction is as follows:
$n(HO-CH_2-CH_2-OH) + n(HOOC-C_6H_4-COOH) \xrightarrow{\Delta, -nH_2O} [-O-CH_2-CH_2-O-CO-C_6H_4-CO-]_n$
This reaction takes place at approximately $425-475 \ K$.

(D) Nylon$-66$ is a polyamide fibre which is manufactured by the condensation polymerisation of adipic acid and hexamethylene diamine. Since it is both a condensation polymer and a polyamide,the statement 'Nylon$-66$ is not a' is false for both options $(a)$ and $(b)$. Therefore,the correct answer is $(d)$.

(A) The Freundlich adsorption isotherm is represented by the equation:
$x/m = k p^{1/n}$
Where:
$x$ is the mass of the adsorbate adsorbed on mass $m$ of the adsorbent.
$p$ is the equilibrium pressure.
$k$ and $n$ are constants that depend on the nature of the adsorbent and the adsorbate at a particular temperature.
Note: The Freundlich isotherm is an empirical relation and is not applicable at very high pressure.

(C) The colour of a colloidal solution depends on the wavelength of the light scattered by the dispersed particles,which in turn depends on the size and the nature of the particle.
The blue colour of water in the sea is primarily due to the scattering of light by water molecules.