TS EAMCET 2005 Chemistry Question Paper with Answer and Solution

(B) 'Natalite' is a mixture of ethanol and ether,which is used as a substitute for petrol in internal combustion engines.

(C) Primary alcohols are dehydrogenated by hot copper at $300^{\circ} C$ to form aldehydes.
The reaction is: $C_2H_5OH \xrightarrow[300^{\circ} C]{Cu} CH_3CHO + H_2$.
The product $X$ is acetaldehyde,which has the molecular formula $C_2H_4O$.

(D) Alkyl halides react with sodium alkoxide to form an ether. This reaction is known as the Williamson ether synthesis.
$C_2H_5Cl + C_2H_5ONa \xrightarrow{\Delta} C_2H_5OC_2H_5 + NaCl$
In this reaction,$C_2H_5Cl$ (ethyl chloride) reacts with $C_2H_5ONa$ (sodium ethoxide) to produce $C_2H_5OC_2H_5$ (diethyl ether).

(D) Rosenmund's reduction involves the catalytic hydrogenation of an acid chloride (acyl chloride) to an aldehyde using $Pd$ supported on $BaSO_4$ (Lindlar's catalyst).
The reaction is: $CH_3COCl + H_2 \xrightarrow{Pd / BaSO_4} CH_3CHO + HCl$.

(C) The inductive reactance $X_L$ is given by the formula $X_L = \omega L$.
Since the angular frequency $\omega = 2 \pi \nu$,where $\nu$ is the frequency of the $AC$ source,we have $X_L = 2 \pi \nu L$.
Given: Inductance $L = 1 \ H$ and frequency $\nu = 50 \ Hz$.
Substituting these values into the formula:
$X_L = 2 \pi \times 50 \times 1 = 100 \pi \ \Omega$.

(B) Initial charge on the capacitor: $q = C_1 V_1 = 4 \times 10^{-6} \times 200 = 800 \times 10^{-6} \ C$.
Initial energy stored: $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 4 \times 10^{-6} \times (200)^2 = 8 \times 10^{-2} \ J$.
When connected to an uncharged $2 \mu F$ capacitor,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{800 \times 10^{-6} + 0}{4 \times 10^{-6} + 2 \times 10^{-6}} = \frac{800}{6} \ V$.
Final energy stored: $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times (6 \times 10^{-6}) \times (\frac{800}{6})^2 = 3 \times 10^{-6} \times \frac{640000}{36} = \frac{64}{12} \times 10^{-2} \approx 5.33 \times 10^{-2} \ J$.
Loss in energy: $\Delta U = U_i - U_f = 8 \times 10^{-2} - 5.33 \times 10^{-2} = 2.67 \times 10^{-2} \ J$.

(A) Let the masses be $m_1=1 ~kg, m_2=2 ~kg, m_3=3 ~kg$ and $m_4=4 ~kg$.
The initial centre of mass $(X_{CM}, Y_{CM}, Z_{CM})$ of the first three particles is $(2,2,2)$.
The formula for the centre of mass is $X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$.
For the first three particles,the sum of masses is $M = 1+2+3 = 6 ~kg$.
Thus,$\sum m_i x_i = X_{CM} \times M = 2 \times 6 = 12$. Similarly,$\sum m_i y_i = 12$ and $\sum m_i z_i = 12$.
Let the position of the fourth mass $m_4$ be $(x_4, y_4, z_4)$.
The new centre of mass is $(0,0,0)$.
Using the formula for the new centre of mass: $X_{CM}^{\prime} = \frac{\sum m_i x_i + m_4 x_4}{M + m_4} = 0$.
Substituting the values: $0 = \frac{12 + 4x_4}{6 + 4} \implies 12 + 4x_4 = 0 \implies 4x_4 = -12 \implies x_4 = -3$.
Similarly,$y_4 = -3$ and $z_4 = -3$.
Therefore,the position of the fourth mass is $(-3,-3,-3)$.

(A) In $SF_6$,the sulfur atom is bonded to $6$ fluorine atoms by $6$ single covalent bonds. Each bond consists of $2$ electrons,so the total number of electrons in the valence shell of $S$ is $6 \times 2 = 12$. This is an example of an expanded octet,which does not obey the octet rule.
Ionic reactions are generally very fast because they involve the interaction of ions already present in solution.
According to $VSEPR$ theory,$SnCl_2$ has $2$ bond pairs and $1$ lone pair on the central $Sn$ atom,resulting in a bent (angular) geometry,not linear.
The ability to form ionic compounds is related to the electropositivity of the metal. $Na$ is more electropositive than $Mg$,which is more electropositive than $Al$. Therefore,the order of ability to form ionic compounds is $Na^{+} > Mg^{2+} > Al^{3+}$.
Thus,the correct statement is $A$.

(A) In $BeCl_2$,the central atom $Be$ has $2$ valence electrons and forms $2$ bonds with $Cl$ atoms. It has no lone pairs on the central atom.
According to $VSEPR$ theory,the molecule adopts a linear geometry with a bond angle of $180^{\circ}$ due to $sp$ hybridisation.
$H_2O$ is angular (bent) due to $2$ lone pairs on oxygen.
$SO_2$ is angular (bent) due to $1$ lone pair on sulfur.
$CH_4$ is tetrahedral due to $sp^3$ hybridisation.

(D) Benzene is a non-polar solvent,while water is a polar solvent. According to the principle of 'like dissolves like',non-polar compounds dissolve in non-polar solvents.
$C_2H_5OH$,$CH_3COOH$,and $CH_3CHO$ are polar and can form hydrogen bonds with water,making them soluble in water.
$C_6H_5NO_2$ (Nitrobenzene) is a non-polar organic compound,which makes it soluble in benzene but almost insoluble in water.

(A) The equilibrium constant expression for the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ is $K_c = \frac{[Z]}{[X][Y]}$.
Given that $K_c = 10^4 \ L \ mol^{-1}$ and at equilibrium $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$.
From the given relations,we have $[X] = \frac{1}{2}[Z]$ and $[Y] = [Z]$.
Substituting these values into the $K_c$ expression:
$10^4 = \frac{[Z]}{(\frac{1}{2}[Z])([Z])}$
$10^4 = \frac{[Z]}{\frac{1}{2}[Z]^2} = \frac{2}{[Z]}$
$[Z] = \frac{2}{10^4} = 2 \times 10^{-4} \ mol \ L^{-1}$.

(C) For the reaction $2 A + B \longrightarrow C$,the rate of reaction is given by:
Rate $= -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt}$
Given that the rate of formation of $C$ is $\frac{d[C]}{dt} = 2.2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.
Equating the terms for $A$ and $C$:
$-\frac{1}{2} \frac{d[A]}{dt} = \frac{d[C]}{dt}$
$-\frac{d[A]}{dt} = 2 \times \frac{d[C]}{dt}$
$-\frac{d[A]}{dt} = 2 \times (2.2 \times 10^{-3}) = 4.4 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.

(B) For $A$,$t_{1/2} = 1 \ min$.
After $4 \ min$,the number of half-lives $n_A = \frac{4}{1} = 4$.
Fraction of $A$ remaining $= (1/2)^4 = 1/16$.
Fraction of $A$ disintegrated $= 1 - 1/16 = 15/16$.
For $B$,$t_{1/2} = 2 \ min$.
After $4 \ min$,the number of half-lives $n_B = \frac{4}{2} = 2$.
Fraction of $B$ remaining $= (1/2)^2 = 1/4$.
Fraction of $B$ disintegrated $= 1 - 1/4 = 3/4$.
Ratio of disintegrated weights of $A$ and $B = \frac{15/16}{3/4} = \frac{15}{16} \times \frac{4}{3} = \frac{5}{4}$ or $5:4$.

(D) Vitamin $B_{12}$ (cyanocobalamine) contains a cobalt ion $(Co^{3+})$ at its center,not magnesium $(Mg^{2+})$. Therefore,the statement in option $D$ is incorrect.

(B) The central metal ion is $Co^{3+}$ and its coordination number is $6$. The molecular formula is $CoCl_3 \cdot yNH_3$.
Since the complex gives $3$ ions in water,it must dissociate as $[CoCl(NH_3)_5]Cl_2 \rightleftharpoons [CoCl(NH_3)_5]^{2+} + 2Cl^-$.
Here,the primary valency is satisfied by $3$ chloride ions ($1$ inside the coordination sphere and $2$ outside).
The secondary valency is satisfied by $5$ $NH_3$ molecules and $1$ $Cl^-$ ion.
Therefore,only $1$ $Cl^-$ ion satisfies both primary and secondary valencies.

(C) Let $i_1$ be the current from the $6 \, V$ cell and $i_2$ be the current from the $10 \, V$ cell. The total current through the $12 \, \Omega$ resistor is $(i_1 + i_2)$.
Applying Kirchhoff's Voltage Law $(KVL)$ to the loop containing both cells:
$10 - i_2(1) + i_1(0.5) - 6 = 0$
$0.5 i_1 - i_2 = -4$ --- (Equation $i$)
Applying $KVL$ to the loop containing the $10 \, V$ cell and the external resistor:
$10 - i_2(1) - (i_1 + i_2)(12) = 0$
$10 - i_2 - 12 i_1 - 12 i_2 = 0$
$12 i_1 + 13 i_2 = 10$ --- (Equation $ii$)
From Equation $i$, $i_1 = 2 i_2 - 8$. Substituting this into Equation $ii$:
$12(2 i_2 - 8) + 13 i_2 = 10$
$24 i_2 - 96 + 13 i_2 = 10$
$37 i_2 = 106$
$i_2 = 106 / 37 \approx 2.87 \, A$.

(C) For a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$,where $R_1$ is the resistance in the left gap and $R_2$ is the resistance in the right gap.
Case $I$: $P$ and $Q$ are in series,so $R_2 = P + Q$. Given $l = 37.5 \text{ cm}$.
$\frac{30}{P+Q} = \frac{37.5}{100-37.5} = \frac{37.5}{62.5} = \frac{3}{5}$
$P+Q = 30 \times \frac{5}{3} = 50 \Omega$ ... $(i)$
Case $II$: $P$ and $Q$ are in parallel,so $R_2 = \frac{PQ}{P+Q}$. Given $l = 71.4 \text{ cm}$.
$\frac{30}{\frac{PQ}{P+Q}} = \frac{71.4}{100-71.4} = \frac{71.4}{28.6} \approx 2.5$
$\frac{30(P+Q)}{PQ} = 2.5$
Since $P+Q = 50$,we have $\frac{30 \times 50}{PQ} = 2.5$
$PQ = \frac{1500}{2.5} = 600 \Omega^2$ ... (ii)
From $(i)$ and (ii),$P$ and $Q$ are roots of the quadratic equation $x^2 - (P+Q)x + PQ = 0$,which is $x^2 - 50x + 600 = 0$.
$(x-30)(x-20) = 0$
Thus,the values are $30 \Omega$ and $20 \Omega$.

(B) The number of moles of $AgNO_3$ is calculated as: $n = M \times V(L) = 1 \ M \times 0.25 \ L = 0.25 \ mol$.
The reduction reaction is: $Ag^+ + e^- \rightarrow Ag(s)$.
Since $1 \ mol$ of $Ag^+$ requires $1 \ mol$ of electrons $(1 \ Faraday = 96500 \ C)$ to deposit $1 \ mol$ of $Ag$,the electricity required for $0.25 \ mol$ is:
$Q = n \times F = 0.25 \ mol \times 96500 \ C/mol = 24125 \ C$.

(C) The induced electromotive force $(e)$ across the axle of the train moving through the Earth's magnetic field is given by the formula: $e = Bvl$, where $B$ is the vertical component of the Earth's magnetic field, $v$ is the velocity of the train, and $l$ is the distance between the rails.
Given values:
$B = 2 \times 10^{-5} \, T$
$v = 72 \, km/h = 72 \times \frac{5}{18} \, m/s = 20 \, m/s$
$l = 1 \, m$
Substituting these values into the formula:
$e = (2 \times 10^{-5} \, T) \times (20 \, m/s) \times (1 \, m)$
$e = 40 \times 10^{-5} \, V$
$e = 4 \times 10^{-4} \, V$
To convert this into millivolts $(mV)$, we multiply by $10^3$:
$e = 4 \times 10^{-4} \times 10^3 \, mV = 0.4 \, mV$.

(D) Initial force between charges $q_1 = 2 C$ and $q_2 = 6 C$ is given by Coulomb's law: $F_1 = k \frac{q_1 q_2}{r^2} = k \frac{(2)(6)}{r^2} = \frac{12k}{r^2}$.
Given $F_1 = 12 \times 10^3 ~N$,so $\frac{k}{r^2} = 10^3$.
After adding $-4 C$ to each charge,the new charges are $q_1' = 2 - 4 = -2 C$ and $q_2' = 6 - 4 = 2 C$.
The new force is $F_2 = k \frac{q_1' q_2'}{r^2} = k \frac{(-2)(2)}{r^2} = -4 \frac{k}{r^2}$.
Substituting $\frac{k}{r^2} = 10^3$,we get $F_2 = -4 \times 10^3 ~N$.
The negative sign indicates that the force is attractive. Thus,the force is $4 \times 10^3 ~N$ (attraction).

(A) Nitrogen gas $(N_2)$ makes up approximately $78 \%$ of the Earth's atmosphere by volume.
It is an inert gas under normal conditions and does not cause air pollution.
In contrast,$N_2O$,$NO$,and $CO$ are known air pollutants.

(A) The correct matching is:
$(A)$ Benzene matches with $4$. $(4n + 2) \pi$ electrons: Benzene has $6 \pi$ electrons,which follows $H$ückel's rule $(4n + 2) \pi$ electrons.
$(B)$ Ethylene matches with $3$. Mustard gas: Ethylene reacts with $S_2Cl_2$ to produce mustard gas (a chemical warfare agent).
$(C)$ Acetaldehyde matches with $2$. Silver mirror: Acetaldehyde gives a positive silver mirror test with Tollen's reagent due to the presence of an aldehyde group.
$(D)$ Chloroform matches with $1$. Phosgene: Chloroform on oxidation in the presence of air and light gives phosgene $(COCl_2)$,which is a poisonous gas.
Therefore,the correct sequence is $A-4, B-3, C-2, D-1$.

(B) Functional isomerism occurs when compounds have the same molecular formula but different functional groups.
Carboxylic acids and esters are functional isomers of each other.
For example,$C_2H_5COOH$ (propanoic acid) and $CH_3COOCH_3$ (methyl acetate) both have the molecular formula $C_3H_6O_2$ but contain different functional groups (carboxyl group vs. ester group).
Therefore,the pair $C_2H_5CO_2H$ and $CH_3CO_2CH_3$ represents functional isomers.

(NONE) $SiO_2$ is an acidic oxide,so it acts as an acidic flux to remove basic impurities like $FeO$ by forming slag $(FeSiO_3)$.
$FeO + SiO_2 \rightarrow FeSiO_3$ (slag).
Option $A$ is correct.
The distance between layers in graphite is $335 \ pm$ or $3.35 \times 10^{-8} \ cm$. Option $B$ is correct.
$SiO_2$ reacts with $Na_2CO_3$ as follows:
$Na_2CO_3 + SiO_2 \rightarrow Na_2SiO_3 + CO_2 \uparrow$.
Option $C$ is correct.
In graphite,each carbon atom is $sp^2$ hybridized. Option $D$ is correct.
Since all given statements are correct,there is no incorrect statement among the options provided.

(D) $I$. Incorrect: In Hall's process,silica is removed as sodium silicate $(Na_2SiO_3)$ by treatment with $Na_2CO_3$.
$II$. Correct: Baeyer's process is used for red bauxite which is contaminated with $Fe_2O_3$.
$III$. Correct: In Serpeck's process,bauxite is heated with coke and nitrogen to form aluminium nitride $(AlN)$: $Al_2O_3 + 3C + N_2 \xrightarrow{1800^{\circ} C} 2AlN + 3CO$.

(D) white dwarf is a stellar remnant supported against gravitational collapse by electron degeneracy pressure.
According to the Chandrasekhar limit,there is a maximum mass that a stable white dwarf can have.
If the mass of a white dwarf exceeds this limit,the electron degeneracy pressure is insufficient to counteract the gravitational force,leading to further collapse.
The Chandrasekhar limit is approximately $1.4$ times the mass of the Sun $(M_{\odot})$.
Therefore,if the mass of the white dwarf becomes $n = 1.4$ times the solar mass,it will collapse.

(D) The reaction shows the conversion of a haloalkane into an alcohol.
$C_2H_5Cl$ reacts with aqueous $NaOH$ (alkaline hydrolysis) to form ethanol $(C_2H_5OH)$.
Similarly,$C_2H_5Cl$ reacts with moist silver oxide $(AgOH)$ to form ethanol $(C_2H_5OH)$.
Therefore,both $A$ and $B$ are $C_2H_5Cl$.

(A) Alkyl halides can be reduced to alkanes using a $Zn-Cu$ couple in the presence of ethanol $(C_2H_5OH)$.
Specifically,$C_2H_5I + 2[H] \xrightarrow{Zn-Cu, C_2H_5OH} C_2H_6 + HI$.
This is a standard method for the reduction of alkyl halides to alkanes.

(A) Acetophenone can be prepared from benzene by its electrophilic substitution (Friedel-Crafts acylation) reaction as follows:
$C_6H_6 + CH_3COCl \xrightarrow{\text{Anhyd. } AlCl_3} C_6H_5COCH_3 + HCl$
In this reaction,the hydrogen atom of the benzene ring is substituted by the acetyl group $(CH_3CO-)$.

(C) $H_2O_2$ acts as a reducing agent towards strong oxidizing agents like $Cl_2$.
The chemical reaction is: $H_2O_2 + Cl_2 \longrightarrow 2HCl + O_2$.
In this reaction,$HCl$ is produced,which is a strong acid.
The formation of $HCl$ increases the concentration of $H^+$ ions in the solution.
Consequently,the $pH$ of the resultant solution decreases (becomes less than $6.0$) and oxygen gas $(O_2)$ is evolved.

(C) In the given reactions,we analyze the products formed:
$1$. $PbO_2 + H_2O_2 \longrightarrow PbO + H_2O + O_2$ (Oxygen gas is released).
$2$. $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \longrightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$ (Oxygen gas is released).
$3$. $PbS + 4H_2O_2 \longrightarrow PbSO_4 + 4H_2O$ (Lead sulphate is a solid,and water is a liquid; no gas is formed).
$4$. $Cl_2 + H_2O_2 + 2OH^- \longrightarrow 2Cl^- + 2H_2O + O_2$ (Oxygen gas is released).
Therefore,the reaction between $PbS$ and $H_2O_2$ does not produce a gaseous product.

(B) Both $(A)$ and $(R)$ are true statements,but $(R)$ is not the correct explanation of $(A)$.
According to the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$
Since the moles of acetic acid and sodium acetate are equal,$[\text{salt}] = [\text{acid}]$,so $\log(1) = 0$.
Therefore,$pH = pK_a = 4.8$. Thus,$(A)$ is true.
The ionic product of water $(K_w)$ at $25^{\circ} C$ is indeed $10^{-14} \ mol^2 \cdot L^{-2}$. Thus,$(R)$ is true.
However,the value of $K_w$ does not explain why the $pH$ of the buffer is $4.8$.

(A) Given:
Initial temperature $T_1 = 15^{\circ} C = 15 + 273 = 288 \ K$.
Final temperature $T_2 = 35^{\circ} C = 35 + 273 = 308 \ K$.
Since the volume of the tyre remains constant (ignoring expansion),we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Rearranging for the ratio of pressures: $\frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308}{288}$.
The percentage increase in pressure is given by $\frac{P_2 - P_1}{P_1} \times 100$.
Substituting the values: $\left( \frac{308}{288} - 1 \right) \times 100 = \left( \frac{308 - 288}{288} \right) \times 100 = \frac{20}{288} \times 100$.
Calculation: $\frac{2000}{288} \approx 6.94 \%$.
Rounding to the nearest integer,the approximate percentage increase is $7 \%$.

(C) The minimum force required to move a body up a rough inclined plane is given by $F_1 = mg(\sin \theta + \mu \cos \theta)$.
The minimum force required to prevent the body from sliding down the rough inclined plane is given by $F_2 = mg(\sin \theta - \mu \cos \theta)$ if the force is applied parallel to the plane. However,if we consider the force required to just prevent sliding (static equilibrium),it is $F_2 = mg(\sin \theta - \mu \cos \theta)$. Given the standard interpretation for this problem type where $F_2$ is the force to prevent sliding down,we use $F_2 = mg(\sin \theta - \mu \cos \theta)$.
Given $F_1 = 3F_2$,we have $mg(\sin \theta + \mu \cos \theta) = 3mg(\sin \theta - \mu \cos \theta)$.
Dividing by $mg$,we get $\sin \theta + \mu \cos \theta = 3\sin \theta - 3\mu \cos \theta$.
Rearranging terms,$4\mu \cos \theta = 2\sin \theta$.
Thus,$\tan \theta = 2\mu$.
Substituting $\mu = \frac{1}{2\sqrt{3}}$,we get $\tan \theta = 2 \times \frac{1}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Therefore,$\theta = 30^{\circ}$.

(B) By definition,a particle and its corresponding anti-particle have the same mass and the same spin,but opposite charges and opposite magnetic moments.
Since the magnetic moment $\mu$ is related to the angular momentum (spin) $S$ by the relation $\mu = g \frac{q}{2m} S$,where $q$ is the charge and $m$ is the mass,changing the sign of the charge $q$ while keeping mass $m$ and spin $S$ constant results in an opposite magnetic moment.
Therefore,particles and their anti-particles have the same masses but opposite magnetic moments.

(A) Given: Mass $m = 1 \times 10^{-26} \ kg$,charge $q = 1.6 \times 10^{-19} \ C$,velocity $\vec{v} = 1.28 \times 10^6 \ \hat{i} \ ms^{-1}$.
Electric field $\vec{E} = -102.4 \times 10^3 \ \hat{k} \ NC^{-1}$.
Magnetic field $\vec{B} = 8 \times 10^{-2} \ \hat{j} \ Wbm^{-2}$.
The Lorentz force on the particle is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
First,calculate the magnetic force component: $\vec{v} \times \vec{B} = (1.28 \times 10^6 \ \hat{i}) \times (8 \times 10^{-2} \ \hat{j}) = (1.28 \times 8 \times 10^4) \ (\hat{i} \times \hat{j}) = 10.24 \times 10^4 \ \hat{k} = 1.024 \times 10^5 \ \hat{k} \ Vm^{-1}$.
Comparing the magnitudes: $|E| = 102.4 \times 10^3 = 1.024 \times 10^5 \ Vm^{-1}$.
Since $\vec{E} = -1.024 \times 10^5 \ \hat{k} \ Vm^{-1}$ and $\vec{v} \times \vec{B} = 1.024 \times 10^5 \ \hat{k} \ Vm^{-1}$,the net force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) = q(-1.024 \times 10^5 \ \hat{k} + 1.024 \times 10^5 \ \hat{k}) = 0$.
Since the net force is zero,the particle will continue to move in its original direction,which is along the positive $X$-axis.

(B) The magnetic moment is a vector quantity directed from the South pole to the North pole.
When two identical magnets of magnetic moment $M$ are placed perpendicular to each other,their magnetic moment vectors $\vec{M_1}$ and $\vec{M_2}$ are also perpendicular.
The resultant magnetic moment $M^{\prime}$ is given by the vector sum:
$M^{\prime} = \sqrt{M_1^2 + M_2^2 + 2M_1M_2 \cos(90^{\circ})}$
Since $M_1 = M_2 = M$ and $\cos(90^{\circ}) = 0$,we get:
$M^{\prime} = \sqrt{M^2 + M^2} = M\sqrt{2}$
Given that the magnetic moment $M = ml$,we substitute this into the expression:
$M^{\prime} = ml\sqrt{2}$

(D) The function $f(n)$ represents the sum of all positive divisors of $n$.
For a number $n = p_1^{a_1} \cdot p_2^{a_2} \cdots$,the sum of divisors is given by $\sigma(n) = \frac{p_1^{a_1+1}-1}{p_1-1} \cdot \frac{p_2^{a_2+1}-1}{p_2-1} \cdots$.
Given $n = 2^k \cdot 3^1$,the sum of divisors is:
$f(2^k \cdot 3) = (1 + 2 + 2^2 + \cdots + 2^k) \cdot (1 + 3)$.
Using the geometric series sum formula $\sum_{i=0}^{k} 2^i = \frac{2^{k+1}-1}{2-1} = 2^{k+1}-1$.
Thus,$f(2^k \cdot 3) = (2^{k+1}-1) \cdot (4) = 4(2^{k+1}-1)$.

(C) Let the three-digit number be represented as $abc$,where $a, b, c$ are digits.
Given that the last two digits are equal,we have $b = c$.
Also,the last two digits differ from the first digit,so $b \neq a$.
The first digit $a$ can be any digit from $1$ to $9$ (since it is a three-digit number,$a \neq 0$).
For each choice of $a$,the digit $b$ can be any digit from $0$ to $9$ except $a$.
There are $9$ possible values for $a$ $(1, 2, 3, 4, 5, 6, 7, 8, 9)$.
For each $a$,there are $9$ possible values for $b$ ($0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ excluding $a$).
Since $c$ must be equal to $b$,there is only $1$ choice for $c$ once $b$ is chosen.
Therefore,the total number of such $n$'s is $9 \times 9 = 81$.

(A) Given that $1$ is a root of $ax^2+bx+c=0$.
Substituting $x=1$,we get $a(1)^2+b(1)+c=0$,which implies $a+b+c=0$.
Therefore,$E_1$ is true.
Given that $\sin \theta$ and $\cos \theta$ are the roots of $ax^2+bx+c=0$.
From the properties of roots,we have:
$\sin \theta + \cos \theta = -\frac{b}{a}$ $(i)$
$\sin \theta \cos \theta = \frac{c}{a}$ (ii)
Squaring equation $(i)$:
$(\sin \theta + \cos \theta)^2 = \left(-\frac{b}{a}\right)^2$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = \frac{b^2}{a^2}$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$1 + 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2}$
Multiplying by $a^2$:
$a^2 + 2ac = b^2$
$b^2 - a^2 = 2ac$.
Therefore,$E_2$ is true.
Thus,both $E_1$ and $E_2$ are true.

(A) Given equation is $x^3-3x-2=0$.
Testing $x=-1$:
$(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$.
Since $x=-1$ is a root,$(x+1)$ is a factor.
Dividing $x^3-3x-2$ by $(x+1)$ using synthetic division or polynomial division:
$x^3-3x-2 = (x+1)(x^2-x-2) = 0$.
Factoring the quadratic part:
$x^2-x-2 = (x+1)(x-2)$.
Thus,the equation becomes $(x+1)(x+1)(x-2) = 0$.
The roots are $x = -1, -1, 2$.

(B) Given the cubic equation $x^3+2x^2-3x-1=0$,let $\alpha, \beta, \gamma$ be its roots.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = -2$ $(i)$
$\alpha\beta+\beta\gamma+\gamma\alpha = -3$ $(ii)$
$\alpha\beta\gamma = 1$ $(iii)$
We need to find $\alpha^{-2}+\beta^{-2}+\gamma^{-2} = \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \frac{\beta^2\gamma^2+\alpha^2\gamma^2+\alpha^2\beta^2}{(\alpha\beta\gamma)^2}$.
First,calculate $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$ using the identity $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2\alpha\beta\gamma(\alpha+\beta+\gamma)$.
Substituting the values from $(i), (ii),$ and $(iii)$:
$(-3)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(1)(-2)$
$9 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 - 4$
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = 9+4 = 13$.
Now,$\alpha^{-2}+\beta^{-2}+\gamma^{-2} = \frac{13}{(1)^2} = 13$.

(C) The increase in length $\Delta l$ is given by the formula $\Delta l = \frac{Fl}{AY}$,where $F$ is the force,$l$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi r^2$,we have $\Delta l = \frac{Fl}{\pi r^2 Y}$.
The percentage increase in length is $\frac{\Delta l}{l} \times 100 = \frac{F}{\pi r^2 Y} \times 100$.
Let $\Delta x$ be the percentage increase in length. Since $F$ is constant,$\Delta x \propto \frac{1}{r^2 Y}$.
Given ratios: $\frac{r_A}{r_B} = \frac{2}{1}$ and $\frac{Y_A}{Y_B} = \frac{1}{2}$.
Therefore,$\frac{\Delta x_A}{\Delta x_B} = \left(\frac{r_B}{r_A}\right)^2 \times \left(\frac{Y_B}{Y_A}\right)$.
Substituting the values: $\frac{1}{\Delta x_B} = \left(\frac{1}{2}\right)^2 \times \left(\frac{2}{1}\right) = \frac{1}{4} \times 2 = \frac{1}{2}$.
Thus,$\Delta x_B = 2\%$. The correct option is $C$.

(C) The standard equation of a projectile trajectory is given by $y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2$.
Comparing this with the given equation $y = 10x - \left(\frac{5}{9}\right) x^2$,we get:
$\tan \theta = 10$
$\frac{g}{2 u^2 \cos^2 \theta} = \frac{5}{9}$
Given $g = 10 \ m/s^2$,we substitute it into the second equation:
$\frac{10}{2 u^2 \cos^2 \theta} = \frac{5}{9}$
$\frac{5}{u^2 \cos^2 \theta} = \frac{5}{9}$
$u^2 \cos^2 \theta = 9$
The range $R$ of a projectile is given by $R = \frac{2 u^2 \sin \theta \cos \theta}{g}$.
We can rewrite this as $R = \frac{2 (u^2 \cos^2 \theta) \tan \theta}{g}$.
Substituting the known values:
$R = \frac{2 \times 9 \times 10}{10} = 18 \ m$.

(C) Let the body be projected with initial velocity $u$. The equation of motion is $h = ut - \frac{1}{2}gt^2$.
Rearranging,we get $\frac{1}{2}gt^2 - ut + h = 0$.
This is a quadratic equation in $t$ with roots $t_1$ and $t_2$.
The sum of roots is $t_1 + t_2 = \frac{u}{g/2} = \frac{2u}{g}$,so $u = \frac{g(t_1+t_2)}{2}$.
The maximum height $H$ reached by the body is given by $H = \frac{u^2}{2g}$.
Substituting the value of $u$,we get $H = \frac{1}{2g} \left[ \frac{g(t_1+t_2)}{2} \right]^2 = \frac{1}{2g} \cdot \frac{g^2(t_1+t_2)^2}{4} = \frac{g(t_1+t_2)^2}{8}$.
Alternatively,$H = 2g \left( \frac{t_1+t_2}{4} \right)^2 = 2g \cdot \frac{(t_1+t_2)^2}{16} = \frac{g(t_1+t_2)^2}{8}$.
Thus,the correct option is $C$.

(B) The relationship between linear velocity $\vec{v}$,angular velocity $\vec{\omega}$,and position vector $\vec{r}$ for a particle in circular motion is given by $\vec{v} = \vec{\omega} \times \vec{r}$.
Since $\vec{\omega}$ is perpendicular to both $\vec{r}$ and $\vec{v}$,we can use the relation $\vec{\omega} = \frac{\vec{r} \times \vec{v}}{|\vec{r}|^2}$.
Given $\vec{r} = \hat{i} + 9 \hat{j} - 8 \hat{k}$ and $\vec{v} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$.
First,calculate the cross product $\vec{r} \times \vec{v}$:
$\vec{r} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 9 & -8 \\ 3 & -4 & 5 \end{vmatrix} = \hat{i}(45 - 32) - \hat{j}(5 - (-24)) + \hat{k}(-4 - 27) = 13 \hat{i} - 29 \hat{j} - 31 \hat{k}$.
Next,calculate $|\vec{r}|^2 = 1^2 + 9^2 + (-8)^2 = 1 + 81 + 64 = 146$.
Therefore,$\vec{\omega} = \frac{13 \hat{i} - 29 \hat{j} - 31 \hat{k}}{146}$.

(C) The time period $T$ of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the horizontal component of the Earth's magnetic field.
For the original magnet: $I_1 = \frac{ml^2}{12}$ and $M_1 = M$.
When the magnet is cut into four equal pieces normal to its length,each piece has length $l' = l/4$ and mass $m' = m/4$.
The new moment of inertia is $I_2 = \frac{m'(l')^2}{12} = \frac{(m/4)(l/4)^2}{12} = \frac{ml^2}{12 \times 4 \times 16} = \frac{I_1}{64}$.
The new magnetic moment is $M_2 = M/4$.
Using the ratio: $\frac{T_2}{T_1} = \sqrt{\frac{I_2}{I_1} \cdot \frac{M_1}{M_2}} = \sqrt{\frac{I_1/64}{I_1} \cdot \frac{M}{M/4}} = \sqrt{\frac{1}{64} \cdot 4} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Given $T_1 = 4 \ s$,then $T_2 = T_1 \times \frac{1}{4} = 4 \times \frac{1}{4} = 1 \ s$.

(B) Option $A$ is correct: Ammonia is used as a refrigerant due to its high latent heat of vaporization.
Option $B$ is incorrect: Nitrolim is a mixture of calcium cyanamide $(CaCN_2)$ and carbon $(C)$,not calcium cyanide $(Ca(CN)_2)$. The reaction is: $CaC_2 + N_2 \xrightarrow{1100^{\circ}C} CaCN_2 + C$.
Option $C$ is correct: Superphosphate of lime is a mixture of $Ca(H_2PO_4)_2$ and $CaSO_4 \cdot 2H_2O$.
Option $D$ is correct: Hydrolysis of $NCl_3$ yields $NH_3$ and $HOCl$ $(NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl)$.

(A) In Fischer-Ringe's method,air free from moisture and $CO_2$ is passed over a heated mixture $(800^{\circ} C)$ of $90 \% CaC_2 + 10 \% CaCl_2$ in an iron tube.
The following reactions take place:
$CaC_2 + N_2 \xrightarrow{800^{\circ} C} CaCN_2 + C$
$2 C + O_2 \longrightarrow 2 CO$
$C + O_2 \longrightarrow CO_2$
$2 CaC_2 + 3 CO_2 \longrightarrow 2 CaCO_3 + 5 C$
$CuO + CO \longrightarrow Cu + CO_2$
Finally,$CO_2$ gas is absorbed by $KOH$ solution,leaving behind a mixture of noble gases.

(C) Let $S = \sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$.
We can write $2n^2+n+1$ in terms of $n(n-1)$,$n$,and constants:
$2n^2+n+1 = 2n(n-1) + 3n + 1$.
Thus,the sum becomes:
$S = \sum_{n=1}^{\infty} \frac{2n(n-1) + 3n + 1}{n!} = \sum_{n=1}^{\infty} \frac{2n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{3n}{n!} + \sum_{n=1}^{\infty} \frac{1}{n!}$.
For $n=1$,the first term is $0$. For $n=1, 2$,the second term is $3/1! = 3$. For $n=1$,the third term is $1/1! = 1$.
$S = \sum_{n=2}^{\infty} \frac{2}{(n-2)!} + \sum_{n=1}^{\infty} \frac{3}{(n-1)!} + \sum_{n=1}^{\infty} \frac{1}{n!}$.
Using the series $e = \sum_{k=0}^{\infty} \frac{1}{k!}$:
$S = 2e + 3e + (e-1) = 6e-1$.

(D) $3$-hydroxybutanal is an aldol addition product.
It is formed by the self-aldol condensation of two molecules of acetaldehyde $(CH_3CHO)$ in the presence of a dilute base like $NaOH$.
The reaction is as follows:
$2CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO$
Therefore,$X = CH_3CHO$,$Y = CH_3CHO$,and $Z = NaOH$.

(C) For the reaction $2 A + B \longrightarrow C$,the rate of reaction is given by: $-\frac{1}{2} \frac{d[A]}{d t} = -\frac{d[B]}{d t} = \frac{d[C]}{d t}$.
Given that the rate of formation of $C$ is $\frac{d[C]}{d t} = 2.2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.
From the rate expression,we have $-\frac{1}{2} \frac{d[A]}{d t} = \frac{d[C]}{d t}$.
Therefore,$-\frac{d[A]}{d t} = 2 \times \frac{d[C]}{d t}$.
Substituting the value: $-\frac{d[A]}{d t} = 2 \times 2.2 \times 10^{-3} = 4.4 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.

(NONE) The electrochemical equivalent $(Z)$ is defined by the relation $w = Z \cdot I \cdot t$,where $w$ is the mass in grams,$I$ is current in Amperes,and $t$ is time in seconds.
Since $1 \ Coulomb = 1 \ Ampere \cdot 1 \ second$,the unit of $Z$ is $g/Coulomb$.
All statements $A$,$B$,$C$,and $D$ are scientifically correct.
However,if this is a multiple-choice question where one must be incorrect,there might be a context-specific error. Given the standard definitions:
$A$ is correct as $NaCl$ dissociates into ions.
$B$ is correct as $Z = w / (I \cdot t)$.
$C$ is correct as $n$ is the stoichiometric coefficient of electrons.
$D$ is correct by $IUPAC$ convention.
Since all are correct,the question is technically flawed.

(C) Primary alcohols,when passed over heated copper at $300^{\circ}C$,undergo dehydrogenation to form aldehydes.
The reaction is: $C_2H_5OH \xrightarrow[300^{\circ}C]{Cu} CH_3CHO + H_2$.
The product $X$ is acetaldehyde,which has the molecular formula $C_2H_4O$.

(D) Alkyl halides react with sodium alkoxide to give ether. This is known as Williamson's ether synthesis.
$C_2H_5Cl + C_2H_5ONa \xrightarrow{\Delta} C_2H_5OC_2H_5 + NaCl$
In option $D$,the reaction between ethyl chloride and sodium ethoxide produces diethyl ether,which is an ether.

(D) Rosenmund's reduction involves the catalytic hydrogenation of an acid chloride (acid halide) to an aldehyde using $Pd$ supported on $BaSO_4$ (Lindlar's catalyst).
The reaction is: $CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl$.

(D) $3$-hydroxybutanal is an aldol product.
It is formed by the aldol condensation of two molecules of acetaldehyde $(CH_3CHO)$ in the presence of a dilute base like $NaOH$.
Reaction: $2 CH_3CHO \xrightarrow{\text{dilute } NaOH} CH_3-CH(OH)-CH_2-CHO$.
Therefore,$X = CH_3CHO$,$Y = CH_3CHO$,and $Z = NaOH$.

(D) Benzene is a non-polar solvent,while water is a polar solvent.
Compounds that are non-polar or have low polarity tend to dissolve in benzene.
$C_2H_5OH$ (ethanol),$CH_3CO_2H$ (acetic acid),and $CH_3CHO$ (acetaldehyde) are polar and can form hydrogen bonds with water,making them soluble in water.
$C_6H_5NO_2$ (nitrobenzene) is a non-polar organic compound,which makes it insoluble in water but soluble in non-polar solvents like benzene.

(B) For $A$,$t_{1/2} = 1 \ min$. After $4 \ min$ ($4$ half-lives),the fraction of $A$ remaining is $(1/2)^4 = 1/16$.
Therefore,the fraction of $A$ disintegrated is $1 - 1/16 = 15/16$.
For $B$,$t_{1/2} = 2 \ min$. After $4 \ min$ ($2$ half-lives),the fraction of $B$ remaining is $(1/2)^2 = 1/4$.
Therefore,the fraction of $B$ disintegrated is $1 - 1/4 = 3/4$.
The ratio of disintegrated weights of $A$ and $B$ is $(15/16) : (3/4) = 15/16 : 12/16 = 15:12 = 5:4$.

(D) Vitamin $B_{12}$ (cyanocobalamine) contains a central cobalt ion $(Co^{3+})$,not magnesium. Magnesium is the central metal ion in chlorophyll.

(B) The central metal ion is $Co^{3+}$ and the coordination number of $Co^{3+}$ is typically $6$. The molecular formula is $CoCl_3 \cdot yNH_3$ because $Co$ is in the $+3$ oxidation state.
When the complex dissolves in water,it produces $3$ ions. This implies the ionization sphere contains $2$ chloride ions,represented as $[CoCl(NH_3)_5]Cl_2$.
The dissociation is: $[CoCl(NH_3)_5]Cl_2 \rightleftharpoons [CoCl(NH_3)_5]^{2+} + 2Cl^-$.
In this structure,the $Cl^-$ ion inside the coordination sphere acts as a ligand (satisfying secondary valency) and also balances the charge of the metal (satisfying primary valency).
Thus,only $1$ $Cl^-$ ion satisfies both primary and secondary valencies.

(B) The unit of electrochemical equivalent $(Z)$ is $g/Coulomb$.
From Faraday's law,$w = Z \cdot I \cdot t$.
Therefore,$Z = \frac{w}{I \cdot t}$,which has units of $g/Coulomb$.
Thus,the statement in option $B$ is incorrect as it states the units are $g \cdot Coulomb$ instead of $g/Coulomb$.

(B) The number of moles of $AgNO_3$ is calculated as: $\text{Moles} = \text{Molarity} \times \text{Volume in Litres} = 1 \text{ M} \times 0.250 \text{ L} = 0.25 \text{ mol}$.
The reduction reaction for silver is: $Ag^+ + e^- \rightarrow Ag$.
Since $1 \text{ mol}$ of $Ag^+$ requires $1 \text{ Faraday}$ $(96500 \text{ C})$ of electricity,
$0.25 \text{ mol}$ of $Ag^+$ requires $0.25 \times 96500 \text{ C} = 24125 \text{ C}$.

(A) The correct matching is:
$(A)$ Benzene matches with $4$. $(4n + 2) \pi$ electrons: Benzene has $6 \pi$ electrons,which follows the Huckel rule $(4n + 2) \pi$ electrons.
$(B)$ Ethylene matches with $3$. Mustard gas: Ethylene reacts with $S_2Cl_2$ to form mustard gas.
$(C)$ Acetaldehyde matches with $2$. Silver mirror: Acetaldehyde gives a positive silver mirror test with Tollen's reagent.
$(D)$ Chloroform matches with $1$. Phosgene: Chloroform on oxidation gives phosgene,which is a poisonous gas.
Therefore,the correct sequence is $A-4, B-3, C-2, D-1$.

(D) Statement $I$ is incorrect because in Hall's process,silica is removed as sodium silicate $(Na_2SiO_3)$ by treatment with $Na_2CO_3$.
Statement $II$ is correct because Baeyer's process is used for red bauxite,which is contaminated with $Fe_2O_3$.
Statement $III$ is correct because in Serpeck's process,$Al_2O_3$ is heated with $C$ and $N_2$ to form aluminum nitride $(AlN)$.
The reactions in Serpeck's process are:
$Al_2O_3 + 3C + N_2 \xrightarrow{1800^{\circ}C} 2AlN + 3CO$
$AlN + 3H_2O \rightarrow Al(OH)_3 + NH_3$
$2Al(OH)_3 \rightarrow Al_2O_3 + 3H_2O$

(D) The reaction of haloalkanes with aqueous $NaOH$ or moist $AgOH$ (which acts as $AgOH$ or $Ag_2O + H_2O$) leads to the nucleophilic substitution of the halogen atom with a hydroxyl group $(-OH)$ to form an alcohol.
In the given reaction,$C_2H_5Cl$ reacts with aqueous $NaOH$ to form $C_2H_5OH$ (ethanol).
Similarly,$C_2H_5Cl$ reacts with moist $AgOH$ to form $C_2H_5OH$ (ethanol).
Therefore,both $A$ and $B$ are $C_2H_5Cl$.

(B) Let's analyze each option:
$1$. Ammonia $(NH_3)$ is widely used as a refrigerant due to its high latent heat of vaporization. This statement is correct.
$2$. Nitrolim is a mixture of calcium cyanamide $(CaCN_2)$ and carbon $(C)$. The option states $Ca(CN)_2$ (calcium cyanide),which is incorrect. Thus,this statement is incorrect.
$3$. Superphosphate of lime is indeed a mixture of $Ca(H_2PO_4)_2$ and $CaSO_4 \cdot 2H_2O$. This statement is correct.
$4$. The hydrolysis of nitrogen trichloride $(NCl_3)$ proceeds as follows: $NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl$. This statement is correct.
Therefore,the incorrect statement is option $B$.

(C) The reactions are as follows:
$(A)$ $2F_2 + 4NaOH \longrightarrow 4NaF + O_2 + 2H_2O$ (Forms $O_2$)
$(B)$ $2F_2 + 2H_2O \longrightarrow 4HF + O_2$ (Forms $O_2$)
$(C)$ $Cl_2 + 2NaOH \text{ (cold, dil)} \longrightarrow NaCl + NaClO + H_2O$ (Does not form $O_2$)
$(D)$ $CaOCl_2 + H_2SO_4 \longrightarrow CaSO_4 + H_2O + Cl_2 + \frac{1}{2}O_2$ (Forms $O_2$)
Therefore,the pair that does not form oxygen is $Cl_2$ and cold,dilute $NaOH$.

(A) In Fischer-Ringe's method,air that is free from moisture and $CO_2$ is passed over a heated mixture $(800^{\circ}C)$ of $90 \% CaC_2 + 10 \% CaCl_2$ in an iron tube.
This process removes nitrogen and oxygen from the air mixture.
The chemical reactions involved are:
$CaC_2 + N_2 \xrightarrow{800^{\circ}C} CaCN_2 + C$
$2C + O_2 \rightarrow 2CO$
$2CaC_2 + 3CO_2 \rightarrow 2CaCO_3 + 5C$
$CuO + CO \rightarrow Cu + CO_2$
The remaining $CO_2$ is then absorbed by a $KOH$ solution,leaving behind the mixture of noble gases.

(B) Given: Vapour pressure of pure water $P^{\circ} = 19.8 \ mm$.
Moles of solute (glucose) $n_A = 0.1 \ mol$.
Moles of solvent (water) $n_B = \frac{178.2 \ g}{18 \ g/mol} = 9.9 \ mol$.
According to Raoult's law for non-volatile solutes: $\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_A}{n_A + n_B}$.
Substituting the values: $\frac{19.8 - P_s}{19.8} = \frac{0.1}{0.1 + 9.9} = \frac{0.1}{10} = 0.01$.
$19.8 - P_s = 19.8 \times 0.01 = 0.198$.
$P_s = 19.8 - 0.198 = 19.602 \ mm$.

(C) In a heterogeneous catalysis reaction,the catalyst and the reactants exist in different phases.
In the reaction $2 H_2 O_{2(l)} \xrightarrow{Pt_{(s)}} 2 H_2 O_{(l)} + O_{2(g)}$,the reactant $H_2 O_2$ is in the liquid phase $(l)$ while the catalyst $Pt$ is in the solid phase $(s)$.
Since they are in different phases,this is an example of heterogeneous catalysis.