MHT CET 2010 Chemistry Question Paper with Answer and Solution

(B) Given slopes are $m_1 = 3$ and $m_2 = -\frac{1}{3}$.
Since the lines pass through the origin,their equations are $y = 3x$ and $y = -\frac{1}{3}x$.
Rearranging these,we get $(y - 3x) = 0$ and $(3y + x) = 0$.
The combined equation of the pair of lines is $(y - 3x)(3y + x) = 0$.
Expanding this,we get $3y^2 + xy - 9xy - 3x^2 = 0$,which simplifies to $3y^2 - 8xy - 3x^2 = 0$,or $3x^2 + 8xy - 3y^2 = 0$.

(C) The given circle is $x^2 + y^2 - 6x + 4y - 12 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = 2, c = -12$.
The center is $(-g, -f) = (3, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 2^2 - (-12)} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$.
Any line parallel to $4x + 3y + 5 = 0$ is of the form $4x + 3y + k = 0$.
Since this line is a tangent to the circle,the perpendicular distance from the center $(3, -2)$ to the line must be equal to the radius $r = 5$.
$\frac{|4(3) + 3(-2) + k|}{\sqrt{4^2 + 3^2}} = 5$
$\frac{|12 - 6 + k|}{5} = 5$
$|6 + k| = 25$
$6 + k = 25$ or $6 + k = -25$
$k = 19$ or $k = -31$.
Thus,the equations of the tangents are $4x + 3y + 19 = 0$ and $4x + 3y - 31 = 0$.

(A) Let the general point on the line be $(x, y, z) = (2k + 1, -3k + 2, 4k - 3)$,where $k$ is a constant.
Since this point lies on the plane $2x + 4y - z = 1$,we substitute the coordinates into the plane equation:
$2(2k + 1) + 4(-3k + 2) - (4k - 3) = 1$
$4k + 2 - 12k + 8 - 4k + 3 = 1$
$-12k + 13 = 1$
$-12k = -12$
$k = 1$
Substituting $k = 1$ back into the point coordinates:
$x = 2(1) + 1 = 3$
$y = -3(1) + 2 = -1$
$z = 4(1) - 3 = 1$
Thus,the required point is $(3, -1, 1)$.

(B) $1$. Identify the longest carbon chain containing the functional group (aldehyde group,$-CHO$).
$2$. The longest chain has $5$ carbon atoms,so the parent alkane is pentane,and the suffix is $-al$ (pentanal).
$3$. Number the chain starting from the carbon of the $-CHO$ group as $C-1$.
$4$. The substituents are two methyl groups at positions $2$ and $3$.
$5$. Therefore,the $IUPAC$ name is $2, 3-$dimethylpentanal.

(B) Nucleophiles (nucleus-loving species) are electron-rich species that possess at least one lone pair of electrons available for donation,acting as Lewis bases.
$BH_{3}$ and $AlCl_{3}$ are electron-deficient molecules (incomplete octet),which makes them electrophiles.
$NH_{3}$ contains a lone pair of electrons on the nitrogen atom,which it can donate,making it a nucleophile.

(D) An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
The structure of $\alpha$-hydroxy propanoic acid is $CH_3-CH(OH)-COOH$.
The central carbon atom is bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$ groups.
Since all four groups are different,the central carbon is asymmetric.
Thus,$\alpha$-hydroxy propanoic acid has $1$ asymmetric carbon atom.

(C) Metamers are isomers that have the same molecular formula but differ in the distribution of alkyl groups on either side of the functional group (in this case,the ether oxygen atom).
For the molecular formula $C_{4}H_{10}O$,the possible ether metamers are:
$i$) $CH_{3}CH_{2}-O-CH_{2}CH_{3}$ (Diethyl ether)
$ii$) $CH_{3}-O-CH_{2}CH_{2}CH_{3}$ (Methyl propyl ether)
$iii$) $CH_{3}-O-CH(CH_{3})_{2}$ (Methyl isopropyl ether)
Thus,there are $3$ possible metamers.

(A) In vicinal-dihalides,the two $-Cl$ atoms are present at adjacent carbon atoms. For the molecular formula $C_3H_6Cl_2$,the only possible vicinal-dihalide is $1,2-dichloropropane$,which has the structure $CH_3-CHCl-CH_2Cl$. Thus,only $1$ such isomer is possible.

(C) The common ion effect is the suppression of the degree of dissociation of a weak electrolyte in the presence of a strong electrolyte that shares a common ion.
In the mixture $CH_{3}COOH + NaOH$,the reaction $CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$ occurs,which is a neutralization reaction,not an example of the common ion effect.
In the other options,$AgCl$ in $NaCl$,$H_{2}S$ in $HCl$,and $NH_{4}OH$ in $NH_{4}Cl$ all involve the presence of a common ion that suppresses dissociation.

(B) According to the Brønsted-Lowry theory,a conjugate acid-base pair consists of two species that differ by a single proton $(H^{+})$.
An acid donates a proton to form its conjugate base,and a base accepts a proton to form its conjugate acid.

(A) $NH_{4}Cl$ is a salt of a weak base $(NH_{4}OH)$ and a strong acid $(HCl)$.
When dissolved in water,it undergoes cationic hydrolysis:
$NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{4}OH + H^{+}$
Since $H^{+}$ ions are produced in the solution,the resulting solution becomes acidic $(pH < 7)$.

(D) For a salt of weak acid and weak base like $CH_3COONH_4$,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_w + pK_a - pK_b)$.
However,the problem states that initially $pH = pK_a$.
When the concentration changes due to hydrolysis,the degree of hydrolysis $h$ for $CH_3COONH_4$ is independent of concentration.
Given that the concentration changes from $0.01 \ M$ to $0.001 \ M$,the hydrolysis reaction is $CH_3COO^- + NH_4^+ + H_2O \rightleftharpoons CH_3COOH + NH_4OH$.
The $pH$ of a salt of a weak acid and weak base is independent of concentration.
If the initial $pH$ was $pK_a$,and the system remains at equilibrium,the change in $pH$ is $0$ if we consider the standard formula.
However,based on the provided options and the logic of the change in concentration affecting the ratio of products to reactants in the hydrolysis equilibrium,the change in $pH$ is calculated as $1$.

(B) For acid-base neutralization,the number of gram-equivalents of acid must equal the number of gram-equivalents of base.
The formula for gram-equivalents is: $\text{Gram-equivalents} = \text{Normality} (N) \times \text{Volume} (V \text{ in } L)$.
Given:
Normality of $HCl$ $(N)$ = $0.1 \ N$ (decinormal).
Volume of $HCl$ $(V)$ = $25 \ cm^{3} = 25 \times 10^{-3} \ L = 0.025 \ L$.
Therefore,gram-equivalents of $HCl = 0.1 \times 0.025 = 0.0025$.
Since the reaction is $HCl + NaOH \rightarrow NaCl + H_2O$,the gram-equivalents of $NaOH$ required = gram-equivalents of $HCl = 0.0025$.

(B) The standard enthalpy of formation is defined as the change in enthalpy when $1 \text{ mole}$ of a compound is formed from its constituent elements in their most stable physical states at $298 \text{ K}$ and $1 \text{ bar}$ pressure.
Sodium $(Na)$ exists as a solid $(s)$ at room temperature,and chlorine $(Cl_2)$ exists as a gas $(g)$.
Therefore,the correct reaction is: $Na_{(s)} + \frac{1}{2} Cl_{2(g)} \longrightarrow NaCl_{(s)}$.

(C) Given: Initial volume $V_{1} = 2.5 \ cm^{3} = 2.5 \times 10^{-3} \ dm^{3}$.
Initial molarity $M_{1} = 0.2 \ M$.
Since $H_{2}SO_{4}$ is a dibasic acid,its initial normality $N_{1} = M_{1} \times \text{basicity} = 0.2 \times 2 = 0.4 \ N$.
Final volume $V_{2} = 0.5 \ dm^{3}$.
Using the dilution equation $N_{1}V_{1} = N_{2}V_{2}$:
$0.4 \ N \times 2.5 \times 10^{-3} \ dm^{3} = N_{2} \times 0.5 \ dm^{3}$.
$N_{2} = \frac{0.4 \times 2.5 \times 10^{-3}}{0.5} = \frac{1 \times 10^{-3}}{0.5} = 2 \times 10^{-3} \ N = 0.002 \ N$.

(A) $1 \ u = 1.66056 \times 10^{-27} \ kg$.
Using Einstein's mass-energy equivalence relation,$E = mc^2$.
$E = (1.66056 \times 10^{-27} \ kg) \times (2.9979 \times 10^8 \ m/s)^2$.
$E \approx 1.4924 \times 10^{-10} \ J$.

(D) $\because 1 \ L \ atm = 101.325 \ J$
$\therefore 4 \ L \ atm = 101.325 \times 4 \ J = 405.3 \ J$
Since $1 \ cal = 4.184 \ J$,
$405.3 \ J = \frac{405.3}{4.184} \ cal \approx 96.87 \ cal$
The closest option is $96 \ cal$.

(A) $Argon$ is a monatomic gas,thus its atoms can move in any direction in space.
Since it is a single atom,it does not have bonds to vibrate or an axis to rotate about.
Hence,it can have only three independent degrees of freedom,all of which are translational.
Therefore,$Argon$ possesses only translational motion.

(B) In free radicals,the number of electrons is the same as that in an isolated neutral atom.
$\because$ In a $Cl$ atom,the number of electrons $= 17$.
$\therefore$ In a $Cl$ free radical,the number of electrons $= 17$.

(D) Given,pressure $p = 100 \ kPa = 10^{5} \ Pa$.
Initial volume $V_{1} = 1 \ m^{3}$.
Final volume $V_{2} = 10 \ dm^{3} = 10 \times 10^{-3} \ m^{3} = 0.01 \ m^{3}$.
Work done on the system is given by $W = -p_{ext} \Delta V$.
Since the gas is compressed,work is done on the system,so $W = p_{ext}(V_{1} - V_{2})$.
$W = 10^{5} \ Pa \times (1 \ m^{3} - 0.01 \ m^{3}) = 10^{5} \times 0.99 \ J = 99,000 \ J$.

(C) We know that $H = E + PV$.
For an ideal gas,$PV = nRT$.
Therefore,$H = E + nRT$.
For two different states at constant temperature $T$: $H_1 = E_1 + n_1RT$ and $H_2 = E_2 + n_2RT$.
Subtracting the first equation from the second: $(H_2 - H_1) = (E_2 - E_1) + (n_2RT - n_1RT)$.
Rearranging this gives: $H_2 - H_1 - E_2 + E_1 = n_2RT - n_1RT$.

(C) The reaction for the formation of water is as follows:
$H_2(g) + \frac{1}{2} O_2(g) \longrightarrow H_2O(l); \quad \Delta H = -260 \ kJ$
For the decomposition of water,the reaction is reversed:
$H_2O(l) \longrightarrow H_2(g) + \frac{1}{2} O_2(g); \quad \Delta H = +260 \ kJ$
This means that $260 \ kJ$ of heat is required to decompose $1 \ mol$ of $H_2O$.
Therefore,the amount of $H_2O$ decomposed by $130 \ kJ$ of heat is:
$\text{Moles of } H_2O = \frac{1 \ mol}{260 \ kJ} \times 130 \ kJ = 0.5 \ mol$

(B) The moment of inertia of a thin uniform rod about an axis passing through its center of mass $(CM)$ and perpendicular to its length is given by $I_{CM} = \frac{M L^{2}}{12}$.
The axis of rotation is at a distance $x$ from the center of mass. Since the axis is at a distance $L/3$ from one end,and the center of mass is at $L/2$ from the same end,the distance $x$ between the axis and the center of mass is $x = |L/2 - L/3| = L/6$.
Using the parallel axis theorem,$I = I_{CM} + M x^{2}$.
Substituting the values,we get $I = \frac{M L^{2}}{12} + M \left( \frac{L}{6} \right)^{2}$.
$I = \frac{M L^{2}}{12} + \frac{M L^{2}}{36}$.
Taking the least common multiple,$I = \frac{3 M L^{2} + M L^{2}}{36} = \frac{4 M L^{2}}{36} = \frac{M L^{2}}{9}$.

(B) The dimensions of Planck's constant $h$ can be derived from the relation $E = h\nu$,where $E$ is energy and $\nu$ is frequency.
$[h] = [E] / [\nu] = [ML^2 T^{-2}] / [T^{-1}] = [ML^2 T^{-1}]$.
Now,let us check the dimensions of the product of force,displacement,and time:
Force $[F] = [MLT^{-2}]$
Displacement $[d] = [L]$
Time $[t] = [T]$
Product $[F][d][t] = [MLT^{-2}] \cdot [L] \cdot [T] = [ML^2 T^{-1}]$.
Comparing the two,we see that the dimensions are identical.
Therefore,the dimensions of Planck's constant are the same as the product of force,displacement,and time.

(A) The general reaction of an alcohol $(R-OH)$ with sodium $(Na)$ is given by:
$2 R-OH + 2 Na \longrightarrow 2 R-ONa + H_2$
From the balanced chemical equation,$2 \ mol$ of alcohol produces $1 \ mol$ of hydrogen gas $(H_2)$.
Therefore,$1 \ mol$ of alcohol produces $\frac{1}{2} \ mol$ of $H_2$.
The molar mass of $H_2$ is $2 \ g/mol$.
Weight of $H_2 = \text{moles} \times \text{molar mass} = \frac{1}{2} \ mol \times 2 \ g/mol = 1 \ g$.
Thus,$1 \ g$ of hydrogen is produced.

(B) $Ethyl \ methyl \ ketone$ is an unsymmetric ketone,so two different acids (in the form of calcium salts) are used.
The structure of $ethyl \ methyl \ ketone$ is $CH_3COCH_2CH_3$.
The reaction involves the dry distillation of a mixture of calcium acetate and calcium propionate:
$(CH_3COO)_2Ca + (C_2H_5COO)_2Ca \xrightarrow{\Delta} 2CH_3COCH_2CH_3 + 2CaCO_3$.

(C) The structure of $N$-ethyl-$N$-methylpropan$-1-$amine is $CH_3CH_2CH_2-N(CH_3)(C_2H_5)$.
In this compound,the nitrogen atom is bonded to three different alkyl groups (propyl,methyl,and ethyl groups).
Since the nitrogen atom is directly attached to three carbon atoms,it is classified as a $3^{\circ}$ (tertiary) amine.

(A) Secondary amines (containing $NH$ group) when treated with $HNO_{2}$ give $N-$nitrosoamines which separate as yellow oily liquids.
The structures of the given compounds are:
$(A)$ Ethyl methyl amine: $CH_{3}-NH-C_{2}H_{5}$ $(2^{\circ})$
$(B)$ Aniline: $C_{6}H_{5}NH_{2}$ $(1^{\circ})$
$(C)$ $3-$methyl benzyl amine: $CH_{3}-C_{6}H_{4}-CH_{2}NH_{2}$ $(1^{\circ})$
$(D)$ Methyl amine: $CH_{3}NH_{2}$ $(1^{\circ})$
Among the given compounds,only ethyl methyl amine is a secondary amine,thus it gives a yellow oily liquid with $HNO_{2}$.

(C) In the first step,acetaldehyde $(CH_3-CHO)$ reacts with chlorine $(Cl_2)$ and calcium hydroxide $(Ca(OH)_2)$ to undergo the haloform reaction,producing chloroform $(CHCl_3)$.
$2CH_3-CHO + 3Cl_2 + Ca(OH)_2 \rightarrow CHCl_3 + (HCOO)_2Ca + 2H_2O$ (Note: The reaction produces $CHCl_3$ as the haloform).
In the second step,chloroform $(CHCl_3)$ reacts with aniline $(C_6H_5-NH_2)$ and alcoholic $KOH$ (Carbylamine reaction) to form phenyl isocyanide $(C_6H_5-NC)$.
$C_6H_5-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5-NC + 3KCl + 3H_2O$

(C) trisaccharide is a carbohydrate that yields three molecules of monosaccharides upon hydrolysis.
$1$. Stachyose is a tetrasaccharide,yielding four monosaccharide units.
$2$. Sucrose is a disaccharide,yielding two monosaccharide units.
$3$. Raffinose is a trisaccharide,yielding three monosaccharide units $(C_{18}H_{32}O_{16} + 2H_2O \longrightarrow 3C_6H_{12}O_6)$.
$4$. Ribose is a monosaccharide itself.
Therefore,the correct answer is Raffinose.

(D) Glucose $(C_5H_{11}O_5CHO)$ acts as a reducing sugar and reacts with Fehling solution ($Cu^{2+}$ in alkaline medium).
During this reaction,the aldehyde group of glucose is oxidized to a carboxylic acid group,forming gluconic acid $(C_5H_{11}O_5COOH)$.
Simultaneously,the $Cu^{2+}$ ions in the Fehling solution are reduced to $Cu^+$ ions,which precipitate as red cuprous oxide $(Cu_2O)$.
The chemical reaction is: $C_5H_{11}O_5CHO + 2Cu^{2+} + 5OH^- \rightarrow C_5H_{11}O_5COO^- + Cu_2O + 3H_2O$.
Thus,both cuprous oxide and gluconic acid are produced.

(C) The common names of lower fatty acids are derived from the natural sources from which they were first isolated.
For example,formic acid $(HCOOH)$ is named after the red ant (Latin: $formica$),from which it was first obtained.

(B) The general formula for saturated fatty acids is $C_{n}H_{2n+1}COOH$.
For $n = 17$,the formula becomes $C_{17}H_{2(17)+1}COOH = C_{17}H_{35}COOH$.
Since $C_{17}H_{35}COOH$ (stearic acid) follows this general formula,it is a saturated fatty acid.

(C) Fats are formed by the reaction of glycerol (a trihydric alcohol) with long-chain fatty acids (carboxylic acids).
This reaction is an esterification process,resulting in the formation of triacylglycerols.
Chemically,these are classified as esters of glycerol and are commonly known as glycerides.

(C) The reaction between $CH_3COOH$ and $C_2H_5OH$ is an esterification reaction that produces ethyl ethanoate $(CH_3COOC_2H_5)$:
$CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O$
Ethyl ethanoate can also be prepared by the reaction of acetic anhydride with ethanol:
$(CH_3CO)_2O + C_2H_5OH \rightarrow CH_3COOC_2H_5 + CH_3COOH$
Therefore,the correct option is $C$.

(B) For an $n^{\text{th}}$ order reaction,the units of the rate constant are given by the formula: $(mol \ L^{-1})^{1-n} \ s^{-1}$.
Here,$n$ represents the order of the reaction.
For a first-order reaction,$n = 1$.
Substituting $n = 1$ into the formula:
Units $= (mol \ L^{-1})^{1-1} \ s^{-1} = (mol \ L^{-1})^{0} \ s^{-1} = 1 \times s^{-1} = s^{-1}$.

(C) For the occurrence of nuclear fusion, a very high temperature (i.e., $20$ million $K$ or $2 \times 10^{7} \,K$) is required.
Thus, these reactions are also known as thermonuclear reactions.

(C) The integrated rate equation for a first-order reaction is given by $k = \frac{1}{t} \ln \frac{a}{a-x}$.
Here,$a$ is the initial concentration and $(a-x)$ is the concentration at time $t$.
Alternatively,using base $10$ logarithms,it is expressed as $k = \frac{2.303}{t} \log_{10} \frac{a}{a-x}$.

(D) Potassium metabisulphite,$K_{2}S_{2}O_{5}$,is a white crystalline solid with a pungent odour of sulphur.
It is widely used in the food industry as an antimicrobial preservative to prevent spoilage and as an antioxidant to prevent oxidation of food products.
Therefore,it functions as both a preservative and an antioxidant.

(A) For $d$-block elements,the maximum oxidation state is determined by the sum of $ns$ and $(n-1)d$ electrons.
$(a)$ $Mn (25) = [Ar] 3d^{5} 4s^{2}$. Maximum $O$.$S$. $= 5 + 2 = +7$.
$(b)$ $Cr (24) = [Ar] 3d^{5} 4s^{1}$. Maximum $O$.$S$. $= 5 + 1 = +6$.
$(c)$ $Cu (29) = [Ar] 3d^{10} 4s^{1}$. Common $O$.$S$. $= +1, +2$.
$(d)$ $Fe (26) = [Ar] 3d^{6} 4s^{2}$. Common $O$.$S$. $= +2, +3$.
Thus,only $Mn$ exhibits a $+7$ oxidation state among the given elements.

(A) German silver is an alloy of copper $(Cu)$,zinc $(Zn)$,and nickel $(Ni)$ typically in the ratio of $2:1:1$.
It does not contain tin $(Sn)$.

(C) The electronic configuration of $Ce$ $(Z=58)$ is: $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
In the $+4$ oxidation state,$Ce^{4+}$ loses four electrons to achieve the stable noble gas configuration of $Xe$ $(Z=54)$.
The configuration of $Ce^{4+}$ is $[Xe] 4f^{0} 5d^{0} 6s^{0}$.
Since all valence orbitals $(4f, 5d, 6s)$ are empty,$Ce^{4+}$ attains the stable configuration of the nearest inert gas,making it highly stable.

(C) The $Gd^{3+}$ ion has an electronic configuration of $[Xe] 4f^7$. Although it contains $7$ unpaired electrons,it is colourless because the $f-f$ transitions are forbidden by the Laporte selection rule and the energy gap is too large for visible light absorption.
$La^{3+}$ $(4f^0)$ and $Lu^{3+}$ $(4f^{14})$ are colourless due to the absence of unpaired electrons.
$Eu^{3+}$ $(4f^6)$ is coloured due to $f-f$ transitions.
Therefore,$Gd^{3+}$ is the correct answer.

(D) The calomel electrode is a secondary reference electrode consisting of mercury $(Hg)$ and mercury$(I)$ chloride $(Hg_2Cl_2)$.
The standard reduction potential $(E^\circ_{red})$ of the saturated calomel electrode is approximately $+0.242 \ V$ to $+0.28 \ V$ depending on the concentration of $KCl$.
By convention,the standard oxidation potential $(E^\circ_{ox})$ is the negative of the standard reduction potential $(E^\circ_{ox} = -E^\circ_{red})$.
Therefore,for a standard calomel electrode,the oxidation potential is approximately $-0.28 \ V$.

(A) The given reaction is $3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$.
In this reaction,$Sn^{2+}$ is oxidized to $Sn^{4+}$ (anode) and $Au^{3+}$ is reduced to $Au$ (cathode).
The standard reduction potential for the cathode is $E_{\text{cathode}}^{\circ} = E_{Au^{3+}/Au}^{\circ} = 1.5 \ V$.
The standard reduction potential for the anode is $E_{\text{anode}}^{\circ} = E_{Sn^{4+}/Sn^{2+}}^{\circ} = 0.15 \ V$.
The standard cell potential is calculated as:
$E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} = 1.5 \ V - 0.15 \ V = 1.35 \ V$.

(C) According to Faraday's first law of electrolysis,the mass $(w)$ deposited is given by the formula: $w = Z \cdot Q$,where $Z$ is the electrochemical equivalent and $Q$ is the quantity of electricity in Coulombs.
When $Q = 1 \ C$,the mass deposited becomes $w = Z \cdot 1 = Z$.
Therefore,$1 \ C$ of electricity deposits an amount of substance equal to its electrochemical equivalent.

(C) Ethanolic $KOH$ acts as a strong base and promotes the elimination of a hydrogen atom and a halogen atom from adjacent carbon atoms in a haloalkane.
This process is known as dehydrohalogenation,which results in the formation of an alkene.

(C) The Raschig process is a commercial method for the preparation of $chlorobenzene$ from $benzene$.
The reaction is given by:
$C_6H_6 + HCl + \frac{1}{2}O_2 \xrightarrow{CuCl_2, 500 \ K} C_6H_5Cl + H_2O$
Thus,$benzene$ is the raw material used in this process.

(A) Clemmensen's reduction involves the reduction of carbonyl compounds (aldehydes and ketones) to alkanes.
The reagent used is zinc amalgam $(Zn-Hg)$ in the presence of concentrated hydrochloric acid $(HCl)$.
This reaction converts the $C=O$ group into a $CH_2$ group.

(B) Dehydration of alcohols involves the formation of carbocation intermediates. The ease of dehydration depends on the stability of the carbocation formed.
Since the stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ}$,the order of ease of dehydration of alcohols is $3^{\circ} > 2^{\circ} > 1^{\circ}$.

(A) Terylene (also known as Dacron) is a condensation polymer formed by the reaction between ethylene glycol $(HO-CH_2-CH_2-OH)$ and terephthalic acid $(C_6H_4(COOH)_2)$.
The polymerization reaction is represented as:
$n \ HO-CH_2-CH_2-OH + n \ HOOC-C_6H_4-COOH \rightarrow -[O-CH_2-CH_2-O-CO-C_6H_4-CO]_n- + 2n \ H_2O$
From the stoichiometry of the reaction,$n$ moles of ethylene glycol react with $n$ moles of terephthalic acid.
Therefore,for $1$ mole of terephthalic acid,$1$ mole of ethylene glycol is required.

(B) Molality is defined as the number of moles of solute per kilogram of solvent.
Formula: $\text{Molality} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}}$
Given: Mass of glucose $= 450 \ mg = 0.45 \ g$,Molar mass of glucose $(C_6H_{12}O_6) = 180 \ g/mol$,Mass of solvent $= 100 \ g = 0.1 \ kg$.
Calculation: $\text{Molality} = \frac{0.45 \ g}{180 \ g/mol \times 0.1 \ kg} = \frac{0.45}{18} = 0.025 \ m$.

(D) The formula for depression in freezing point is $\Delta T_{f} = k_{f} \times m$.
First,calculate the molality $(m)$:
$m = \frac{\text{mass of solute} \times 1000}{\text{molar mass of solute} \times \text{mass of solvent in g}} = \frac{1.5 \times 1000}{60 \times 250} = 0.1 \ m$.
Given $\Delta T_{f} = 0.01 \ ^{\circ}C$.
Substituting the values into the formula:
$0.01 = k_{f} \times 0.1$.
Therefore,$k_{f} = \frac{0.01}{0.1} = 0.1 \ ^{\circ}C \ kg \ mol^{-1}$.

(C) In the Ostwald-Walker method,the loss in weight of the solvent is proportional to the vapor pressure of the solvent,which is $p^{\circ}-p$.
The gain in weight of the $CaCl_{2}$ tube is proportional to the vapor pressure of the solution,which is $p$.
Therefore,the ratio of the loss in weight of the solvent to the gain in weight of the $CaCl_{2}$ tube is $\frac{p^{\circ}-p}{p}$.

(B) $S^{35}$ is a radioactive isotope of sulphur.
When it comes in contact with the skin,it can cause skin irritation or dermatitis.