MHT CET 2019 Chemistry Question Paper with Answer and Solution

(A) For a telescope adjusted for parallel rays (normal adjustment),the magnifying power is given by $m = \frac{f_o}{f_e} = 9$,which implies $f_o = 9f_e$ .........$(i)$
The length of the telescope tube is the sum of the focal lengths of the objective and the eye-piece: $L = f_o + f_e = 20\, cm$ .............$(ii)$
Substituting equation $(i)$ into equation $(ii)$:
$9f_e + f_e = 20\, cm$
$10f_e = 20\, cm$
$f_e = 2\, cm$
Now,substituting $f_e$ back into equation $(i)$:
$f_o = 9 \times 2\, cm = 18\, cm$
Therefore,the focal lengths are $18\, cm$ and $2\, cm$.

(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.
For the Balmer series,$n_1 = 2$. The first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \Rightarrow \lambda = \frac{36}{5R}$.
The second line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$.
$\frac{1}{\lambda'} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \Rightarrow \lambda' = \frac{16}{3R}$.
Dividing $\lambda'$ by $\lambda$:
$\frac{\lambda'}{\lambda} = \frac{16}{3R} \times \frac{5R}{36} = \frac{16 \times 5}{3 \times 36} = \frac{80}{108} = \frac{20}{27}$.
Therefore,$\lambda' = \frac{20}{27}\lambda$.

(A) The magnifying power $m$ of a telescope adjusted for parallel rays (normal adjustment) is given by $m = \frac{f_o}{f_e} = 9$,where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.
From this,we get $f_o = 9f_e$.
The length of the telescope tube in normal adjustment is $L = f_o + f_e = 20 \ cm$.
Substituting $f_o = 9f_e$ into the length equation: $9f_e + f_e = 20 \ cm$.
$10f_e = 20 \ cm \Rightarrow f_e = 2 \ cm$.
Now,calculating $f_o$: $f_o = 9 \times 2 \ cm = 18 \ cm$.
Thus,the focal lengths are $18 \ cm$ and $2 \ cm$.

(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$.
The first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$.
Thus,$\lambda = \frac{36}{5R}$.
The second line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$:
$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = R \left( \frac{3}{16} \right)$.
Thus,$\lambda_2 = \frac{16}{3R}$.
Now,express $\lambda_2$ in terms of $\lambda$:
$\lambda_2 = \frac{16}{3R} = \frac{16}{3} \times \left( \frac{5 \lambda}{36} \right) = \frac{16 \times 5 \lambda}{3 \times 36} = \frac{80 \lambda}{108} = \frac{20 \lambda}{27}$.

(A) The reaction proceeds via the electrophilic addition of $H_2SO_4$ to propene according to Markovnikov's rule.
$CH_3-CH=CH_2 + HOSO_3H \rightarrow CH_3-CH(OSO_3H)-CH_3$ (Isopropyl hydrogen sulphate).
On heating with water (hydrolysis),the isopropyl hydrogen sulphate yields propan$-2-$ol.
$CH_3-CH(OSO_3H)-CH_3 + H_2O \xrightarrow{\Delta} CH_3-CH(OH)-CH_3 + H_2SO_4$.

(A) Alkali metals react with liquid ammonia to form amides $(MNH_2)$ and hydrogen gas,except for lithium.
Lithium reacts with ammonia to form a complex compound known as tetraamminelithium,$[Li(NH_3)_4]$.
The chemical equation for this reaction is:
$Li + 4NH_3 \rightarrow [Li(NH_3)_4]$
Therefore,$Li$ does not form an amide under these conditions.

(B) The central atom in $BrF_5$ is Bromine $(Br)$,which has $7$ valence electrons.
It forms $5$ bonds with $5$ Fluorine $(F)$ atoms and has $1$ lone pair of electrons.
The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization.
According to $VSEPR$ theory,a molecule with $5$ bond pairs and $1$ lone pair has a square pyramidal geometry.

(D) The chemical formula of $2-$formylbenzoic acid is $C_8H_6O_3$.
Its structure consists of a benzene ring substituted with a carboxylic acid group $(-COOH)$ and an aldehyde group $(-CHO)$ at the ortho position.
Counting the bonds:
- There are $12$ $\sigma$ bonds in the benzene ring and its substituents (including $C-H$,$C-C$,$C=O$,$C-O$,and $O-H$ bonds).
- Specifically,the structure contains $17$ $\sigma$ bonds and $5$ $\pi$ bonds (three from the benzene ring,one from the $C=O$ of the aldehyde,and one from the $C=O$ of the carboxylic acid).

(B) In $H_2S$,the $H-S-H$ bond angle is found to be $92.1^{\circ}$.
This is significantly smaller than the $104.5^{\circ}$ bond angle observed in $H_2O$.
As we move down the group in the hydrides of the oxygen family $(H_2O, H_2S, H_2Se, H_2Te)$,the electronegativity of the central atom decreases.
This leads to less hybridization and the bond angles approach $90^{\circ}$ as the bonds are formed primarily by pure $p$-orbitals.

(C) Bond length is approximately the sum of the covalent radii of the bonded atoms.
Using the approximate covalent radii: $r_H \approx 0.37 \ \mathring{A}$,$r_C \approx 0.77 \ \mathring{A}$,$r_N \approx 0.75 \ \mathring{A}$,$r_O \approx 0.73 \ \mathring{A}$.
Calculating the bond lengths:
$C-C \approx 0.77 + 0.77 = 1.54 \ \mathring{A}$
$C-N \approx 0.77 + 0.75 = 1.52 \ \mathring{A}$
$C-O \approx 0.77 + 0.73 = 1.50 \ \mathring{A}$
$C-H \approx 0.77 + 0.37 = 1.14 \ \mathring{A}$
Comparing these values,the $C-C$ bond has the maximum bond length.

(C) The chemical formula for the manganite ion is $MnO_4^{2-}$ and for the permanganate ion is $MnO_4^-$.
Thus,the ionic charge of the manganite ion is $-2$ and the ionic charge of the permanganate ion is $-1$.

(B) The structure of benzoic acid $(C_6H_5COOH)$ consists of a benzene ring attached to a carboxylic acid group.
In the benzene ring,there are $3$ $\pi-$bonds due to the alternating double bonds.
In the carboxylic acid group $(-COOH)$,there is one $C=O$ double bond,which contains $1$ $\pi-$bond.
Therefore,the total number of $\pi-$bonds in a benzoic acid molecule is $3 + 1 = 4$.

(C) The $IUPAC$ name $1-$iodo$-2,3-$dimethylpentane indicates a pentane chain ($5$ carbons) with an iodine atom at position $1$,and methyl groups at positions $2$ and $3$.
In the bond line notation,the carbon chain is represented by a zigzag line.
Starting from the end with the iodine atom (position $1$),the next carbon (position $2$) has a methyl group attached,and the third carbon (position $3$) also has a methyl group attached.
Analyzing the given options:
Option $A$ shows iodine at position $2$.
Option $B$ shows iodine at position $2$.
Option $C$ shows the iodine at the end of the chain (position $1$),with methyl groups at positions $2$ and $3$.
Option $D$ shows iodine at position $3$.
Therefore,the correct structure is represented by option $C$.

(D) nucleophile is a chemical species that donates an electron pair to form a chemical bond.
Species with at least one lone pair of electrons or a $\pi-$bond can act as nucleophiles.
In option $D$,$CN^{-}$ has a lone pair on carbon,$H_{2}O$ has two lone pairs on oxygen,and $ROH$ has two lone pairs on oxygen.
Therefore,$CN^{-}, H_{2}O, ROH$ are all nucleophiles.

(B) There are $3$ possible isomers for an alkane with the molecular formula $C_5H_{12}$. These are:
$1$. $n$-pentane: $CH_3CH_2CH_2CH_2CH_3$
$2$. Isopentane ($2$-methylbutane): $CH_3-CH(CH_3)-CH_2-CH_3$
$3$. Neopentane ($2,2$-dimethylpropane): $C(CH_3)_4$

(C) In a molecule of ethyl magnesium bromide $(CH_3CH_2MgBr)$,the $Mg - Br$ bond is ionic in nature.
This is because the electronegativity difference between magnesium $(Mg)$ and bromine $(Br)$ is significantly large compared to the other bonds present in the molecule.
Consequently,the $Mg - Br$ bond exhibits the highest ionic character.

(C) The reaction of $HCl$ with an unsymmetrical alkene like propene follows Markovnikov's rule,even in the presence of peroxide.
Peroxide effect (Kharasch effect) is only applicable to $HBr$ and not to $HCl$ or $HI$.
According to Markovnikov's rule,the negative part of the reagent $(Cl^-)$ attaches to the carbon atom with fewer hydrogen atoms.
Thus,$CH_3-CH=CH_2 + HCl \rightarrow CH_3-CHCl-CH_3$ ($2-$chloropropane).

(C) The reaction of sodium butanoate with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
In this process,the carboxyl group $(-COO^-Na^+)$ is removed as $Na_2CO_3$,and the resulting alkane contains one carbon atom less than the parent carboxylic acid salt.
Sodium butanoate $(CH_3CH_2CH_2COONa)$ has $4$ carbon atoms.
Upon decarboxylation,it yields propane $(CH_3CH_2CH_3)$,which has $3$ carbon atoms.
Reaction: $CH_3CH_2CH_2COONa + NaOH \xrightarrow[\Delta]{CaO} CH_3CH_2CH_3 + Na_2CO_3$

(D) The Wurtz reaction between methyl chloride and sodium metal in the presence of dry ether is represented by the equation:
$2 CH_3Cl + 2 Na \xrightarrow{\text{Dry ether}} CH_3-CH_3 + 2 NaCl$
From the stoichiometry of the reaction,$2$ moles of sodium $(Na)$ are required to produce $1$ mole of ethane $(CH_3-CH_3)$.
The molar mass of sodium is $23 \ g/mol$.
Therefore,the mass of $2$ moles of sodium is $2 \times 23 \ g = 46 \ g$.
Thus,$46 \ g$ of sodium is required to prepare one mole of ethane.

(D) The given reaction is the Wurtz reaction.
$2n \ R-X + 2n \ Na \xrightarrow{\text{Dry ether}} n \ R-R + 2n \ NaX$
Here,$R-R$ represents an alkane.
Thus,the product obtained is $n \ \text{alkane}$.

(C) The hydroboration-oxidation of $1-$butylene $(CH_2=CH-CH_2-CH_3)$ follows anti-Markovnikov addition of water.
$CH_2=CH-CH_2-CH_3 + (i) B_2H_6, THF / (ii) H_2O_2, OH^- \rightarrow CH_2(OH)-CH_2-CH_2-CH_3$.
The product formed is $n-$butyl alcohol (butan-$1-$ol).

(D) The structure of $H_2O_2$ in the gaseous phase is non-planar (open book structure).
In this structure,the $H-O-O$ bond angle is $94.8^{\circ}$,and the dihedral angle is $111.5^{\circ}$.
Therefore,the correct bond angle $H-O-O$ is $94.8^{\circ}$.

(C) The density of water vapour depends on temperature and pressure.
At the boiling point of water $(100^{\circ}C)$ and standard atmospheric pressure $(1 \ atm)$,the density of water vapour is approximately $6 \times 10^{-4} \ g \ cm^{-3}$.

(D) According to Fajan's rules,the covalent character of an ionic bond increases with an increase in the charge on the cation.
$SbCl_5$ contains the $Sb^{5+}$ cation,which has the highest positive charge among the given options ($Sn^{2+}$,$Pb^{2+}$,$Sb^{3+}$,$Sb^{5+}$).
$A$ higher charge on the cation results in greater polarising power,which distorts the electron cloud of the anion more effectively,leading to increased covalent character.
Therefore,$SbCl_5$ is the most covalent compound among the choices.

(D) The oxidation number of nitrogen in $HNO_3$ is calculated as follows:
Let the oxidation number of $N$ be $x$.
For $HNO_3$: $1 + x + 3(-2) = 0$
$1 + x - 6 = 0$
$x - 5 = 0$
$x = +5$
Therefore,the oxidation number of nitrogen is $ +5 $ in $HNO_3$.

(A) The chemical formula for nitric oxide is $NO$.
Let the oxidation state of nitrogen be $x$.
The oxidation state of oxygen in oxides is generally $-2$.
For the neutral molecule $NO$,the sum of oxidation states must be zero:
$x + (-2) = 0$
$x = +2$.
Therefore,the oxidation state of nitrogen in nitric oxide is $+2$.

(D) The structure of the ozone $(O_3)$ molecule consists of a central oxygen atom double-bonded to one terminal oxygen atom and single-bonded to another terminal oxygen atom.
The central oxygen atom has $6$ valence electrons.
In the resonance structure,the central oxygen atom forms $3$ bonds (total $6$ bonding electrons) and has $1$ lone pair ($2$ electrons).
Using the formula: $\text{Formal charge} = V - L - \frac{1}{2}B$,where $V$ is the number of valence electrons,$L$ is the number of lone pair electrons,and $B$ is the number of bonding electrons.
For the central oxygen atom: $V = 6$,$L = 2$,$B = 6$.
$\text{Formal charge} = 6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1$.

(B) Let the oxidation state of sulphur be $x$. The oxidation state of hydrogen is $+1$ and oxygen is $-2$. In $H_2S_2O_7$,the sum of oxidation states is $2(+1) + 2(x) + 7(-2) = 0$. Solving for $x$: $2 + 2x - 14 = 0$,$2x = 12$,$x = +6$. Thus,the oxidation state of sulphur in $H_2S_2O_7$ is $+6$.

(B) The oxidation number of an element in its elemental form or in a homoatomic molecule is always $0$.
Since $S_8$ is a homoatomic molecule consisting only of sulphur atoms,the oxidation number of sulphur in $S_8$ is $0$.

(B) $SO_2$ can act both as an oxidising agent and a reducing agent because the oxidation state of sulphur in $SO_2$ is $+4$,which is an intermediate oxidation state (between $-2$ and $+6$).
As a reducing agent: $SO_2$ reduces $Fe^{3+}$ to $Fe^{2+}$ and decolourises acidified $KMnO_4$ solution.
$2Fe^{3+} + SO_2 + 2H_2O \rightarrow 2Fe^{2+} + SO_4^{2-} + 4H^+$
As an oxidising agent: $SO_2$ reacts with $H_2S$ to produce sulphur and water,where the oxidation state of sulphur in $SO_2$ decreases from $+4$ to $0$.
$SO_2 + 2H_2S \rightarrow 3S + 2H_2O$

(D) The given reaction is $MnO_{4(aq)}^{-} + Br_{(aq)}^{-} \rightarrow MnO_{2(s)} + BrO_{3(aq)}^{-}$.
In $MnO_{4}^{-}$,the oxidation state of $Mn$ is $x + 4(-2) = -1$,so $x = +7$.
In $MnO_{2}$,the oxidation state of $Mn$ is $x + 2(-2) = 0$,so $x = +4$.
In $Br^{-}$,the oxidation state of $Br$ is $-1$.
In $BrO_{3}^{-}$,the oxidation state of $Br$ is $x + 3(-2) = -1$,so $x = +5$.
Therefore,the oxidation number of bromine changes from $Br^{-1}$ to $Br^{+5}$.

(C) The Solvay process is used for the industrial production of sodium carbonate $(Na_2CO_3)$.
The raw materials required are sodium chloride $(NaCl)$,ammonia $(NH_3)$,and limestone $(CaCO_3)$.
The chemical reactions involved are:
$2 NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3$
$(NH_4)_2CO_3 + H_2O + CO_2 \rightarrow 2 NH_4HCO_3$
$NH_4HCO_3 + NaCl \rightarrow NaHCO_3 + NH_4Cl$
$2 NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2$
Thus,the correct set of raw materials is $NaCl, NH_3, CaCO_3$.

(C) The diagonal relationship between $Be$ and $Al$ is due to their similar ionic size and charge/radius ratio (ionic potential).
Diagonal relationships occur because the increase in ionic charge/radius ratio when moving across a period is compensated by the decrease in ionic charge/radius ratio when moving down a group.
This results in similar polarizing power,leading to similar chemical properties.

(D) When an ionic hydride of $s$-block elements (e.g.,$NaH$) is electrolyzed in its molten state,the following reactions occur:
At the cathode: $Na^+ + e^- \rightarrow Na$
At the anode: $2H^- \rightarrow H_2 + 2e^-$
Thus,dihydrogen gas $(H_2)$ is liberated at the anode.

(C) The Nicol prism is an optical device made from a crystal of $CaCO_3$ (calcite).
It is used to produce and analyze plane-polarized light due to its property of double refraction.

(D) homogeneous mixture is a mixture in which the composition is uniform throughout the mixture.
$Ethyl \ alcohol$ and $water$ are polar in nature and can form hydrogen bonds with each other,making them miscible in all proportions.
Therefore,the $(Ethyl \ alcohol + Water)$ system forms a homogeneous mixture.

(D) The combining ratios of hydrogen and oxygen in water $(H_2O)$ and hydrogen peroxide $(H_2O_2)$ are $1: 8$ and $1: 16$ respectively.
This is an example of the law of multiple proportions.
According to this law,if two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
Here,for a fixed mass of hydrogen $(1 \ g)$,the masses of oxygen are $8 \ g$ and $16 \ g$,which are in the ratio $8:16$ or $1:2$.

(B) The chemical formula for aluminium phosphate is $AlPO_4$.
Atomicity of a molecule is the total number of atoms present in one molecule.
In $AlPO_4$,there is $1$ atom of $Al$,$1$ atom of $P$,and $4$ atoms of $O$.
$\therefore$ Atomicity $= 1 + 1 + 4 = 6$.

(A) The molar volume of any ideal gas at standard temperature and pressure $(STP)$ is $22.414 \ L$.
Since $1 \ m^3 = 1000 \ L$,we have $1 \ L = 10^{-3} \ m^3$.
Therefore,$22.414 \ L = 22.414 \times 10^{-3} \ m^3 = 0.022414 \ m^3$.

(A) The chemical formula of urea is $NH_2CONH_2$.
The molar mass of urea is $(2 \times 14) + (4 \times 1) + 12 + 16 = 60 \ g/mol$.
The number of carbon atoms in one molecule of urea is $1$.
The percentage of carbon by mass is calculated as: $\frac{\text{Mass of Carbon}}{\text{Molar mass of Urea}} \times 100 = \frac{12}{60} \times 100 = 20 \%$.

(D) The number of moles of $CO_2$ dissolved is calculated as:
$n = M \times V(L)$
$n = 0.1 \times \frac{200}{1000} = 0.02 \ mol$
At $S.T.P$,$1 \ mol$ of an ideal gas occupies $22.4 \ L$.
Therefore,the volume of $0.02 \ mol$ of $CO_2$ at $S.T.P$ is:
$V = n \times 22.4 \ L/mol$
$V = 0.02 \times 22.4 \ L = 0.448 \ L$

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
For an isobaric process (constant pressure),the heat exchanged is $q_p = \Delta H = \Delta U + P\Delta V$.
Since work done $w = -P\Delta V$,we have $P\Delta V = -w$.
Substituting this,$q_p = \Delta U - w$,which means $\Delta U = q_p + w$.
Thus,option $A$ is correct.
For an adiabatic process,$q = 0$,so $\Delta U = w$.
For an isochoric process,$\Delta V = 0$,so $w = 0$ and $\Delta U = q_v$.
For an isothermal process of an ideal gas,$\Delta U = 0$,so $q = -w$.

(C) According to the first law of thermodynamics,$\Delta U = q + W$.
Since the gas performs work on the surrounding,the work done is negative: $W = -0.320 \ kJ = -320 \ J$.
The heat absorbed by the system is positive: $q = 120 \ J$.
Substituting these values into the equation: $\Delta U = 120 \ J + (-320 \ J) = -200 \ J$.

(A) The temperature in $^{\circ} F$ and $^{\circ} C$ are related to each other by the following relationship:
$^{\circ} F = \frac{9}{5}(^{\circ} C) + 32$
Substituting the value of $32^{\circ} C$ into the formula:
$^{\circ} F = \frac{9 \times 32}{5} + 32 = 57.6 + 32 = 89.6^{\circ} F$
Therefore,the temperature of $32^{\circ} C$ is equivalent to $89.6^{\circ} F$.

(C) The statement “The mass and energy both are conserved in an isolated system” refers to the modified first law of thermodynamics.
According to the original first law of thermodynamics,energy can neither be created nor destroyed,only transformed.
However,considering the mass-energy equivalence principle $(E = mc^2)$,the law is modified to state that the total mass and energy of an isolated system remain constant.

(A) The work done in an irreversible isothermal expansion against a constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Given: $P_{ext} = 1.9 \ atm$,$V_1 = 300 \ cm^3 = 0.3 \ L$,$V_2 = 2.5 \ L$.
Change in volume: $\Delta V = V_2 - V_1 = 2.5 \ L - 0.3 \ L = 2.2 \ L$.
Work done: $W = -1.9 \ atm \times 2.2 \ L = -4.18 \ atm \cdot L$.
Converting to Joules using the conversion factor $1 \ atm \cdot L = 101.325 \ J$:
$W = -4.18 \times 101.325 \ J = -423.56 \ J$.

(B) To find the enthalpy of formation of $CO_{(g)}$,we need the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$.
Given equations:

$(i) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$	$\Delta H = -X$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$	$\Delta H = -Y$

Subtracting equation $(ii)$ from equation $(i)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \rightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2} O_{2(g)} - CO_{(g)} \rightarrow 0$
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
The enthalpy change is $\Delta_f H = (-X) - (-Y) = Y - X$.

(D) For an ideal gas,the enthalpy $H$ is a function of temperature only,i.e.,$H = f(T)$.
Since the process is isothermal,the temperature remains constant,so $\Delta T = 0$.
The change in enthalpy is given by $\Delta H = n C_p \Delta T$.
Since $\Delta T = 0$,the enthalpy change $\Delta H$ is $0 \ kJ$.

(B) The relationship between the change in enthalpy $(\Delta H)$ and the change in internal energy $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For $\Delta H$ to be equal to $\Delta U$,the term $\Delta n_g$ must be equal to $0$.
$\Delta n_g$ is the difference between the sum of the stoichiometric coefficients of gaseous products and gaseous reactants.
For option $(B)$: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta n_g = 1 - 1 = 0$.
Therefore,$\Delta H = \Delta U$ for this reaction.

(C) The combustion reaction for carbon monoxide is: $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$
The change in the number of gaseous moles is calculated as: $\Delta n_{g} = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = 1 - 1.5 = -0.5$ or $-\frac{1}{2}$.
The relationship between enthalpy change (heat at constant pressure,$\Delta H$) and internal energy change (heat at constant volume,$\Delta E$) is given by: $\Delta H - \Delta E = \Delta n_{g} RT$.
Given $T = 27^{\circ} C = 27 + 273 = 300 \ K$ and $R = 2 \ cal \ K^{-1} \ mol^{-1}$.
Substituting the values: $\Delta H - \Delta E = -\frac{1}{2} \times 2 \times 300 = -300 \ cal$.

(A) The reaction shown is the reduction of $p$-toluenediazonium chloride to toluene. This transformation is commonly achieved using ethanol $(C_2H_5OH)$ or hypophosphorous acid $(H_3PO_2)$ in the presence of water. Ethanol acts as a reducing agent and gets oxidized to acetaldehyde,while the diazonium group is replaced by a hydrogen atom.

(D) Hydroboration-oxidation of $Isobutylene$ $(CH_2=C(CH_3)_2)$ involves the addition of $H_2O$ across the double bond in an anti-$Markovnikov$ fashion.
In this reaction,the $OH$ group attaches to the less substituted carbon atom.
The reaction proceeds as follows:
$CH_2=C(CH_3)_2 + (H-BH_2)_2 \xrightarrow{H_2O_2, OH^-} (CH_3)_2CH-CH_2OH$
The product formed is isobutyl alcohol ($2-$methylpropan$-1-$ol).

(A) The reaction of anisole $(C_6H_5OCH_3)$ with hydrogen iodide $(HI)$ involves the cleavage of the $C-O$ bond.
Since the $C_6H_5-O$ bond has partial double bond character due to resonance,the $O-CH_3$ bond is cleaved.
This results in the formation of phenol $(C_6H_5OH)$ and iodomethane $(CH_3I)$.
The reaction is: $C_6H_5OCH_3 + HI \xrightarrow{\Delta} C_6H_5OH + CH_3I$.

(C) Carbolic acid (also known as phenol) is oxidised by acidified sodium dichromate $(Na_2Cr_2O_7 / H_2SO_4)$ to give $p$-benzoquinone.
The reaction is as follows:
$C_6H_5OH + [O] \xrightarrow{Na_2Cr_2O_7 / H_2SO_4} C_6H_4O_2$ (Benzoquinone)
In the presence of air,phenols are slowly oxidised to dark-coloured mixtures containing quinones.

(D) The given reaction is the $Clemmensen$ reduction,which reduces a carbonyl group $(C=O)$ to a methylene group $(CH_2)$.
$C_6H_5COCH_3 + 4[H] \xrightarrow[Zn-Hg / \text{conc. } HCl] C_6H_5CH_2CH_3 + H_2O$.
The product $X$ is $C_6H_5CH_2CH_3$,which is ethylbenzene.

(B) Pentan-$3$-one does not give a yellow solid on treatment with sodium hypoiodite because this reaction is given only by methyl ketones.
Ketones that have at least one methyl group linked to the carbonyl carbon atom (i.e.,$CH_3-CO-$ group) are oxidized by sodium hypoiodite to form iodoform $(CHI_3)$,which is a yellow solid.
This reaction is known as the iodoform reaction.
Since pentan-$3$-one $(CH_3CH_2COCH_2CH_3)$ lacks a methyl group attached to the carbonyl carbon,it does not undergo the iodoform reaction.

(A) Hinsberg's reagent is benzene sulphonyl chloride.
Its molecular formula is $C_6H_5SO_2Cl$.
This reagent is used to distinguish primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ amines.

(C) The reaction of $ethyltrimethyl$ ammonium iodide with silver hydroxide $(AgOH)$ produces $ethyltrimethyl$ ammonium hydroxide.
Upon strong heating,this undergoes the $Hofmann$ elimination reaction.
Since the ethyl group has $\beta$-hydrogens,it is eliminated as ethene $(CH_2=CH_2)$,while the nitrogen retains the three methyl groups to form trimethylamine,$(CH_3)_3N$.

(B) $1$. The reduction of acetaldoxime $(CH_3CH=NOH)$ with $Na$ in alcohol gives ethylamine $(CH_3CH_2NH_2)$ as product $A$.
$2$. The reaction of ethylamine $(CH_3CH_2NH_2)$ with nitrous acid $(NaNO_2 + HCl)$ is a diazotization reaction followed by hydrolysis,which produces ethanol $(C_2H_5OH)$ as product $B$,along with the evolution of nitrogen gas $(N_2 \uparrow)$ and water $(H_2O)$.

(D) Acid hydrolysis of ethanenitrile $(CH_3CN)$ involves the reaction with water in the presence of an acid catalyst $(H^+)$.
Initially,it forms acetamide,which further hydrolyzes to form acetic acid $(CH_3COOH)$ and ammonium salt $(NH_4^+)$.
The overall reaction is:
$CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_4^+$

(B) Mandelonitrile is a cyanohydrin formed by the nucleophilic addition of hydrogen cyanide $(HCN)$ to benzaldehyde $(C_6H_5CHO)$.
The reaction is as follows:
$C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN$ (Mandelonitrile).

(A) The enzyme that converts maltose into glucose is $Maltase$.
During digestion,starch is partially transformed into maltose by pancreatic or salivary enzymes called amylases.
$Maltase$,secreted by the intestine,then converts maltose into glucose.
The glucose so produced is either utilised by the body or stored in the liver as glycogen.

(D) The monomers used in the preparation of Dextron are lactic acid and glycolic acid.
$n CH_3-CH(OH)-COOH + n HO-CH_2-COOH \rightarrow -[O-CH(CH_3)-CO-O-CH_2-CO]_n- + n H_2O$
Thus,Dextron is a copolymer of lactic acid and glycolic acid.

(C) Saccharin is an artificial sweetening agent with the chemical formula $C_7H_5NO_3S$.
It contains Carbon $(C)$,Hydrogen $(H)$,Nitrogen $(N)$,Oxygen $(O)$,and Sulfur $(S)$.
Phosphorus $(P)$ is not present in its structure.
The structure is as follows:
(Structure of Saccharin: $A$ benzene ring fused to a five-membered ring containing $CO$,$NH$,and $SO_2$ groups).

(B) Zwitter ion is a dipolar molecule that contains both a positively charged group and a negatively charged group within the same molecule,resulting in a net charge of zero.
Amino acids,such as glycine $(H_2NCH_2COOH)$,contain both a basic amino group $(-NH_2)$ and an acidic carboxylic group $(-COOH)$.
In an aqueous solution,the proton from the $-COOH$ group is transferred to the $-NH_2$ group,forming a dipolar ion known as a Zwitter ion: $H_2NCH_2COOH \rightleftharpoons ^+H_3NCH_2COO^-$.
Among the given options,only $H_2NCH_2COOH$ is an amino acid capable of forming a Zwitter ion.

(D) Uracil is not present in $DNA$. It is present in $RNA$.
$DNA$ contains four bases,i.e.,adenine $(A)$,guanine $(G)$,cytosine $(C)$,and thymine $(T)$.
In $RNA$,thymine $(T)$ is replaced by uracil $(U)$.

(B) The vitamin that belongs to the aromatic series is vitamin $K$.
Vitamin $K$ contains a naphthoquinone ring system,which is an aromatic structure.
The structure of vitamin $K$ is shown below:

(A) flavouring agent found in oil of wintergreen is $Methyl \ salicylate$.
It is an organic compound with the formula $C_6H_4(OH)(CO_2CH_3)$.
It is the methyl ester of salicylic acid.
It is a colourless,viscous liquid.
The structure of $Methyl \ salicylate$ is as follows:

(B) For the elementary reaction: $3 H_{2(g)} + N_{2(g)} \rightarrow 2 NH_{3(g)}$
The rate of reaction is expressed as:
Rate $= -\frac{1}{3} \frac{d[H_{2(g)}]}{dt} = -\frac{d[N_{2(g)}]}{dt} = \frac{1}{2} \frac{d[NH_{3(g)}]}{dt}$
To find the relation between $\frac{d[H_{2(g)}]}{dt}$ and $\frac{d[NH_{3(g)}]}{dt}$,we equate their respective parts:
$-\frac{1}{3} \frac{d[H_{2(g)}]}{dt} = \frac{1}{2} \frac{d[NH_{3(g)}]}{dt}$
Multiplying both sides by $2$:
$-\frac{2}{3} \frac{d[H_{2(g)}]}{dt} = \frac{d[NH_{3(g)}]}{dt}$
Thus,option $(B)$ is correct.

(B) The initial rate of the reaction is given by: $\text{Rate}_1 = k[A]^2[B]$.
When the concentration of $A$ is doubled,the new concentration is $[A]' = 2[A]$.
The new rate of the reaction is: $\text{Rate}_2 = k[2A]^2[B]$.
Simplifying the expression: $\text{Rate}_2 = k \times 4[A]^2[B] = 4 \times \text{Rate}_1$.
Therefore,the rate of reaction increases by a factor of $4$.

(C) The rate of reaction for $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$ is expressed as:
$Rate = -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = +\frac{1}{2} \frac{d[SO_3]}{dt}$
Comparing the terms for $SO_3$ and $O_2$:
$+\frac{1}{2} \frac{d[SO_3]}{dt} = -\frac{d[O_2]}{dt}$
Multiplying both sides by $2$,we get:
$+\frac{d[SO_3]}{dt} = -2 \frac{d[O_2]}{dt}$
Thus,option $C$ is correct.

(A) For a first-order reaction,$A \rightarrow \text{product}$,the rate is given by: $\text{Rate} = -\frac{d[A]}{dt} = k[A]$.
Rearranging and integrating the equation: $\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -\int_0^t k dt$.
This yields: $\ln \frac{[A]_t}{[A]_0} = -kt$.
Thus,the integrated rate equation is $k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t}$.
Alternatively,using base $10$ logarithms,$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Both options $A$ and $B$ represent the correct integrated rate equation for a first-order reaction.

(B) zero-order reaction is one where the rate of reaction is independent of the concentration of the reactants.
The decomposition of ammonia $(NH_3)$ on a hot platinum $(Pt)$ surface is a classic example of a zero-order reaction because the metal surface becomes saturated with ammonia molecules,making the rate independent of the concentration of $NH_3$.

(C) The Arrhenius equation is given by: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$.
Given that the activation energy $E_a = 0$.
Substituting $E_a = 0$ into the equation:
$\log \frac{k_2}{k_1} = \frac{0}{2.303 \ R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right] = 0$.
This implies $\frac{k_2}{k_1} = 10^0 = 1$,so $k_2 = k_1$.
Given $k_1 = 1.6 \times 10^{-6} \ s^{-1}$ at $280 \ K$,the rate constant at $300 \ K$ will also be $1.6 \times 10^{-6} \ s^{-1}$.

(A) Antibiotics are classified as broad-spectrum or narrow-spectrum based on their range of activity.
Broad-spectrum antibiotics kill or inhibit a wide range of Gram-positive and Gram-negative bacteria.
Examples of broad-spectrum antibiotics include $Chloramphenicol$,$Ampicillin$,and $Amoxicillin$.
$Penicillin$ $G$ is a narrow-spectrum antibiotic as it is primarily effective against Gram-positive bacteria.
Therefore,$Penicillin$ is not a broad-spectrum antibiotic.

(D) Veronal is a trade name for a barbiturate drug. It is used as a tranquilizer. Tranquilizers are a class of chemical compounds used for the treatment of stress,anxiety,irritability,or excitement by inducing a sense of well-being. The chemical structure of Veronal is shown below:
[Image: $231626-$s]

(D) Antiseptics are chemical substances that prevent the growth of microorganisms or kill them and are applied to living tissues.
$Boric \ acid$ (in dilute aqueous solution) and $Iodoform$ are well-known antiseptics.
$Hydrogen \ peroxide$ $(H_2O_2)$ is also used as an antiseptic.
$Potassium \ sulphite$ $(K_2SO_3)$ is not used as an antiseptic compound.

(A) Antibiotics are classified as either bactericidal or bacteriostatic based on their action on bacteria.
Bactericidal antibiotics kill the bacteria,whereas bacteriostatic antibiotics inhibit the growth of bacteria.
$Tetracycline$ is a well-known bacteriostatic antibiotic.
$Aminoglycosides$,$Penicillin$,and $Ofloxacin$ are examples of bactericidal antibiotics.

(D) Soaps are the sodium or potassium salts of long-chain fatty acids,typically containing more than $12$ carbon atoms,such as stearic,oleic,and palmitic acids.
These are prepared by heating fats or oils with an aqueous sodium hydroxide solution,a process known as saponification.
The reaction is as follows:
$CH_2(OCOC_{17}H_{35})CH(OCOC_{17}H_{35})CH_2(OCOC_{17}H_{35}) + 3NaOH$ $\rightarrow 3C_{17}H_{35}COONa + CH_2(OH)CH(OH)CH_2(OH)$

(B) $2,4-D$ichlorophenoxyacetic acid (commonly known as $2,4-D$) is a synthetic auxin used as a selective herbicide.
It mimics the plant growth hormone auxin,causing uncontrolled and disorganized growth in broadleaf weeds,which leads to their death while leaving grasses and crops relatively unaffected.

(D) The highest oxidation state in plutonium (Atomic number = $94$) is $+7$.
This is because the $5f$,$6d$,and $7s$ energy levels have comparable energies,allowing for a wide range of oxidation states.
Plutonium $(Pu)$ exhibits oxidation states of $+3$,$+4$,$+5$,$+6$,and $+7$ in its various compounds.

(D) The correct answer is $(d)$.
$Se$ $(selenium)$ does not belong to group $15$; it belongs to group $16$ of the periodic table.
$Se$ is located in the $4^{th}$ period and $16^{th}$ group with an atomic number of $34$.
The electronic configuration of $Se$ is $[Ar] 3d^{10} 4s^2 4p^4$.

(C) Group $15$ elements have a valence shell configuration of $ns^2 np^3$.
In the $-3$ oxidation state,the central atom possesses a lone pair of electrons,which can be donated to an electron-deficient species.
Therefore,they act as Lewis bases in the $-3$ oxidation state.

(C) Key Idea: The total number of electrons which the central metal ion appears to possess in the complex,including those gained by it in bonding,is called the effective atomic number $(EAN)$ of the central metal ion.
In the given complex ion $[Fe(CN)_6]^{3-}$,the $Fe$ is in $+3$ oxidation state.
Atomic number of $Fe = 26$.
The number of electrons in $Fe^{3+} = 26 - 3 = 23$.
Each of the six $CN^-$ ligands donates a pair of electrons (total $6 \times 2 = 12$ electrons).
Therefore,$EAN = 23 + 12 = 35$.

(C) neutral complex is a coordination compound that carries no net electrical charge.
In the given options:
$(A)\ [Fe(H_2O)_6]Cl_3$ dissociates into $[Fe(H_2O)_6]^{3+}$ and $3Cl^-$,making it a cationic complex.
$(B)\ [Ni(NH_3)_6]Cl_2$ dissociates into $[Ni(NH_3)_6]^{2+}$ and $2Cl^-$,making it a cationic complex.
$(C)\ [Pt(NH_3)_2Cl_2]$ has no ionizable groups outside the coordination sphere and carries a net charge of $0$,making it a neutral complex.
$(D)\ K[Ag(CN)_2]$ dissociates into $K^+$ and $[Ag(CN)_2]^-$,making it an anionic complex.

(B) $cis-[PtCl_2(NH_3)_2]$,commonly known as cisplatin,is used in the treatment of cancer.

(C) Barium chloride $(BaCl_2)$ reacts with sulfate ions $(SO_4^{2-})$ to form a white precipitate of barium sulfate $(BaSO_4)$.
For a complex to give this precipitate,the $SO_4^{2-}$ ion must be present as a counter-ion outside the coordination sphere.
In the complex $[Co(NH_3)_5NO_2]SO_4$,the sulfate ion is present as a counter-ion.
The reaction is: $SO_4^{2-} (aq) + Ba^{2+} (aq) \rightarrow BaSO_4 (s) \text{ (White precipitate)}$.

(A) The complex $Ba[CuCl_4]$ consists of a barium cation $Ba^{2+}$ and a complex anion $[CuCl_4]^{2-}$.
In the anion $[CuCl_4]^{2-}$,let the oxidation state of $Cu$ be $x$.
$x + 4(-1) = -2$,which gives $x = +2$.
Thus,the oxidation state of copper is $(II)$.
According to $IUPAC$ nomenclature,the cation is named first,followed by the anion.
The ligand $Cl^-$ is named as 'chloro' and since there are four,it is 'tetrachloro'.
Since the complex is an anion,the metal name ends in '-ate',so 'copper' becomes 'cuprate'.
Therefore,the $IUPAC$ name is barium tetrachlorocuprate $(II)$.

(B) The permanganate ion $(MnO_4^-)$ has a tetrahedral geometry.
In this ion,the manganese atom is in the $+7$ oxidation state,meaning it has a $d^0$ electronic configuration.
The $\pi -$ bonding occurs due to the overlap of $p-$ orbitals of oxygen with $d-$ orbitals of manganese.
Since there are no unpaired electrons in the $d^0$ configuration,the ion is diamagnetic.
The structure is shown below:
$Mn$ is at the center bonded to four oxygen atoms in a tetrahedral arrangement.

(C) Werner’s theory was used to describe the structure and formation of complex compounds or coordination compounds.
According to this theory,the primary valency corresponds to the oxidation number,while the secondary valency corresponds to the coordination number.
The geometry of the complex is determined by the number and position of the secondary valences in space,as the ligands satisfying the secondary valencies are always directed towards fixed positions in space.

(C) When a mixture of manganese dioxide $(MnO_2)$,potassium hydroxide $(KOH)$,and potassium chlorate $(KClO_3)$ is fused,the potassium chlorate acts as an oxidizing agent and decomposes to provide oxygen.
The overall reaction is:
$2KClO_{3(s)} \xrightarrow{\Delta} 2KCl_{(s)} + 3O_{2(g)}$
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
The final product obtained is potassium manganate $(K_2MnO_4)$.

(A) The chemical formula for the chromate ion is $CrO_4^{2-}$,which carries a charge of $-2$.
The chemical formula for the dichromate ion is $Cr_2O_7^{2-}$,which also carries a charge of $-2$.
Therefore,the ionic charges are $-2$ and $-2$ respectively.

(B) The electrochemical equivalent $(Z)$ of a substance is defined as the mass of the substance liberated or deposited during electrolysis by the passage of $1 \ C$ of charge.
According to Faraday's laws of electrolysis,the mass $(m)$ deposited is given by $m = Z \times Q$,where $Q$ is the charge in Coulombs $(C)$.
Rearranging for $Z$,we get $Z = \frac{m}{Q}$.
Since the mass $(m)$ is measured in kilograms $(Kg)$ in the $SI$ system and charge $(Q)$ is measured in Coulombs $(C)$,the $SI$ unit of electrochemical equivalent is $Kg C^{-1}$.

(B) The general form of the Nernst equation is:
$E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Product]}{[Reactant]}$
For the given cell reaction:
Oxidation: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} (0.01 \ M) + 2e^-$
Reduction: $Cu^{2+}_{(aq)} (1.0 \ M) + 2e^- \rightarrow Cu_{(s)}$
Here,$n = 2$.
Applying the Nernst equation:
$E_{cell} = 2.0 - \frac{0.0591}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
$E_{cell} = 2.0 - \frac{0.0591}{2} \log \left( \frac{0.01}{1.0} \right)$
$E_{cell} = 2.0 - \frac{0.0591}{2} \log (10^{-2})$
$E_{cell} = 2.0 - \frac{0.0591}{2} \times (-2)$
$E_{cell} = 2.0 + 0.0591 = 2.0591 \ V \approx 2.0592 \ V$

(C) In the electrolysis of aqueous sodium chloride $(NaCl)$ with inert electrodes,the solution contains $Na^{+}$,$Cl^{-}$,$H^{+}$,and $OH^{-}$ ions.
At the cathode,the reduction of water is preferred over the reduction of $Na^{+}$ ions because the standard reduction potential of water is higher:
$2H_2O(l) + 2e^{-} \rightarrow H_2(g) + 2OH^{-}(aq) ; E^{\circ} = -0.83 \ V$
$Na^{+}(aq) + e^{-} \rightarrow Na(s) ; E^{\circ} = -2.71 \ V$
Thus,$H_2$ gas is produced at the cathode.
At the anode,although the oxidation potential of water is higher,the oxidation of $Cl^{-}$ ions is preferred due to overpotential effects:
$2Cl^{-}(aq) \rightarrow Cl_2(g) + 2e^{-} ; E^{\circ} = -1.36 \ V$
$2H_2O(l) \rightarrow O_2(g) + 4H^{+}(aq) + 4e^{-} ; E^{\circ} = -1.23 \ V$
Therefore,$Cl_2$ gas is produced at the anode.

(C) In the electrolysis of brine solution ($NaCl$ aqueous),the solution contains $Na^{+}$,$Cl^{-}$,$H^{+}$,and $OH^{-}$ ions.
At the cathode,$H^{+}$ ions (from water) have a higher reduction potential than $Na^{+}$ ions,so $H_2$ gas is liberated.
Reaction at cathode: $2H_2O(l) + 2e^{-} \rightarrow H_2(g) + 2OH^{-}(aq)$.
At the anode,$Cl^{-}$ ions are oxidized to $Cl_2$ gas.
Reaction at anode: $2Cl^{-}(aq) \rightarrow Cl_2(g) + 2e^{-}$.
Therefore,$H_2$ gas is liberated at the cathode.

(A) Standard hydrogen electrode $(SHE)$ is a primary reference electrode.
$SHE$ is represented by $Pt_{(s)} | H_{2(g)} | H^{+}_{(aq)}$ and is assigned a potential of $0.00 \ V$ at all temperatures.
It serves as the fundamental standard for measuring electrode potentials of other half-cells.
The half-cell reaction is: $H^{+}_{(aq)} + e^- \rightarrow \frac{1}{2} H_{2(g)}$

(A) The general Nernst equation for a half-cell reaction is given by:
$E = E^{o} - \frac{0.0592}{n} \log \frac{[\text{Product}]}{[\text{Reactant}]}$
For the given half-cell reaction:
$Cu^{2+}_{(aq)} + e^{-} \rightarrow Cu^{+}_{(aq)}$
The number of electrons involved is $n = 1$.
Substituting the values into the Nernst equation:
$E_{Cu^{2+}/Cu^{+}} = E^{o}_{Cu^{2+}/Cu^{+}} - \frac{0.0592}{1} \log \frac{[Cu^{+}]}{[Cu^{2+}]}$
Thus,the correct option is $A$.

(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through cells in series,the number of equivalents deposited is equal.
$n_{Cu} \times (n\text{-factor})_{Cu} = n_{Al} \times (n\text{-factor})_{Al}$
For $Cu^{2+} + 2e^- \rightarrow Cu$,the $n\text{-factor}$ is $2$.
For $Al^{3+} + 3e^- \rightarrow Al$,the $n\text{-factor}$ is $3$.
Given $n_{Cu} = 0.4$ moles.
$0.4 \times 2 = n_{Al} \times 3$
$n_{Al} = \frac{0.8}{3} \approx 0.266$ moles.
Rounding to two decimal places,we get $0.27$ moles.

(B) hydrogen-oxygen fuel cell is an electrochemical cell that converts the chemical energy of hydrogen,which acts as a fuel,and oxygen,which acts as an oxidising agent,into electricity through a pair of redox reactions.
In this cell,the cathode reaction involves the reduction of oxygen: $O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)$.
Since oxygen gains electrons,it acts as the oxidising agent.
Thus,option $B$ is correct.