KCET 2010 Chemistry Question Paper with Answer and Solution

(D) The force on a moving charge in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the velocity vector $\vec{v}$ is directed vertically upwards (let this be the $+z$-axis).
The magnetic field $\vec{B}$ is directed towards the north (let this be the $+y$-axis).
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$:
Point your fingers in the direction of $\vec{v}$ (upwards) and curl them towards $\vec{B}$ (north).
The thumb points towards the west.
Therefore,the force on the charge $+Q$ will be towards the west.

(A) For the Lyman series,the frequency is given by $v = RC \left[ \frac{1}{1^2} - \frac{1}{n^2} \right]$,where $n = 2, 3, 4, \dots$
For the series limit of the Lyman series,$n = \infty$.
Therefore,$v_1 = RC \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = RC$ ..... $(i)$
For the first line of the Lyman series,$n = 2$.
Therefore,$v_2 = RC \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = RC \left( 1 - \frac{1}{4} \right) = \frac{3}{4} RC$ ..... $(ii)$
For the Balmer series,the frequency is given by $v = RC \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$
For the series limit of the Balmer series,$n = \infty$.
Therefore,$v_3 = RC \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = \frac{RC}{4}$ ..... $(iii)$
From equations $(i)$,$(ii)$,and $(iii)$,we observe that $v_2 + v_3 = \frac{3}{4} RC + \frac{1}{4} RC = RC = v_1$.
Thus,$v_1 = v_2 + v_3$,which implies $v_1 - v_2 = v_3$.

(A) In the given circuit,the first three capacitors are connected in parallel. Let their equivalent capacitance be $C_1$. Since they are in parallel,$C_1 = C + C + C = 3C$.
Similarly,the last three capacitors are also connected in parallel. Let their equivalent capacitance be $C_2$. Thus,$C_2 = C + C + C = 3C$.
Now,these two groups (each of capacitance $3C$) are connected in series between points $A$ and $B$.
The effective capacitance $C_{eq}$ is given by the series combination formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{3C} + \frac{1}{3C} = \frac{2}{3C}$.
Therefore,$C_{eq} = \frac{3C}{2} = 1.5C$.

(A) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Here,$k$ is the rate constant,$A$ is the Arrhenius factor (frequency factor),$E_a$ is the activation energy,$R$ is the gas constant,and $T$ is the temperature.
We are given $E_a = 2.303 \ RT \ J \ mol^{-1}$.
Substituting the value of $E_a$ into the Arrhenius equation:
$k = A e^{-(2.303 \ RT) / RT}$
$k = A e^{-2.303}$
Since $e^{-2.303} = 10^{-1}$ (because $\ln(10) = 2.303$,so $e^{2.303} = 10$,which implies $e^{-2.303} = 1/10 = 10^{-1}$),
$k = A \times 10^{-1}$.
Therefore,the ratio of the rate constant to the Arrhenius factor is $k/A = 10^{-1}$.

(B) Coagulation is the process of aggregating together the colloidal particles so as to change them into large sized particles which ultimately settle as a precipitate.
$A$,$C$,and $D$ are examples of coagulation.
$B$ Peptization is the process of converting a freshly prepared precipitate into a colloidal sol by adding a suitable electrolyte,which is the reverse of coagulation.

(D) In Young's double slit experiment,the phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the path difference $\Delta x = \lambda/6$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} = 60^{\circ}$.
The intensity at any point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Assuming $I_1 = I_2 = I'$,the maximum intensity is $I_0 = I' + I' + 2\sqrt{I' I'} \cos(0^{\circ}) = 4I'$.
Substituting the values for the given point: $I = I' + I' + 2\sqrt{I' I'} \cos(60^{\circ}) = 2I' + 2I'(0.5) = 2I' + I' = 3I'$.
Therefore,the ratio is $\frac{I}{I_0} = \frac{3I'}{4I'} = 0.75$.

(B) Habitat loss and fragmentation is the most important cause driving animals and plants to extinction.
Due to various human activities,when large habitats are destroyed,various animals are badly affected,leading to population declines.

(D) The letter '$D$' before the name of a monosaccharide indicates the configuration at the highest numbered chiral carbon (the penultimate carbon atom).
In $D$-glucose,the $-OH$ group at the $C-5$ position is on the right side in the Fischer projection.
Therefore,it specifies the configuration at a particular chiral carbon.

(B) The $-Cl$ atom is an electron-withdrawing group that exhibits a negative inductive effect ($-I$ effect).
When attached to the $\alpha$-carbon of acetic acid,it withdraws electron density through the sigma bond network.
This withdrawal decreases the electron density on the oxygen atom of the $-OH$ group,which increases the polarity of the $O-H$ bond.
Consequently,the release of the $H^{+}$ ion becomes easier,thereby increasing the acidity of chloroacetic acid compared to acetic acid.

(C) The peroxide ion is $O_{2}^{2-}$.
Electronic configuration: $O_{2}^{2-} (18 \text{ electrons}) = \sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} \approx \pi^{*} 2p_{y}^{2}$.
Bond order $= \frac{N_{b} - N_{a}}{2} = \frac{10 - 8}{2} = 1$.
It contains four completely filled antibonding molecular orbitals $(\sigma^{*} 1s, \sigma^{*} 2s, \pi^{*} 2p_{x}, \pi^{*} 2p_{y})$.
Since all electrons are paired,$O_{2}^{2-}$ is diamagnetic.
It has $18$ electrons,so it is isoelectronic with argon $(18)$,not neon $(10)$.

(B) For the first reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$; $K_{1} = \frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$
For the second reaction: $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$; $K_{2} = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$
Comparing the two expressions,we can see that $K_{2} = \left( \frac{[SO_{2}][O_{2}]^{1/2}}{[SO_{3}]} \right)^{2} = \left( \frac{1}{K_{1}} \right)^{2} = \frac{1}{K_{1}^{2}}$
Therefore,$K_{1}^{2} = \frac{1}{K_{2}}$.

(A) For the reaction,
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$
Equilibrium constant,$K_{1} = \frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$ ... $(I)$
For the reaction,
$2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$
Equilibrium constant,$K_{2} = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$ ... $(II)$
On squaring both sides in Eq $(I)$,we get
$K_{1}^{2} = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}$ ... $(III)$
Comparing Eq $(II)$ and Eq $(III)$:
$K_{2} = \frac{1}{K_{1}^{2}}$
Therefore,$K_{1}^{2} = \frac{1}{K_{2}}$.

(C) Let us analyze each conversion:
$(a)$ $CH_{4}$ ($sp^{3}$,tetrahedral) $\longrightarrow C_{2}H_{6}$ ($sp^{3}$,tetrahedral). No change in hybridisation or shape.
$(b)$ $NH_{3}$ ($sp^{3}$,trigonal pyramidal) $\longrightarrow NH_{4}^{+}$ ($sp^{3}$,tetrahedral). Change in shape,but no change in hybridisation.
$(c)$ $BF_{3}$ ($sp^{2}$,trigonal planar) $\longrightarrow BF_{4}^{-}$ ($sp^{3}$,tetrahedral). Both hybridisation and shape change.
$(d)$ $H_{2}O$ ($sp^{3}$,bent) $\longrightarrow H_{3}O^{+}$ ($sp^{3}$,trigonal pyramidal). Change in shape,but no change in hybridisation.
Thus,the conversion of $BF_{3}$ into $BF_{4}^{-}$ involves a change in both hybridisation and shape.

(A) $1$. Identify the longest carbon chain containing the functional group (aldehyde). The chain has $5$ carbon atoms,so the parent alkane is pentane,and the aldehyde suffix is $-al$,making it pentanal.
$2$. Number the chain starting from the aldehyde carbon as $C-1$. The $C-2$ position has a methyl group,and the $C-3$ position has a bromine atom.
$3$. Arrange the substituents alphabetically: bromo comes before methyl.
$4$. The $IUPAC$ name is $3-$bromo$-2-$methylpentanal.

(B) Benzene exhibits resonance,and all carbon-carbon bonds are equivalent due to the delocalization of $\pi$ electrons.
It does not contain distinct single and double bonds as suggested in option $B$.
Therefore,the statement that there are three carbon-carbon single bonds and three carbon-carbon double bonds is incorrect.

(A) The stability order of cyclohexane conformations is $Chair > Twist-boat > Boat > Half-chair$.
The $Half-chair$ conformation is the least stable due to high torsional strain and angle strain.

(A) The formula for the percentage of nitrogen in Kjeldahl's method is given by:
$Percentage \ of \ N = \frac{1.4 \times N \times V}{W}$
Where:
$N = 0.1 \ N$ (Normality of acid)
$V = 30 \ cm^{3}$ (Volume of acid used)
$W = 5 \ g$ (Weight of the food sample)
Substituting the values:
$Percentage \ of \ N = \frac{1.4 \times 0.1 \times 30}{5} = \frac{4.2}{5} = 0.84 \%$

(D) Chromite ore is $FeCr_2O_4$,which can be represented as $FeO \cdot Cr_2O_3$.
In this compound,the oxidation number of $Fe$ is $+2$ and the oxidation number of $Cr$ is $+3$.

(D) Chromite or chrome iron ore is $FeCr_{2}O_{4}$. Its actual composition is $FeO \cdot Cr_{2}O_{3}$.
In $FeO$,the oxidation state of $Fe$ is calculated as:
$x + (-2) = 0 \implies x = +2$
Similarly,in $Cr_{2}O_{3}$,the oxidation state of $Cr$ is calculated as:
$2x + 3(-2) = 0 \implies 2x = +6 \implies x = +3$
Thus,the oxidation states of $Fe$ and $Cr$ are $+2$ and $+3$ respectively.

(D) When $\omega$-dihalides (terminal dihalides) react with sodium metal in the presence of dry ether,an intramolecular Wurtz reaction occurs,leading to the formation of a cyclic hydrocarbon.
For example,$1, 4-$dibromobutane reacts with $2Na$ to form cyclobutane and $2NaBr$ as a byproduct.
The reaction is: $Br-CH_2-CH_2-CH_2-CH_2-Br + 2Na \rightarrow \text{Cyclobutane} + 2NaBr$.

(A) For a solution to have $pH = 1$,the concentration of $[H^+]$ must be $0.1 \ M$.
$(A)$ $0.1 \ M \ CH_3COOH$ is a weak acid and does not dissociate completely,so $[H^+] < 0.1 \ M$,hence $pH > 1$.
$(B)$ $0.1 \ M \ HNO_3$ is a strong acid,$[H^+] = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
$(C)$ $0.05 \ M \ H_2SO_4$ is a strong acid,$[H^+] = 2 \times 0.05 = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
$(D)$ For the mixture: $n(H^+) = 50 \times 0.4 = 20 \ mmol$,$n(OH^-) = 50 \times 0.2 = 10 \ mmol$. Remaining $n(H^+) = 20 - 10 = 10 \ mmol$. Total volume = $100 \ cm^3$. $[H^+] = 10 / 100 = 0.1 \ M$,so $pH = 1$.

(D) The chemical reaction is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$
Number of moles of $Na = \frac{0.023 \ g}{23 \ g/mol} = 1 \times 10^{-3} \ mol$.
According to the stoichiometry,$2 \ mol$ of $Na$ produces $2 \ mol$ of $NaOH$.
Therefore,$1 \times 10^{-3} \ mol$ of $Na$ produces $1 \times 10^{-3} \ mol$ of $NaOH$.
The volume of the solution is $100 \ cm^3 = 0.1 \ L$.
Concentration of $[OH^-] = \frac{1 \times 10^{-3} \ mol}{0.1 \ L} = 1 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(1 \times 10^{-2}) = 2$.
Since $pH + pOH = 14$,$pH = 14 - 2 = 12$.

(C) For $0.1 \ M \ HNO_3$,$[H^+] = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
For $0.05 \ M \ H_2SO_4$,$[H^+] = 2 \times 0.05 = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
For $0.1 \ M \ CH_3COOH$,it is a weak acid and does not ionize completely,so $[H^+] < 0.1 \ M$ and $pH > 1$.
For the mixture,$n(HCl) = 0.05 \ L \times 0.4 \ M = 0.02 \ mol$ and $n(NaOH) = 0.05 \ L \times 0.2 \ M = 0.01 \ mol$.
Remaining $n(HCl) = 0.02 - 0.01 = 0.01 \ mol$ in $100 \ mL$ $(0.1 \ L)$ solution.
$[H^+] = 0.01 \ mol / 0.1 \ L = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
Thus,$0.1 \ M \ CH_3COOH$ does not have a $pH$ of $1$.

(D) Initially,the amount of salt $[CH_{3}COONa] = 0.1 \ mol$ and the amount of acid $[CH_{3}COOH] = 0.1 \ mol$.
When an additional $0.1 \ mol$ of $CH_{3}COONa$ is added,the total amount of salt becomes $0.1 + 0.1 = 0.2 \ mol$.
The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[salt]}{[acid]}$.
Substituting the values: $pH = pK_{a} + \log \frac{0.2}{0.1}$.
Therefore,$pH = pK_{a} + \log 2$.

(C) For precipitation to occur,the ionic product must exceed the solubility product $(K_{sp})$.
For $Ag_{2}CrO_{4}$:
Ionic Product $(IP) = [Ag^{+}]^{2}[CrO_{4}^{2-}] = (10^{-4})^{2}(10^{-5}) = 10^{-13}$.
Given $K_{sp}(Ag_{2}CrO_{4}) = 4 \times 10^{-12}$.
Since $IP < K_{sp}$,$Ag_{2}CrO_{4}$ will not precipitate.
For $AgCl$:
Ionic Product $(IP) = [Ag^{+}][Cl^{-}] = (10^{-4})(10^{-5}) = 10^{-9}$.
Given $K_{sp}(AgCl) = 1 \times 10^{-10}$.
Since $IP > K_{sp}$,$AgCl$ will precipitate.
Therefore,silver chloride gets precipitated first.

(B) The first equation is $y_{1} = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$.
We can rewrite this by multiplying and dividing by $2$: $y_{1} = 10 [\frac{1}{2} \sin 2 \pi t + \frac{\sqrt{3}}{2} \cos 2 \pi t]$.
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.
Thus,$y_{1} = 10 [\sin 2 \pi t \cos \frac{\pi}{3} + \cos 2 \pi t \sin \frac{\pi}{3}] = 10 \sin(2 \pi t + \frac{\pi}{3})$.
Therefore,the amplitude $A_{1} = 10$.
For the second equation,$y_{2} = 5 \sin (2 \pi t + \frac{\pi}{4})$,the amplitude $A_{2} = 5$.
The ratio of the amplitudes is $\frac{A_{1}}{A_{2}} = \frac{10}{5} = \frac{2}{1}$.

(C) In general, density increases on moving downward in a group, but the density of potassium $(K)$ is lower than that of sodium $(Na)$.
This is due to an abnormal increase in atomic size from $Na$ $(186 \text{ pm})$ to $K$ $(227 \text{ pm})$.
Therefore, the correct order of density is $Li < K < Na < Rb < Cs$.
Thus, the trend given in option $C$ is incorrect.

(A) The ideal gas equation is $pV = nRT$.
Since $V$ and $T$ are constant,$p \propto n$.
Therefore,the substance with the least number of moles $(n)$ will exert the least pressure.
Note: The molar mass of chlorine gas $(Cl_2)$ is $71 \ g/mol$.
$(A)$ $n = \frac{0.0355 \ g}{71 \ g/mol} = 0.0005 \ mol = 5 \times 10^{-4} \ mol$.
$(B)$ $n = \frac{0.071 \ g}{71 \ g/mol} = 0.001 \ mol = 1 \times 10^{-3} \ mol$.
$(C)$ $n = \frac{6.023 \times 10^{21}}{6.023 \times 10^{23}} = 0.01 \ mol$.
$(D)$ $n = 0.02 \ mol$.
Comparing the values,$5 \times 10^{-4} \ mol$ is the smallest.
Thus,$0.0355 \ g$ of chlorine exerts the least pressure.

(A) The chemical reactions involved are:
$CaCl_{2} + Na_{2}CO_{3} \rightarrow CaCO_{3} + 2NaCl$
$CaCO_{3} \xrightarrow{\Delta} CaO + CO_{2}$
From the stoichiometry,$1 \ mol$ of $CaO$ is obtained from $1 \ mol$ of $CaCO_{3}$,which in turn comes from $1 \ mol$ of $CaCl_{2}$.
Molar mass of $CaO = 40 + 16 = 56 \ g/mol$.
Moles of $CaO = \frac{0.56 \ g}{56 \ g/mol} = 0.01 \ mol$.
Therefore,moles of $CaCl_{2} = 0.01 \ mol$.
Molar mass of $CaCl_{2} = 40 + (2 \times 35.5) = 111 \ g/mol$.
Mass of $CaCl_{2} = 0.01 \ mol \times 111 \ g/mol = 1.11 \ g$.
Mass of $NaCl$ in the mixture $= 4.44 \ g - 1.11 \ g = 3.33 \ g$.
Percentage of $NaCl = \frac{3.33 \ g}{4.44 \ g} \times 100 = 75 \%$.

(D) Given,$T_{1} = 273 + 10 = 283 \ K$ and $T_{2} = 273 + 20 = 293 \ K$.
Average kinetic energy $(KE)$ is directly proportional to temperature $(T)$: $KE = \frac{3}{2} RT$.
Ratio $\frac{KE_{2}}{KE_{1}} = \frac{293}{283} \approx 1.035$.
Root mean square velocity $(v_{rms})$ is proportional to $\sqrt{T}$: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Ratio $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{293}{283}} \approx 1.017$.
Since the temperature change is small relative to the absolute temperature,both the average kinetic energy and the rms velocity increase,but not significantly.

(A) The wave number $\bar{\nu}$ of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula: $\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given that $\bar{\nu} = \frac{8}{9} R_H$,we substitute this into the formula:
$\frac{8}{9} R_H = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Dividing both sides by $R_H$,we get $\frac{8}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$.
For the emission spectrum,$n_1 < n_2$. Since the value is $\frac{8}{9}$,we test $n_1 = 1$:
$\frac{8}{9} = 1 - \frac{1}{n_2^2} \implies \frac{1}{n_2^2} = 1 - \frac{8}{9} = \frac{1}{9}$.
Thus,$n_2^2 = 9$,which means $n_2 = 3$.
Therefore,the electron jumps from $n=3$ to $n=1$.

(C) The electronic configuration of the $Cu$ $(Z=29)$ atom is $[Ar] 3d^{10} 4s^{1}$.
Since the outermost shell is $4s$,the outermost electron resides in the $4s$ orbital.
For the $4s^{1}$ electron:
Principal quantum number $(n)$ = $4$.
Azimuthal quantum number $(l)$ for $s$-orbital = $0$.
Magnetic quantum number $(m_l)$ = $0$.
Spin quantum number $(m_s)$ = $+\frac{1}{2}$.
Thus,the set of quantum numbers is $(4, 0, 0, +\frac{1}{2})$.

(B) Coagulation is the process of converting a colloidal sol into a precipitate.
Peptization is the reverse process,where a fresh precipitate is converted into a colloidal sol by adding an electrolyte.
Therefore,peptization does not involve coagulation.

(A) The first law of thermodynamics is given by $\Delta E = q + w$.
$(A)$ For an isochoric process,$\Delta V = 0$,so the work done $w = -P_{ext} \Delta V = 0$. Therefore,$\Delta E = q$.
$(B)$ For an adiabatic process,$q = 0$,so $\Delta E = w$.
$(C)$ For an isothermal process,$\Delta T = 0$,which implies $\Delta E = 0$ for an ideal gas,so $q = -w$.
$(D)$ For a cyclic process,the change in state function $\Delta E = 0$,therefore $q = -w$.

(A) The neutralization reaction is: $HCl + NaOH \rightarrow NaCl + H_{2}O$
Initial moles of $HCl = \frac{500 \times 0.1}{1000} = 0.05 \text{ mol}$
Initial moles of $NaOH = \frac{200 \times 0.2}{1000} = 0.04 \text{ mol}$
Since $NaOH$ is the limiting reagent,$0.04 \text{ mol}$ of $HCl$ will react with $0.04 \text{ mol}$ of $NaOH$.
The heat of neutralization for $1 \text{ mole}$ of a strong acid and a strong base is $57.3 \text{ kJ/mol}$.
Therefore,the heat evolved for $0.04 \text{ mol}$ is $57.3 \times 0.04 \text{ kJ} = 2.292 \text{ kJ}$.

(B) The first ionization energy generally increases across a period from left to right due to an increase in effective nuclear charge.
Exceptions occur when a more stable electronic configuration (like fully filled or half-filled orbitals) is present in an element to the left of another.
For $N$ $(2s^2 2p^3)$ and $O$ $(2s^2 2p^4)$,$N$ has a higher $IE$ than $O$ due to its stable half-filled $p$-orbital.
For $Be$ $(2s^2)$ and $B$ $(2s^2 2p^1)$,$Be$ has a higher $IE$ than $B$ due to its stable fully filled $s$-orbital.
For $Mg$ $(3s^2)$ and $Al$ $(3s^2 3p^1)$,$Mg$ has a higher $IE$ than $Al$ due to its stable fully filled $s$-orbital.
For $Na$ $(3s^1)$ and $Mg$ $(3s^2)$,the trend follows the general rule where $IE$ of $Mg > IE$ of $Na$,so this is not an exception.

(B) The process is the condensation of benzene vapor to liquid,which is the reverse of vaporization.
The enthalpy of condensation is $\Delta H_{\text{cond}} = -\Delta H_{\text{vap}} = -35.3 \ kJ \ mol^{-1} = -35.3 \times 10^{3} \ J \ mol^{-1}$.
The boiling point is $T_b = 80 + 273 = 353 \ K$.
The entropy change for the phase transition is given by $\Delta S = \frac{\Delta H_{\text{cond}}}{T_b}$.
$\Delta S = \frac{-35.3 \times 10^{3} \ J \ mol^{-1}}{353 \ K} = -100 \ J \ K^{-1} \ mol^{-1}$.

(B) For the given reversible reaction,$\Delta G^{\circ} = -350 \ kJ$.
Since $\Delta G^{\circ} < 0$,the reaction is thermodynamically feasible.
We know the relationship $\Delta G^{\circ} = -RT \ln K$.
Substituting the value,$-350 \times 10^3 = -RT \ln K$,which implies $\ln K > 0$,so $K > 1$.
Therefore,the equilibrium constant is greater than one.
Regarding entropy,the reaction involves the conversion of $1 \ mol$ of solid and $1 \ mol$ of gas into $2 \ mol$ of gas,leading to an increase in disorder,so $\Delta S$ is positive.

(D) The letter '$D$' in $D$-glucose refers to the configuration of the hydroxyl $(-OH)$ group attached to the penultimate chiral carbon (the chiral carbon furthest from the carbonyl group). If the $-OH$ group is on the right side in the Fischer projection,it is designated as '$D$'.

(C) Saponification of oils yields a triol (glycerol).
Drying (hardening) of oil involves hydrogenation.
Refining of oil is done by distillation or other such processes,not by hydrogenation.
Antioxidants are added to prevent the oxidation of oil,thus they minimize rancidity.

(A) The Cannizzaro reaction mechanism involves the following steps:
Step $I$: Nucleophilic attack of $OH^{-}$ on the carbonyl carbon of the first aldehyde molecule to form a dihydroxyalkoxide intermediate.
Step $II$: Transfer of a hydride ion $(H^{-})$ from the intermediate to the carbonyl carbon of a second aldehyde molecule,resulting in the formation of a carboxylate ion and an alkoxide ion.
Step $III$: Rapid proton transfer $(H^{+})$ from the carboxylic acid (or solvent) to the alkoxide ion to form the alcohol.

(D) The molecular formula $C_{3}H_{8}O$ corresponds to the general formula $C_{n}H_{2n+2}O$,which suggests that the compound $A$ is either an alcohol or an ether.
Since the compound reacts with $HI$ to form two different products ($X$ and $Y$),it must be an unsymmetrical ether. The reaction is: $CH_{3}OC_{2}H_{5} + HI \longrightarrow CH_{3}I + C_{2}H_{5}OH$.
Here,$X$ is $CH_{3}I$ and $Y$ is $C_{2}H_{5}OH$ (ethanol).
When ethanol $(Y)$ is treated with aqueous alkali (like $NaOH$) and $I_{2}$,it undergoes the iodoform test because it can be oxidized to acetaldehyde,which contains the $CH_{3}CO-$ group.
The reaction for the iodoform test is: $C_{2}H_{5}OH + 4I_{2} + 6NaOH \longrightarrow CHI_{3} + HCOONa + 5NaI + 5H_{2}O$.
Thus,the compound $A$ is methoxyethane $(CH_{3}OC_{2}H_{5})$.

(D) Benzaldehyde is an aromatic aldehyde,while acetone is a ketone.
$Tollen's$ reagent is used to distinguish between aldehydes and ketones.
Aldehydes like benzaldehyde react with $Tollen's$ reagent to form a silver mirror (precipitate),whereas ketones like acetone do not react with $Tollen's$ reagent.
Fehling's solution is generally used to distinguish between aliphatic aldehydes and ketones,but it does not react with aromatic aldehydes like benzaldehyde.
Therefore,$Tollen's$ reagent is the best choice for distinguishing benzaldehyde and acetone.

(A) Alanine is an amino acid with the structure $CH_3-CH(NH_2)-COOH$.
In an alkaline medium $(pH > 7)$,the carboxylic acid group $(-COOH)$ loses a proton $(H^+)$ to form a carboxylate ion $(-COO^-)$.
Thus,the structure becomes $CH_3-CH(NH_2)-COO^-$,which carries a net negative charge.
Therefore,in an alkaline medium,alanine exists predominantly as an anion.

(D) The relationship between standard $emf$ $(E_{\text{cell}}^{\circ})$ and equilibrium constant $(K_{c})$ is given by the Nernst equation at $298 \ K$:
$E_{\text{cell}}^{\circ} = \frac{0.0591}{n} \log K_{c}$
Given $E_{\text{cell}}^{\circ} = 0.59 \ V$ and $n = 3$:
$0.59 = \frac{0.0591}{3} \log K_{c}$
Approximating $0.0591 \approx 0.059$:
$0.59 = \frac{0.059}{3} \log K_{c}$
$\log K_{c} = \frac{0.59 \times 3}{0.059} = 10 \times 3 = 30$
$K_{c} = 10^{30}$

(C) For a zero order reaction,the integrated rate equation is given by:
$[A] = -kt + [A]_0$
Where $[A]_0$ is the initial concentration $(a)$ and $[A]$ is the concentration at time $t$.
For $100\%$ completion,the remaining concentration $[A] = 0$.
Substituting these values into the equation:
$0 = -kt + a$
$kt = a$
$t = \frac{a}{k}$

(A) For the reaction $2A_{(g)} \longrightarrow B_{(g)} + C_{(s)}$,let the initial pressure of $A$ be $P_0 = 2p$.
At $t = \infty$ (completion),$2A$ is fully consumed,so $P_{total} = P_B + P_C = p + p = 2p = 200 \ Pa$. Thus,$p = 100 \ Pa$ and $P_0 = 200 \ Pa$.
At $t = 10 \ min$,the pressure of $A$ is $2p - x$,$B$ is $x/2$,and $C$ is solid (negligible pressure).
$P_{total} = (2p - x) + x/2 = 2p - x/2 = 300 \ Pa$.
Substituting $2p = 200$,we get $200 - x/2 = 300$,which implies $x/2 = -100$. This suggests the reaction stoichiometry or data provided is inconsistent with standard gas laws.
Assuming the standard model where $P_{total} = P_0 + (n-1)x$,for $2A \rightarrow B + C$,$P_{total} = P_0 + (2-1)x = P_0 + x$.
At $t = \infty$,$P_{\infty} = P_0/2 = 200 \implies P_0 = 400 \ Pa$.
At $t = 10$,$P_t = P_0 + x/2 = 300 \implies 400 + x/2 = 300 \implies x/2 = -100$.
Given the options,the intended calculation is $k = \frac{2.303}{t} \log \frac{P_0}{P_A}$,where $P_A = 2(P_0 - P_t) = 2(400 - 300) = 200$.
$k = \frac{2.303}{10} \log \frac{400}{200} = 0.2303 \times 0.3010 \approx 0.0693 \ min^{-1}$.

(A) The Arrhenius equation is given by $k = A \ e^{-E_{a} / RT}$.
Given that $E_{a} = 2.303 \ RT \ J \ mol^{-1}$.
Substituting the value of $E_{a}$ in the equation:
$k = A \ e^{-(2.303 \ RT) / RT}$
$k = A \ e^{-2.303}$
Since $e^{-2.303} = 10^{-1}$,we have:
$k = A \times 10^{-1}$
Therefore,the ratio of the rate constant to the Arrhenius factor is $\frac{k}{A} = 10^{-1}$.

(A) The complex is $K_{2}[Ni(CN)_{4}]$.
First,determine the oxidation state of $Ni$ in the complex $[Ni(CN)_{4}]^{2-}$.
Let the oxidation state of $Ni$ be $x$.
$x + 4(-1) = -2$
$x - 4 = -2$
$x = +2$.
Since the complex ion $[Ni(CN)_{4}]^{2-}$ is anionic,the metal name ends in '-ate',which is 'nickelate'.
The ligand $CN^-$ is named 'cyano'.
Thus,the $IUPAC$ name is potassium tetracyanonickelate $(II)$.

(C) $(a): \text{In } [Cu(NH_{3})_{4}]^{2+}, Cu \text{ is present as } Cu^{2+}. \ Cu^{2+} = [Ar] 3d^{9} 4s^{0}. \text{ The } NH_{3} \text{ ligand causes } dsp^{2} \text{ hybridisation, resulting in a square planar geometry.}$
$(b): \text{In } [Ni(CO)_{4}], CO \text{ is a neutral ligand.}$
$(c): \text{In } [Fe(CN)_{6}]^{3-}, Fe \text{ is present as } Fe^{3+}. \ Fe^{3+} = [Ar] 3d^{5} 4s^{0}. \text{ Since } CN^{-} \text{ is a strong field ligand, it causes pairing of electrons. Thus, its hybridisation is } d^{2}sp^{3}, \text{ not } sp^{3}d^{2}. \text{ It is an inner orbital complex.}$
$(d): \text{In } [Co(en)_{3}]^{3+}, \text{ the central metal } Co^{3+} \text{ has } 24 \text{ electrons and } 6 \text{ ligands donate } 12 \text{ electrons, total } = 36 \text{ electrons, which follows the EAN rule.}$

(C) In homogeneous catalysis,the reactants and the catalyst are in the same phase.
$(A)$ $2SO_{2(g)} + O_{2(g)} \xrightarrow{Pt_{(s)}} 2SO_{3(g)}$ (Contact process): This is an example of heterogeneous catalysis as the catalyst $(Pt_{(s)})$ is in the solid phase while reactants are in the gaseous phase.
$(B)$ $N_{2(g)} + 3H_{2(g)} \xrightarrow{Fe_{(s)}} 2NH_{3(g)}$ (Haber's process): This is an example of heterogeneous catalysis as the catalyst $(Fe_{(s)})$ is in the solid phase.
$(C)$ $C_{12}H_{22}O_{11(aq)} + H_2O_{(l)} \xrightarrow{HCl_{(aq)}} C_6H_{12}O_{6(aq)} + C_6H_{12}O_{6(aq)}$: Here,the reactant (sucrose) and the catalyst $(HCl_{(aq)})$ are both in the aqueous phase. Thus,it is an example of homogeneous catalysis.
$(D)$ Hydrogenation of oil involves solid catalysts like $Ni_{(s)}$,thus,it is an example of heterogeneous catalysis.

(B) Gas $B$ turns the colour of acidified $K_{2}Cr_{2}O_{7}$ green,thus it is $SO_{2}$. $SO_{2}$ is obtained along with a yellow precipitate of sulfur when thiosulfate is treated with dilute acids. Thus,$A$ is $Na_{2}S_{2}O_{3}$,$B$ is $SO_{2}$,and $C$ is $Cr_{2}(SO_{4})_{3}$.
The reactions are as follows:
$Na_{2}S_{2}O_{3} + 2HCl \rightarrow 2NaCl + H_{2}O + SO_{2} + S \text{ (yellow precipitate)}$
$K_{2}Cr_{2}O_{7} + 3SO_{2} + H_{2}SO_{4} \rightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} \text{ (green solution)} + H_{2}O$

(B) The Nernst equation for the cell reaction is $E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
$(I) E_{1} = E_{\text{cell}}^{\circ} - \frac{0.0591}{2} \log \frac{1}{0.1} = E_{\text{cell}}^{\circ} - \frac{0.0591}{2} \log(10) = E_{\text{cell}}^{\circ} - 0.02955 \ V$.
$(II) E_{2} = E_{\text{cell}}^{\circ} - \frac{0.0591}{2} \log \frac{1}{1} = E_{\text{cell}}^{\circ} - 0 = E_{\text{cell}}^{\circ}$.
$(III) E_{3} = E_{\text{cell}}^{\circ} - \frac{0.0591}{2} \log \frac{0.1}{1} = E_{\text{cell}}^{\circ} - \frac{0.0591}{2} \log(10^{-1}) = E_{\text{cell}}^{\circ} + 0.02955 \ V$.
Comparing the values,we get $E_{3} > E_{2} > E_{1}$.

(B) The oxidation of a primary alcohol $(R-CH_{2}OH)$ to a carboxylic acid $(R-COOH)$ involves the replacement of two hydrogen atoms with one oxygen atom.
The mass difference is calculated as:
Mass of $-COOH$ group = $12 + 16 + 16 + 1 = 45 \ u$.
Mass of $-CH_{2}OH$ group = $12 + 2(1) + 16 + 1 = 31 \ u$.
Difference = $45 - 31 = 14 \ u$.
Thus,the compound is a primary alcohol.

(A) Nucleophiles are species that have an excess of electrons.
In $Pyrrole$,the lone pair of nitrogen is involved in the delocalization of the ring to maintain aromaticity,so it is not available for donation.
In $Aniline$,the lone pair is involved in conjugation with the $\pi$ electrons of the benzene ring,which reduces its availability.
In $Acetanilide$ $(C_6H_5NHCOCH_3)$,the lone pair on nitrogen is involved in resonance with the carbonyl group $(C=O)$,making it even less nucleophilic.
In $Pyridine$,the lone pair on the nitrogen atom is in an $sp^2$ hybridized orbital that is perpendicular to the $\pi$ system of the ring. Therefore,it is not involved in the aromatic sextet and is relatively free for donation.
Thus,the nitrogen of $Pyridine$ is the most nucleophilic.

(D) When acetaldehyde $(CH_3CHO)$ reacts with dilute aqueous sodium hydroxide $(NaOH)$,it undergoes aldol condensation due to the presence of $\alpha-H$ atoms.
The reaction is: $2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO$ ($3$-hydroxybutanal).
This aldol product contains one chiral (asymmetric) carbon atom,which makes it optically active,thus exhibiting optical isomerism.
Upon heating,the aldol loses a water molecule to form an $\alpha,\beta$-unsaturated aldehyde: $CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO$ (but$-2-$enal).
This alkene $(CH_3CH=CHCHO)$ exhibits geometrical isomerism due to the restricted rotation around the $C=C$ double bond.

(D) In the electrolytic refining of zinc,the anode is made of impure zinc,while a strip of pure zinc acts as the cathode. An acidified solution of zinc sulphate is used as the electrolyte. When electricity is passed,the following reactions occur:
At the cathode: $Zn^{2+} + 2e^{-} \longrightarrow Zn_{(pure)}$
At the anode: $Zn_{(impure)} \longrightarrow Zn^{2+} + 2e^{-}$

(B) According to the Ellingham diagram,the line for the formation of $CO_2$ from $C$ and $O_2$ intersects the line for the formation of $Fe_2O_3$ from $Fe$ and $O_2$ at approximately $983 \ K$.
Above $983 \ K$,the Gibbs free energy change $(\Delta G)$ for the formation of $CO_2$ becomes more negative than the $\Delta G$ for the formation of $Fe_2O_3$.
This implies that at temperatures above $983 \ K$,carbon has a higher affinity for oxygen than iron does,allowing carbon to effectively reduce ferric oxide to metallic iron.

(C) Step $1$: $CH_{3}CH_{2}Br$ reacts with $aq. KOH$ (nucleophilic substitution) to form ethanol $(A = CH_{3}CH_{2}OH)$.
Step $2$: Ethanol $(CH_{3}CH_{2}OH)$ is oxidized by $KMnO_{4} / H^{+}$ to form acetic acid $(B = CH_{3}COOH)$.
Step $3$: Acetic acid reacts with $NH_{3}$ followed by heating to form acetamide $(C = CH_{3}CONH_{2})$.
Step $4$: Acetamide reacts with $Br_{2}$ in the presence of alkali (Hofmann bromamide degradation reaction) to form methylamine $(D = CH_{3}NH_{2})$.
Thus,the final product '$D$' is $CH_{3}NH_{2}$.

(D) Methyl bromide $(CH_3Br)$ is a primary alkyl halide.
In the reaction with aqueous sodium hydroxide $(NaOH)$,the nucleophile $(OH^-)$ attacks the electrophilic carbon from the backside.
Since the reaction rate depends on the concentration of both the substrate $(CH_3Br)$ and the nucleophile $(OH^-)$,it follows a second-order kinetics,which is characteristic of an $S_{N}2$ mechanism.
This process involves a single-step transition state and results in the inversion of configuration.

(A) When ethyl bromide $(C_{2}H_{5}Br)$ reacts with alcoholic $KCN$,a nucleophilic substitution reaction occurs where the cyanide ion $(CN^-)$ acts as an ambident nucleophile and attacks through the carbon atom to form an alkyl cyanide (nitrile).
The reaction is: $C_{2}H_{5}Br + KCN (\text{alc.}) \rightarrow C_{2}H_{5}CN + KBr$.
Here,$C_{2}H_{5}CN$ is propane nitrile (also known as propanenitrile).

(B) When phenol is treated with bromine water $(Br_2/H_2O)$,it undergoes electrophilic substitution at all available ortho and para positions due to the highly activating nature of the $-OH$ group.
This reaction results in the formation of $2,4,6$-tribromophenol as a white precipitate.
Therefore,the reagent "$X$" is bromine in water.

(C) The electrolysis reaction is: $MgCl_{2} \longrightarrow Mg^{2+} + 2Cl^{-}$
At the cathode: $Mg^{2+} + 2e^{-} \longrightarrow Mg$
Since $2 \ F$ $(2 \times 96500 \ C)$ of charge deposits $1 \ mol$ of $Mg$,
$9.65 \ C$ of charge will deposit $Mg = \frac{1 \times 9.65}{2 \times 96500} \ mol = 5 \times 10^{-5} \ mol$.
To prepare a Grignard reagent $(R-Mg-X)$,$1 \ mol$ of $Mg$ is required per mole of reagent.
Therefore,$5 \times 10^{-5} \ mol$ of $Mg$ will produce $5 \times 10^{-5} \ mol$ of Grignard reagent.

(A) The conversion of $p$-nitrophenol to quinol proceeds through the following steps:
$1$. Reduction: $p$-nitrophenol is reduced using $Sn/HCl$ to form $p$-aminophenol.
$2$. Diazotization: $p$-aminophenol is treated with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form the corresponding diazonium salt.
$3$. Hydrolysis: The diazonium salt is hydrolyzed by warming with water to yield quinol ($p$-dihydroxybenzene).

(A) Noble gases exist as monoatomic molecules.
Since a monoatomic molecule consists of only a single atom,it has only translational degrees of freedom and lacks rotational or vibrational degrees of freedom.
Therefore,noble gases do not possess vibrational energy.

(B) Initial milliequivalents of $HCl = N \times V = 0.2 \ N \times 50 \ cm^{3} = 10 \ meq$.
Milliequivalents of $NaOH$ added $= 0.1 \ N \times 50 \ cm^{3} = 5 \ meq$.
Remaining milliequivalents of $HCl = 10 \ meq - 5 \ meq = 5 \ meq$.
To neutralize the remaining $5 \ meq$ of $HCl$,we use $0.5 \ N \ KOH$.
Let the volume of $KOH$ be $V_{KOH}$.
$N_{KOH} \times V_{KOH} = 5 \ meq$.
$0.5 \ N \times V_{KOH} = 5 \ meq$.
$V_{KOH} = \frac{5}{0.5} = 10 \ cm^{3}$.

(C) In dry ice (solid $CO_2$),the constituent particles are molecules.
These molecules are held together by weak intermolecular forces such as London dispersion forces or dipole-dipole interactions.
Therefore,it is classified as a molecular crystal.
Rock salt $(NaCl)$ is an ionic crystal,while quartz $(SiO_2)$ and diamond $(C)$ are covalent (network) crystals.

(A) In $(A)$ (para-nitrophenol),intermolecular $H$-bonding exists,which leads to association of molecules and higher boiling point.
In $(B)$ (ortho-nitrophenol),intramolecular $H$-bonding exists,which restricts association and results in a lower boiling point.
Since $(B)$ has a lower boiling point,it is more volatile than $(A)$.
Therefore,the vapour pressure of $(B)$ is higher than that of $(A)$ at a particular temperature.

(C) Let the mass of silver in the zinc layer be $x \ g$. Then the mass of silver in the lead layer is $(1-x) \ g$.
Given the distribution coefficient $K_D = \frac{C_{Zn}}{C_{Pb}} = 300$.
Concentration in zinc $C_{Zn} = \frac{x}{10} \ g/cm^3$.
Concentration in lead $C_{Pb} = \frac{1-x}{100} \ g/cm^3$.
Setting up the equation: $\frac{x/10}{(1-x)/100} = 300$.
$\frac{10x}{1-x} = 300$.
$10x = 300 - 300x$.
$310x = 300$.
$x = \frac{300}{310} \approx 0.9677$.
Thus,the percentage of silver in the zinc layer is approximately $97 \%$.

(D) Since the solutions have the same osmotic pressure at the same temperature,their molar concentrations must be equal.
Concentration of compound = Concentration of glucose = $0.05 \ M$.
Let $M$ be the molar mass of the compound. The concentration is given by $\frac{\text{mass}}{M \times \text{Volume}}$. Assuming $1 \ L$ of solution,$\frac{6 \ g}{M} = 0.05 \ mol/L$.
$M = \frac{6}{0.05} = 120 \ g/mol$.
The empirical formula mass of $CH_{2}O = 12 + (2 \times 1) + 16 = 30 \ g/mol$.
The value of $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{120}{30} = 4$.
Therefore,the molecular formula is $(CH_{2}O)_{4} = C_{4}H_{8}O_{4}$.

(C) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^{5} \ 4s^{2}$.
For $Mn^{4+}$ ion,we remove four electrons,resulting in the configuration $[Ar] \ 3d^{3}$.
This ion contains $n=3$ unpaired electrons.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=3$,we get $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
Rounding to the nearest integer,the value is $4 \ BM$.

(B) Adsorption is an exothermic process,therefore the enthalpy change is negative $(\Delta H < 0)$.
Furthermore,adsorption leads to a more ordered arrangement of gas molecules on the surface,resulting in a decrease in entropy $(\Delta S < 0)$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$,the process is spontaneous when $\Delta G < 0$,which is favored at low temperatures.

(D) Coagulation is the process of aggregating colloidal particles to form a precipitate.
$A$,$B$,and $C$ involve the aggregation of particles (coagulation).
Peptization is the reverse process,where a freshly prepared precipitate is converted into a colloidal sol by adding a suitable electrolyte (peptizing agent).
Therefore,peptization does not involve coagulation.