MHT CET 2009 Chemistry Question Paper with Answer and Solution

(C) The angular speed $\omega$ is defined as $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the minute hand,the time period $T_{\min} = 60 \text{ minutes}$. Thus,$\omega_{\min} = \frac{2\pi}{60} \text{ rad/min}$.
For the hour hand,the time period $T_{hr} = 12 \text{ hours} = 12 \times 60 \text{ minutes}$. Thus,$\omega_{hr} = \frac{2\pi}{12 \times 60} \text{ rad/min}$.
The ratio of angular speeds is $\frac{\omega_{\min}}{\omega_{hr}} = \frac{2\pi / 60}{2\pi / (12 \times 60)} = \frac{12 \times 60}{60} = 12:1$.

(B) Let $L$ be the original length of the string and $K$ be the force constant of the string.
The final length is given by: $\text{Final length} = \text{Original length} + \text{Elongation}$.
Using Hooke's Law,elongation $\Delta L = \frac{F}{K}$,so $L' = L + \frac{F}{K}$.
For the first condition: $a = L + \frac{4}{K}$ ... $(i)$
For the second condition: $b = L + \frac{5}{K}$ ... $(ii)$
Subtracting $(i)$ from $(ii)$:
$b - a = \frac{5}{K} - \frac{4}{K} = \frac{1}{K} \implies K = \frac{1}{b - a}$.
Substituting $K$ into $(i)$:
$a = L + 4(b - a) \implies a = L + 4b - 4a \implies L = 5a - 4b$.
Now,for a tension of $9 \, N$,the length $L_{9}$ is:
$L_{9} = L + \frac{9}{K} = (5a - 4b) + 9(b - a)$
$L_{9} = 5a - 4b + 9b - 9a = 5b - 4a$.

(D) The relationship between Young's modulus $(Y)$,rigidity modulus $(\eta)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
Given that the Young's modulus is $2.4$ times the rigidity modulus,we have $Y = 2.4\eta$.
Substituting this into the formula: $2.4\eta = 2\eta(1 + \sigma)$.
Dividing both sides by $2\eta$,we get $1.2 = 1 + \sigma$.
Solving for $\sigma$,we find $\sigma = 1.2 - 1 = 0.2$.

(A) The root mean square velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we have $T \propto v_{rms}^2$.
Given that the final velocity $v_2 = \frac{1}{2} v_1$,we can write the ratio as $\frac{T_2}{T_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
The initial temperature $T_1 = 327^{\circ}C = 327 + 273 = 600 K$.
Therefore,$T_2 = \frac{T_1}{4} = \frac{600}{4} = 150 K$.
Converting the final temperature back to Celsius: $T_2(^{\circ}C) = 150 - 273 = -123^{\circ}C$.

(A) Let $l_1$ and $l_2$ be the lengths of the air column for the first and second resonance,and $e$ be the end correction.
For the first resonance: $l_1 + e = \frac{\lambda}{4}$ (Equation $1$)
For the second resonance: $l_2 + e = \frac{3\lambda}{4}$ (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$\frac{l_2 + e}{l_1 + e} = 3$
$l_2 + e = 3l_1 + 3e$
$2e = l_2 - 3l_1$
$e = \frac{l_2 - 3l_1}{2}$
Substituting the given values $l_1 = 22.7 \ cm$ and $l_2 = 70.2 \ cm$:
$e = \frac{70.2 - 3(22.7)}{2} = \frac{70.2 - 68.1}{2} = \frac{2.1}{2} = 1.05 \ cm$.

(C) The magnetic field $B$ at the center of a long solenoid is given by the formula $B = \mu_0 n' i$,where $n'$ is the number of turns per unit length.
In this problem,there are $n$ layers of windings,and each layer has $N$ turns over a length $L$.
Therefore,the total number of turns per unit length is $n' = \frac{n \times N}{L}$.
Substituting this into the formula,we get $B = \mu_0 \left( \frac{nN}{L} \right) i$.
As we can see from the expression,the magnetic field $B$ depends on the permeability of free space $\mu_0$,the number of layers $n$,the number of turns $N$,the current $i$,and the length $L$.
It does not depend on the diameter $D$ of the solenoid.
Thus,the magnetic field is independent of $D$.

(C) The given equations are $e = 200 \sin(314t)$ and $i = \sin(314t + \frac{\pi}{3})$.
Comparing these with the standard forms $e = E_0 \sin(\omega t)$ and $i = I_0 \sin(\omega t + \phi)$,we get peak values $E_0 = 200 \text{ V}$ and $I_0 = 1 \text{ A}$,and phase difference $\phi = \frac{\pi}{3}$.
The root mean square values are $V_{rms} = \frac{E_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} \text{ V}$ and $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{1}{\sqrt{2}} \text{ A}$.
The average power consumed is given by $P = V_{rms} I_{rms} \cos \phi$.
Substituting the values: $P = \left(\frac{200}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) \cos(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = 0.5$,we have $P = \frac{200}{2} \times 0.5 = 100 \times 0.5 = 50 \text{ W}$.

(D) For the first circuit (inductor only),the current is given by $i = \frac{V}{X_L} = \frac{V}{\omega L} = \frac{V}{2\pi f L}$.
As the frequency $f$ increases,the inductive reactance $X_L = 2\pi f L$ increases,which causes the current $i$ to decrease.
For the second circuit (capacitor only),the current is given by $i = \frac{V}{X_C} = V\omega C = V(2\pi f)C$.
As the frequency $f$ increases,the capacitive reactance $X_C = \frac{1}{2\pi f C}$ decreases,which causes the current $i$ to increase.
Therefore,the current decreases in the first circuit and increases in the second circuit.

(A) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.
Rearranging this equation,we get $2\pi r = n \left( \frac{h}{mv} \right)$.
From the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
Substituting this into the Bohr's condition,we get $2\pi r = n\lambda$.
For the first orbit,$n = 1$.
Therefore,$2\pi r = 1 \cdot \lambda$,which implies $\lambda = 2\pi r$.

(D) The condition for the $n$-th bright fringe in a Young's double-slit experiment is given by $y = \frac{n \lambda D}{d}$.
Since the position $y$ on the screen is the same for both cases,we have $y = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$.
This simplifies to $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 3$,$\lambda_1 = 700 \, nm$,and $n_2 = 5$,we substitute these values into the equation:
$3 \times 700 = 5 \times \lambda_2$.
$\lambda_2 = \frac{2100}{5} = 420 \, nm$.
Therefore,the required wavelength is $420 \, nm$.

(D) The equation for displacement in simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.
Given $T = 4 \ s$,we have $\omega = \frac{2\pi}{4} = \frac{\pi}{2} \ rad/s$.
We need to find the time $t$ when the displacement $y = \frac{A}{2}$.
Substituting these values into the equation: $\frac{A}{2} = A \sin(\frac{\pi}{2} \cdot t)$.
This simplifies to $\sin(\frac{\pi}{2} \cdot t) = \frac{1}{2}$.
Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$,we equate the arguments: $\frac{\pi}{2} \cdot t = \frac{\pi}{6}$.
Solving for $t$: $t = \frac{1}{3} \ s$.

(A) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc of radius $R$ is $M_1 = \sigma \pi R^2$ and its moment of inertia is $I_1 = \frac{1}{2} M_1 R^2 = \frac{1}{2} (\sigma \pi R^2) R^2 = \frac{1}{2} \sigma \pi R^4$.
The mass of the removed circular portion of radius $r$ is $M_2 = \sigma \pi r^2$ and its moment of inertia is $I_2 = \frac{1}{2} M_2 r^2 = \frac{1}{2} (\sigma \pi r^2) r^2 = \frac{1}{2} \sigma \pi r^4$.
The mass of the remaining annular disc is $M = M_1 - M_2 = \sigma \pi (R^2 - r^2)$,which implies $\sigma \pi = \frac{M}{R^2 - r^2}$.
The moment of inertia of the annular disc is $I = I_1 - I_2 = \frac{1}{2} \sigma \pi (R^4 - r^4)$.
Substituting $\sigma \pi$,we get $I = \frac{1}{2} \left( \frac{M}{R^2 - r^2} \right) (R^4 - r^4) = \frac{1}{2} \left( \frac{M}{R^2 - r^2} \right) (R^2 - r^2)(R^2 + r^2) = \frac{1}{2} M(R^2 + r^2)$.

(C) The angular speed $\omega$ is defined as the rate of change of angular displacement,given by $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the minute hand,the time period $T_m = 1 \text{ hour}$. Thus,$\omega_m = \frac{2\pi}{1} = 2\pi \text{ rad/hr}$.
For the hour hand,the time period $T_h = 12 \text{ hours}$. Thus,$\omega_h = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/hr}$.
The ratio of the angular speed of the minute hand to the hour hand is $\frac{\omega_m}{\omega_h} = \frac{2\pi}{\frac{2\pi}{12}} = \frac{12}{1}$.
Therefore,the ratio is $12 : 1$.

(D) The relationship between Young's modulus $(Y)$, modulus of rigidity $(\eta)$, and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
Given that the Young's modulus is $2.4$ times the modulus of rigidity, we have: $Y = 2.4\eta$.
Substituting this into the formula:
$2.4\eta = 2\eta(1 + \sigma)$
Dividing both sides by $2\eta$:
$1.2 = 1 + \sigma$
Solving for $\sigma$:
$\sigma = 1.2 - 1 = 0.2$.
Thus, the Poisson's ratio is $0.2$.

(D) The current in an inductive circuit is given by $I_{L} = \frac{V}{X_{L}}$,where $X_{L} = \omega L = 2\pi f L$. Thus,$I_{L} = \frac{V}{2\pi f L}$. As frequency $f$ increases,$X_{L}$ increases,so $I_{L}$ decreases.
The current in a capacitive circuit is given by $I_{C} = \frac{V}{X_{C}}$,where $X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. Thus,$I_{C} = V(2\pi f C)$. As frequency $f$ increases,$X_{C}$ decreases,so $I_{C}$ increases.
Therefore,the current decreases in the first circuit (inductor) and increases in the second circuit (capacitor).

(B) Cannizzaro's reaction is given only by those aldehydes that do not contain any $\alpha$-hydrogen atoms.
$CH_{3}CHO$ (acetaldehyde) contains three $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,it undergoes aldol condensation instead of Cannizzaro's reaction.

(B) Hardening of oil involves the addition of hydrogen $(H_2)$ to unsaturated vegetable oils in the presence of a catalyst such as $Ni$,$Pd$,or $Pt$ to form saturated solid fats.
This process is called hydrogenation.

(C) The equation given in option $(A)$ is the Arrhenius equation.
The equation given in option $(B)$ is the Clausius-Clapeyron equation.
The equation given in option $(C)$ is Kirchoff's equation,which relates the change in enthalpy of a reaction to the change in temperature using the heat capacity difference $\Delta C_{p}$.
The equation given in option $(D)$ is the van't Hoff equation.

(C) The general formula for both ethers and alcohols is $C_n H_{2n+2} O$.
In ethers,the functional group is an ether linkage $(-O-)$,while in alcohols,the functional group is a hydroxyl group $(-OH)$.
Since they possess different functional groups despite having the same molecular formula,they are classified as functional isomers.

(C) The structural formula of crotonaldehyde is $CH_3-CH=CH-CHO$.
Numbering the carbon chain starting from the aldehyde group $(-CHO)$ as carbon-$1$,we get:
$CH_3(4)-CH(3)=CH(2)-CHO(1)$.
Since there are $4$ carbon atoms,the parent alkane is butane. The double bond is at position $2$,and the aldehyde group is at position $1$. Therefore,the $IUPAC$ name is but-$2$-en-$1$-al.

(C) The inductive effect is the permanent displacement of sigma electrons along a carbon chain due to the presence of an atom or group with a different electronegativity.
Groups that are electron-withdrawing exhibit a negative inductive effect ($-I$ effect).
Among the given options,$-CH_3$,$-CH_2-CH_3$,and $-(CH_3)_2-CH^{-}$ are electron-donating groups (alkyl groups or anions),which exhibit a positive inductive effect ($+I$ effect).
$-NH_2$ contains a highly electronegative nitrogen atom,which withdraws electron density from the carbon chain,thus exhibiting a negative inductive effect ($-I$ effect).

(C) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In butan$-2-$ol $(CH_3-CH_2-CH(OH)-CH_3)$,the second carbon atom is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$.
Therefore,butan$-2-$ol is optically active.

(B) The reaction of acetylene $(CH \equiv CH)$ with excess of $HCl$ proceeds in two steps following Markovnikov's rule:
$1$. $CH \equiv CH + HCl \rightarrow CH_2 = CHCl$ (Vinyl chloride)
$2$. $CH_2 = CHCl + HCl \rightarrow CH_3CHCl_2$ (Ethylidene dichloride)
Thus,ethylidene dichloride is obtained from acetylene.

(D) The concentration of $H^{+}$ ions is given by $[H^{+}] = C \cdot \alpha = 0.02 \times 0.04 = 8 \times 10^{-4} \ M$.
Using the ionic product of water,$[H^{+}][OH^{-}] = 10^{-14}$.
Therefore,$[OH^{-}] = \frac{10^{-14}}{8 \times 10^{-4}} = 1.25 \times 10^{-11} \ M$.

(C) Given: $[H^{+}] = 0.2 \text{ M}$,Volume $V = 10 \text{ cm}^3 = 10^{-2} \text{ L}$.
At $298 \text{ K}$,the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$.
$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{0.2} = 5 \times 10^{-14} \text{ mol/L}$.
Number of $OH^{-}$ ions $= [OH^{-}] \times V \times N_A$.
$= (5 \times 10^{-14} \text{ mol/L}) \times (10^{-2} \text{ L}) \times (6.022 \times 10^{23} \text{ ions/mol}) \approx 3.011 \times 10^8$.
Note: The calculated value is $3.011 \times 10^8$. The provided option $(c)$ $3 \times 10^{12}$ is mathematically inconsistent with the given parameters.

(B) The degree of dissociation $(\alpha)$ is almost unity for strong electrolytes,which dissociate completely in aqueous solution.
$KCl$ (Potassium chloride) is a strong electrolyte.
$NH_4Cl$ (Ammonium chloride) is a salt of a weak base and a strong acid,which undergoes hydrolysis.
$CH_3COONa$ (Sodium acetate) is a salt of a weak acid and a strong base,which also undergoes hydrolysis.
Therefore,only $KCl$ has a degree of dissociation close to $1$.

(C) The concentration of hydrogen ions $[H^{+}]$ is given by the product of the degree of dissociation $(\alpha)$ and the molar concentration $(C)$.
$[H^{+}] = \alpha \cdot C = 10^{-3} \cdot 0.01 = 10^{-5} \ M$.
Now,calculate the $pH$ using the formula $pH = -\log [H^{+}]$.
$pH = -\log(10^{-5}) = 5$.
Finally,calculate the $pOH$ using the relation $pH + pOH = 14$.
$pOH = 14 - 5 = 9$.

(C) Given: Initial $pH = 3$.
Since $pH = -\log[H^{+}]$,we have $[H^{+}]_1 = 10^{-pH} = 1 \times 10^{-3} \ M$.
Using the dilution formula $M_1 V_1 = M_2 V_2$:
$(1 \times 10^{-3} \ M) \times (20 \ mL) = M_2 \times (100 \ mL)$.
$M_2 = \frac{1 \times 10^{-3} \times 20}{100} = 2 \times 10^{-4} \ M$.
Therefore,the final $H^{+}$ ion concentration is $2 \times 10^{-4} \ M$.

(C) For a solution of the same substance,the dilution formula is used:
$N_1V_1 = N_2V_2$
Here,$N_1 = 2 \ N$ and $V_1 = 0.1 \ dm^3$.
$A$ decinormal solution means $N_2 = \frac{1}{10} \ N = 0.1 \ N$.
Substituting the values:
$2 \ N \times 0.1 \ dm^3 = 0.1 \ N \times V_2$
$V_2 = \frac{2 \times 0.1}{0.1} = 2 \ dm^3$.

(C) $\gamma$-rays are electromagnetic radiations with zero charge and zero rest mass.
Due to their electromagnetic nature,they travel with the speed of light,which is approximately $3 \times 10^8 \ m/s$.

(B) The formula for work done in an isothermal reversible expansion is $W = -2.303 \ nRT \log \frac{V_2}{V_1}$.
Given:
Mass of $O_2 = 16 \ g$,Molar mass of $O_2 = 32 \ g/mol$,so $n = \frac{16}{32} = 0.5 \ mol$.
$T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$V_1 = 10 \ dm^3$,$V_2 = 100 \ dm^3$.
Substituting the values:
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log \frac{100}{10}$
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log(10)$
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times 1$
$W \approx -2872.9 \ J \approx -2875 \ J$ (rounding to the nearest option).

(A) For an ideal gas,the work done in an expansion process is related to the number of moles $n$ by the ideal gas equation $pV = nRT$.
Since $p$,$V$,and $T$ are constant,$n$ must be constant for all gases.
Given that the mass $m$ of each gas is the same,the number of moles is given by $n = \frac{m}{M}$,where $M$ is the molar mass.
For $n$ to be constant for a fixed mass $m$,the molar mass $M$ must be the same,which is not possible for different gases.
However,if we consider the work done per unit mass or the expansion work capacity,$W \propto n = \frac{m}{M}$.
Since $m$ is constant,$W \propto \frac{1}{M}$.
Comparing the molar masses: $M(NH_{3}) = 17 \ g/mol$,$M(N_{2}) = 28 \ g/mol$,$M(Cl_{2}) = 71 \ g/mol$,and $M(H_{2}S) = 34 \ g/mol$.
Since $NH_{3}$ has the lowest molar mass,it will have the highest number of moles for a given mass,resulting in the maximum work done.

(C) According to the first law of thermodynamics:
$\Delta U = q + w$
For an adiabatic process,there is no exchange of heat with the surroundings,so $q = 0$.
Substituting this into the equation,we get $\Delta U = w$.
This implies that the work done $(w)$ is equal to the change in internal energy $(\Delta U)$.
If work is done by the system $(w < 0)$,the internal energy decreases $(\Delta U < 0)$,meaning work is done at the expense of internal energy.

(C) The molar mass of methane $(CH_{4})$ is $16 \ g \ mol^{-1} = 16 \times 10^{-3} \ kg \ mol^{-1}$.
Given that the heat of combustion for $1 \ mol$ $(16 \times 10^{-3} \ kg)$ of $CH_{4}$ is $-800 \ kJ$.
To find the heat of combustion for $4 \times 10^{-4} \ kg$ of $CH_{4}$,we use the unitary method:
$\Delta H = \frac{-800 \ kJ}{16 \times 10^{-3} \ kg} \times (4 \times 10^{-4} \ kg)$
$\Delta H = -800 \times \frac{4 \times 10^{-4}}{16 \times 10^{-3}}$
$\Delta H = -800 \times \frac{4}{16} \times 10^{-1}$
$\Delta H = -800 \times 0.25 \times 0.1 = -20 \ kJ$.

(B) The given voltage is $E = E_0 \sin(\omega t)$,where $E_0 = 200 \sqrt{2} \text{ V}$ and $\omega = 100 \text{ rad/s}$.
The capacitive reactance is $X_C = \frac{1}{\omega C}$.
Given $C = 1 \mu F = 1 \times 10^{-6} \text{ F}$.
$X_C = \frac{1}{100 \times 10^{-6}} = 10^4 \Omega$.
The $RMS$ voltage is $E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \text{ V}$.
The reading of the $AC$ ammeter is the $RMS$ current $I_{rms} = \frac{E_{rms}}{X_C}$.
$I_{rms} = \frac{200}{10^4} = 2 \times 10^{-2} \text{ A} = 20 \text{ mA}$.

(B) Let $L$ be the original length of the string and $k$ be the force constant of the string. The final length $L'$ is given by the sum of the original length and the elongation: $L' = L + \frac{F}{k}$.
For the first condition: $a = L + \frac{4}{k}$ $(i)$
For the second condition: $b = L + \frac{5}{k}$ (ii)
Subtracting equation $(i)$ from (ii): $b - a = \frac{5}{k} - \frac{4}{k} = \frac{1}{k}$,which implies $k = \frac{1}{b-a}$.
Substituting $k$ into equation $(i)$: $L = a - 4(b-a) = a - 4b + 4a = 5a - 4b$.
Now,for a longitudinal tension of $9 \ N$,the length $L_{new}$ is:
$L_{new} = L + \frac{9}{k} = (5a - 4b) + 9(b-a) = 5a - 4b + 9b - 9a = 5b - 4a$.

(D) In a stationary wave, the displacement of a particle is given by $y = A \sin(kx) \cos(\omega t)$.
The velocity of the particle is $v = \frac{\partial y}{\partial t} = -A\omega \sin(kx) \sin(\omega t)$.
The kinetic energy of the string is maximum when the velocity of all particles is maximum.
This occurs when $\sin(\omega t) = \pm 1$, which implies $\cos(\omega t) = 0$.
Substituting $\cos(\omega t) = 0$ into the displacement equation, we get $y = 0$ for all $x$.
Therefore, at the instant of maximum kinetic energy, the string appears as a straight line along the equilibrium position.

(B) According to Hooke's Law,the extension of an elastic string is proportional to the applied tension. Let $L_0$ be the natural length of the string and $k$ be the force constant.
The length $L$ under tension $T$ is given by $L = L_0 + \frac{T}{k}$.
For $T_1 = 4 \ N$,$L_1 = a = L_0 + \frac{4}{k}$ --- $(1)$
For $T_2 = 5 \ N$,$L_2 = b = L_0 + \frac{5}{k}$ --- $(2)$
Subtracting $(1)$ from $(2)$: $b - a = \frac{5}{k} - \frac{4}{k} = \frac{1}{k} \Rightarrow k = \frac{1}{b-a}$.
Substituting $k$ into $(1)$: $L_0 = a - 4(b-a) = a - 4b + 4a = 5a - 4b$.
Now,for $T_3 = 9 \ N$,the length $x$ is:
$x = L_0 + \frac{9}{k} = (5a - 4b) + 9(b - a)$
$x = 5a - 4b + 9b - 9a$
$x = 5b - 4a$.

(C) Only those aldehydes which do not have $\alpha-H$ atoms undergo Cannizzaro's reaction.
$CH_3CHO$ (acetaldehyde) contains $3$ $\alpha-H$ atoms attached to the $\alpha$-carbon.
Therefore,it undergoes aldol condensation instead of Cannizzaro's reaction.
$C_6H_5CHO$,$HCHO$,and $CCl_3CHO$ do not have any $\alpha-H$ atoms and thus undergo Cannizzaro's reaction.

(C) When $Ethyl \ alcohol$ $(C_{2}H_{5}OH)$ is treated with $H_{2}SO_{4}$ at $413 \ K$,the reaction proceeds via an $S_{N}2$ mechanism to form $Diethyl \ ether$ $(C_{2}H_{5}OC_{2}H_{5})$.
Step $1$: Protonation of $Ethyl \ alcohol$ to form protonated $Ethyl \ alcohol$.
Step $2$: Nucleophilic attack of another molecule of $Ethyl \ alcohol$ on the protonated $Ethyl \ alcohol$ with the loss of a water molecule.
Step $3$: Loss of a proton to form $Diethyl \ ether$.

(B) The reduction of aldehydes gives primary alcohols,and the reduction of ketones gives secondary alcohols.
$1$. Butan$-1-$ol (primary alcohol) can be obtained by the reduction of butanal.
$2$. Butan$-2-$ol (secondary alcohol) can be obtained by the reduction of butan$-2-$one.
$3$. $2-$methyl propan$-1-$ol (primary alcohol) can be obtained by the reduction of $2-$methyl propanal.
$4$. $2-$methyl propan$-2-$ol is a tertiary alcohol. Tertiary alcohols cannot be prepared by the reduction of carbonyl compounds (aldehydes or ketones) as they require the addition of a Grignard reagent to a ketone.

(A) $2, 2-$dichloropentane on hydrolysis gives $pentan-2-one$ because the geminal dichloride is at the $C-2$ position.
$3, 3-$dichloropentane on hydrolysis gives $pentan-3-one$.
$Pentan-3-ol$ on oxidation gives $pentan-3-one$.
$Pent-2-yne$ on hydration in the presence of $Hg^{2+}/H^+$ gives $pentan-2-one$ and $pentan-3-one$ as a mixture.
Therefore,$2, 2-$dichloropentane is the correct answer as it exclusively yields $pentan-2-one$.

(B) Since the obtained compound reduces Tollen's reagent,it must be an aldehyde.
Hydrolysis of a gem-dihalide (where both halogen atoms are on the same carbon) yields a gem-diol,which is unstable and loses a water molecule to form an aldehyde or ketone.
For an aldehyde to be formed,the dihaloalkane must be a terminal gem-dihalide (i.e.,the $-Cl$ atoms are at the $C_{1}$ position).
Thus,$CH_{3}CH_{2}CHCl_{2}$ ($1,1$-dichloropropane) on hydrolysis gives propanal $(CH_{3}CH_{2}CHO)$,which reduces Tollen's reagent.
The reaction is:
$CH_{3}CH_{2}CHCl_{2}$ $\xrightarrow{H_{2}O} CH_{3}CH_{2}CH(OH)_{2}$ $\xrightarrow{-H_{2}O} CH_{3}CH_{2}CHO$ $\xrightarrow{\text{Tollen's reagent}} CH_{3}CH_{2}COOH + Ag \downarrow$

(B) The reaction sequence is as follows:
$1$. Ethyl amine $(C_2H_5NH_2)$ reacts with nitrous acid $(HNO_2)$ to form ethanol $(C_2H_5OH)$,which is compound '$A$'.
$2$. Ethanol $(C_2H_5OH)$ reacts with phosphorus pentachloride $(PCl_5)$ to form ethyl chloride $(C_2H_5Cl)$,which is compound '$B$'.
$3$. Ethyl chloride $(C_2H_5Cl)$ reacts with ammonia $(NH_3)$ via nucleophilic substitution to form ethyl amine $(C_2H_5NH_2)$,which is compound '$C$'.
Thus,the compound '$C$' is $C_2H_5NH_2$.

(D) Secondary amines (containing the $R_2NH$ group) react with nitrous acid $(HNO_2)$ to form $N$-nitrosoamines,which are yellow oily liquids.
Among the given options:
$(a)$ $2-$methyl aniline is a primary aromatic amine.
$(b)$ Methyl amine $(CH_3NH_2)$ is a primary aliphatic amine.
$(c)$ Benzyl amine $(C_6H_5CH_2NH_2)$ is a primary aliphatic amine.
$(d)$ Diethyl amine $((C_2H_5)_2NH)$ is a secondary aliphatic amine.
Therefore,diethyl amine reacts with $HNO_2$ to form $N$-nitrosodiethylamine,which is a yellow oily liquid.
Reaction: $(C_2H_5)_2NH + HNO_2 \rightarrow (C_2H_5)_2N-N=O + H_2O$.

(D) Stachyose is a tetrasaccharide consisting of two $\alpha-D$-galactose units,one $\alpha-D$-glucose unit,and one $\beta-D$-fructose unit.
These units are sequentially linked as $gal(\alpha 1$ $\rightarrow 6)gal(\alpha 1$ $\rightarrow 6)glc(\alpha 1$ $\rightarrow 2\beta)fru$.
It is naturally found in numerous vegetables and plants.

(A) In Millon's test,the protein is first treated with $HNO_3$ to form nitrated derivatives.
Upon addition of Millon's reagent (a solution of $Hg(NO_3)_2$ and $Hg(NO_3)_2$ in $HNO_3$),a white precipitate is initially formed due to the coagulation of proteins.
On heating,this precipitate turns red,which is a characteristic test for the presence of the phenolic group in the amino acid tyrosine.

(D) Formic acid $(HCOOH)$ cannot be prepared from methyl magnesium bromide $(CH_{3}MgBr)$ because Grignard reagents react with $CO_{2}$ to form carboxylic acids with at least two carbon atoms (e.g.,acetic acid).
Formic acid can be prepared from the other reagents as follows:
$(a)$ $CH_{3}OH$ $\xrightarrow{[O]} HCHO$ $\xrightarrow{[O]} HCOOH$
$(b)$ $CO + NaOH$ $\longrightarrow HCOONa$ $\xrightarrow{H_{2}SO_{4}} HCOOH + NaHSO_{4}$
$(c)$ Glycerol + Oxalic acid $\xrightarrow{383 \ K} HCOOH + \text{glycerol}$

(B) The reaction of a silver salt of a carboxylic acid with an alkyl halide $(R-X)$ is known as the alkylation of carboxylate ions.
This reaction proceeds via a nucleophilic substitution mechanism $(S_N2)$ where the carboxylate ion $(R'COO^-)$ acts as a nucleophile and attacks the alkyl halide.
The general reaction is: $R'COOAg + R-X \rightarrow R'COOR + AgX \downarrow$.
The product formed is an ester.

(B) The general unit of the rate constant for an $n^{\text{th}}$ order reaction is given by the formula: $\text{mol}^{1-n} \ \text{L}^{n-1} \ \text{s}^{-1}$.
Given that the unit of the rate constant is $\text{time}^{-1}$,which is equivalent to $\text{mol}^{0} \ \text{L}^{0} \ \text{s}^{-1}$.
Comparing the exponents of the units:
For $\text{mol}$: $1 - n = 0 \implies n = 1$.
For $\text{L}$: $n - 1 = 0 \implies n = 1$.
Therefore,the reaction is of the first order.

(B) The standard integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
By using the logarithmic property $\log(\frac{x}{y}) = -\log(\frac{y}{x})$,we can rewrite the equation as $k = \frac{-2.303}{t} \log \frac{a-x}{a}$.
Therefore,option $B$ is a mathematically correct representation of the integrated rate equation for a first-order reaction.

(A) The binding energy $(BE)$ released is related to the mass defect $(\Delta m)$ by the equation: $BE = \Delta m \times 931 \ MeV$.
Given $BE = 306.3 \ MeV$,we calculate the mass defect:
$\Delta m = \frac{306.3}{931} = 0.3290 \ u$.
The mass defect is defined as: $\Delta m = \text{Calculated mass} - \text{Isotopic mass}$.
Therefore,$\text{Isotopic mass} = \text{Calculated mass} - \Delta m$.
$\text{Isotopic mass} = 40.328 - 0.3290 = 39.998 \ u$.

(B) The nuclear reaction is given by: ${ }_{x} A^{y} \longrightarrow { }_{82} B^{207} + 5 { }_{2} \alpha^{4} + 4 { }_{-1} \beta^{0}$
By balancing the atomic number $(x)$: $x = 82 + (5 \times 2) + (4 \times -1) = 82 + 10 - 4 = 88$
By balancing the mass number $(y)$: $y = 207 + (5 \times 4) + (4 \times 0) = 207 + 20 = 227$
Thus,the element $A$ is ${ }_{88} A^{227}$.
Number of protons = $88$
Number of neutrons = $227 - 88 = 139$

(B) The chemicals that are used to bring down body temperature in high fever are called antipyretics. Examples include $Paracetamol$ and $Analgin$.

(B) Ethyl butyrate $(C_6H_{12}O_2)$ has the characteristic flavour of pineapple.
It is widely used in the food and beverage industry to impart a pineapple flavour to various products,including alcoholic beverages.

(A) The electronic configurations of the given elements are as follows:
$Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$. It has $4$ unpaired electrons.
$Cu$ $(Z=29)$: $[Ar] 3d^{10} 4s^1$. It has $1$ unpaired electron.
$Co$ $(Z=27)$: $[Ar] 3d^7 4s^2$. It has $3$ unpaired electrons.
$Ni$ $(Z=28)$: $[Ar] 3d^8 4s^2$. It has $2$ unpaired electrons.
Thus,the maximum number of unpaired electrons is present in $Fe$.

(D) Due to lanthanide contraction,the ionic radius of $M^{3+}$ ions decreases as the atomic number increases from $Ce$ to $Lu$.
As the size of the $M^{3+}$ ion decreases,the covalent character of the $M-OH$ bond increases,which leads to a decrease in the basic strength of the hydroxides.
The order of ionic radii is $Ce^{3+} > Eu^{3+} > Yb^{3+} > Lu^{3+}$.
Therefore,the order of basic strength of the hydroxides is $Ce(OH)_3 > Eu(OH)_3 > Yb(OH)_3 > Lu(OH)_3$.
Thus,$Ce(OH)_3$ is the most basic hydroxide.

(C) Due to the lanthanoid contraction, the atomic radii of elements of the second transition series are very similar to those of the corresponding elements of the third transition series.
This phenomenon occurs because the $4f$-electrons do not screen the nuclear charge effectively.
As a result, pairs like $Zr-Hf$, $Nb-Ta$, $Mo-W$, and $Tc-Re$ are known as 'chemical twins' due to their nearly identical chemical properties and atomic sizes.

(B) The reaction for the liberation of hydrogen gas is: $H^{+} + e^{-} \longrightarrow \frac{1}{2} H_{2}$.
According to Faraday's law, $1 \,F$ $(96500 \,C)$ of charge liberates $0.5 \,mol$ of $H_{2}$ gas, which occupies $11.2 \,L$ at $STP$.
The total charge passed is $Q = I \times t = 10 \,A \times (80 \times 60) \,s = 48000 \,C$.
The volume of $H_{2}$ gas liberated is given by: $V = \frac{11.2 \,L \times 48000 \,C}{96500 \,C} \approx 5.57 \,L$.

(B) The reduction reaction is: $Al^{3+} + 3e^- \rightarrow Al$.
Moles of $Al = \frac{\text{Mass}}{\text{Atomic mass}} = \frac{4.5 \times 10^{-5} \text{ g}}{27 \text{ g/mol}} = 1.666 \times 10^{-6} \text{ mol}$.
Moles of electrons required $= 3 \times \text{moles of } Al = 3 \times 1.666 \times 10^{-6} = 5 \times 10^{-6} \text{ mol}$.
Number of electrons $= \text{moles of electrons} \times N_A = 5 \times 10^{-6} \times 6.022 \times 10^{23} \approx 3.011 \times 10^{18}$.

(D) The standard cell potential is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
In the given reaction,$Mg$ is oxidized to $Mg^{2+}$ (anode) and $Cu^{2+}$ is reduced to $Cu$ (cathode).
Therefore,$E^{\circ}_{cell} = E^{\circ}_{(Cu^{2+}/Cu)} - E^{\circ}_{(Mg^{2+}/Mg)}$
Substituting the given values: $E^{\circ}_{cell} = 0.337 \ V - (-2.37 \ V)$
$E^{\circ}_{cell} = 0.337 + 2.37 = 2.707 \ V \approx 2.7 \ V$.

(A) The structure of iso-propyl amine is $(CH_3)_2CHNH_2$.
In this molecule,the nitrogen atom is attached to only one carbon atom.
Therefore,it is a $primary$ $(1^{\circ})$ amine.

(A) Willemite is a rare zinc silicate mineral with the chemical formula $Zn_{2}SiO_{4}$.
It exhibits trigonal symmetry and is known for its strong green fluorescence under ultraviolet light.

(D) The reduction of chloroform $(CHCl_{3})$ with zinc dust and water $(Zn / H_{2}O)$ leads to the formation of methane $(CH_{4})$.
The chemical reaction is: $CHCl_{3} + 6[H] \xrightarrow{Zn / H_{2}O} CH_{4} + 3HCl$.
Therefore,the correct product is methane.

(B) In Dow's process,chlorobenzene is treated with $NaOH$ at $573-623 \ K$ and $200 \ atm$ pressure to form sodium phenoxide,which upon subsequent acid hydrolysis with $HCl$ yields phenol as the final product.
The reaction sequence is:
$C_6H_5Cl + 2NaOH \xrightarrow{573-623 \ K, 200 \ atm} C_6H_5ONa + NaCl + H_2O$
$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$
Thus,the starting raw material is chlorobenzene.

(C) Nylon-$66$ is a condensation polymer formed by the reaction between hexamethylenediamine and adipic acid.
$n HOOC(CH_2)_4COOH + n H_2N(CH_2)_6NH_2 \rightarrow [NH(CH_2)_6NHCO(CH_2)_4CO]_n + 2n H_2O$

(B) The electrochemical equivalent $(Z)$ is related to the equivalent mass $(E)$ by the formula: $Z = \frac{E}{F}$,where $F = 96500 \ C/mol$ is Faraday's constant.
Given $Z = 3.29 \times 10^{-4} \ g/C$.
$E = Z \times F = 3.29 \times 10^{-4} \times 96500 = 31.7485 \ g/eq$.
Since copper is a divalent metal $(n = 2)$,the atomic mass is given by: $\text{Atomic mass} = n \times E$.
$\text{Atomic mass} = 2 \times 31.7485 = 63.497 \ g/mol \approx 63.5 \ g/mol$.

(D) Colligative properties depend on the number of solute particles in a solution. The four main colligative properties are:
$1$. Relative lowering of vapour pressure
$2$. Elevation in boiling point
$3$. Depression in freezing point
$4$. Osmotic pressure
Since 'Freezing point' is a physical property of a substance and not a colligative property (which is the 'Depression in freezing point'),option $D$ is the correct answer.

(B) Depression in freezing point is a colligative property,which depends on the number of particles in the solution.
$KCl \rightleftharpoons K^{+} + Cl^{-}$ ($2$ particles)
$Na_{2}SO_{4} \rightleftharpoons 2Na^{+} + SO_{4}^{2-}$ ($3$ particles)
$MgSO_{4} \rightleftharpoons Mg^{2+} + SO_{4}^{2-}$ ($2$ particles)
$MgCO_{3} \rightleftharpoons Mg^{2+} + CO_{3}^{2-}$ ($2$ particles)
Since $Na_{2}SO_{4}$ produces the maximum number of particles ($3$ ions per formula unit),it causes the maximum depression in freezing point.

(A) $1.25$ molal solution means $1.25$ moles of $NaOH$ are present in $1000 \ g$ of solvent.
Molar mass of $NaOH = 40 \ g/mol$.
Mass of $NaOH = 1.25 \times 40 = 50 \ g$.
Mass of solvent $= 1000 \ g$.
Total mass of solution $= 1000 + 50 = 1050 \ g$.
Percentage by weight $= \frac{50}{1050} \times 100 = 4.76 \%$.