IIT JEE 1983 Chemistry Question Paper with Answer and Solution

(B) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the surface of the earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
At a height $h = R$ above the surface,the distance from the center is $r = R + h = R + R = 2R$.
Thus,the potential energy at height $h$ is $U_f = -\frac{GMm}{2R}$.
The gain in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR$.

(A) Initial mass of $CO_2 = 200 \ mg = 0.2 \ g$.
Molar mass of $CO_2 = 44 \ g/mol$.
Initial number of moles of $CO_2 = \frac{0.2 \ g}{44 \ g/mol} = 0.004545 \ mol$.
Initial number of molecules $= 0.004545 \times 6.022 \times 10^{23} \approx 2.737 \times 10^{21}$ molecules.
Number of molecules removed $= 10^{21}$.
Remaining molecules $= 2.737 \times 10^{21} - 1 \times 10^{21} = 1.737 \times 10^{21}$ molecules.
Remaining moles $= \frac{1.737 \times 10^{21}}{6.022 \times 10^{23}} \approx 0.288 \times 10^{-2} = 2.88 \times 10^{-3} \ mol$.
Note: Using $6 \times 10^{23}$ as Avogadro's number gives $2.72 \times 10^{21}$ molecules,leading to $1.72 \times 10^{21}$ remaining,which results in $\frac{1.72}{6} \times 10^{-2} \approx 0.286 \times 10^{-2} = 2.86 \times 10^{-3} \ mol$. The closest option is $2.85$.

(A) Rutherford's $\alpha$-particle scattering experiment demonstrated that the positive charge and most of the mass of an atom are concentrated in a very small central region called the nucleus.
Thus,the experiment provided direct evidence for the existence and size of the nucleus.

(A) The principal quantum number,denoted by $n$,determines the main energy level or shell of an electron.
It primarily provides information about the size of the orbital and the average distance of the electron from the nucleus.
Therefore,the correct option is $A$.

(D) According to the Pauli Exclusion Principle,an orbital can hold a maximum of $2$ electrons,and these electrons must have opposite spins. Therefore,the $p_x$ orbital can accommodate $2$ electrons with opposite spins.

(B) $CCl_4$ has no net dipole moment because of its regular tetrahedral structure.
In a regular tetrahedral geometry,the four $C-Cl$ bond dipoles are oriented at an angle of $109.5^{\circ}$ to each other,which results in the cancellation of individual bond moments,leading to a net dipole moment of zero.

(D) The structure of $1, 2-$butadiene is $CH_2=C=CH-CH_3$.
In this molecule:
- The first carbon atom $(CH_2=)$ is $sp^2$ hybridized.
- The second carbon atom $(=C=)$ is $sp$ hybridized.
- The third carbon atom $(-CH=)$ is $sp^2$ hybridized.
- The fourth carbon atom $(-CH_3)$ is $sp^3$ hybridized.
Thus,the compound contains $sp, sp^2,$ and $sp^3$ hybridized carbon atoms.

(D) Hydrogen bonding occurs in compounds where $H$ is covalently bonded to highly electronegative atoms like $F$,$O$,or $N$.
In $Phenol$ $(C_6H_5OH)$,$H$ is bonded to $O$.
In $Liquid \ NH_3$,$H$ is bonded to $N$.
In $Water$ $(H_2O)$,$H$ is bonded to $O$.
In $Liquid \ HCl$,$Cl$ is not sufficiently electronegative to support hydrogen bonding,and there is no $N$,$O$,or $F$ present.
Therefore,$Liquid \ HCl$ does not form hydrogen bonds.

(C) In $CuSO_4 \cdot 5H_2O$,there is an electrovalent (ionic) bond between $Cu^{2+}$ and $SO_4^{2-}$ ions.
Covalent bonds exist within the $SO_4^{2-}$ ion (between $S$ and $O$ atoms) and within the $H_2O$ molecules (between $H$ and $O$ atoms).
Coordinate covalent bonds exist between the $Cu^{2+}$ ion and four $H_2O$ molecules,forming the complex $[Cu(H_2O)_4]^{2+}$.

(A) Geometrical isomerism is shown by compounds that have restricted rotation around a double bond and where each carbon atom of the double bond is attached to two different groups.
In $2-$butene $(CH_3-CH=CH-CH_3)$,the carbon atoms of the double bond are each bonded to a hydrogen atom $(H)$ and a methyl group $(CH_3)$.
Thus,it exists as cis-$2-$butene and trans-$2-$butene,which are geometrical isomers.

(C) $Hexane$ is an alkane,which is also known as paraffin due to its low reactivity.
Alkanes are non-polar and do not react with or dissolve in concentrated $H_2SO_4$.
In contrast,$Ethylene$ (alkene) reacts via addition,$Benzene$ undergoes sulfonation,and $Aniline$ (base) reacts to form a salt,making them soluble in concentrated $H_2SO_4$.

(C) The reaction of propyne with water in the presence of $40\% \ H_2SO_4$ and $1\% \ HgSO_4$ is an example of hydration of alkynes.
The reaction proceeds via the formation of an enol intermediate:
$CH_3-C \equiv CH + H_2O \xrightarrow{H_2SO_4, HgSO_4} CH_3-C(OH)=CH_2$.
This enol intermediate undergoes tautomerization to form the more stable ketone,which is acetone:
$CH_3-C(OH)=CH_2 \rightarrow CH_3-C(=O)-CH_3$ (Acetone).

(B) Atoms of different elements having different atomic numbers but the same mass number are known as isobars.

(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.
For the points to be collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel,i.e.,$\overrightarrow{AB} = k(\overrightarrow{BC})$ for some scalar $k$.
First,calculate $\overrightarrow{AB} = (40 - 60)i + (-8 - 3)j = -20i - 11j$.
Next,calculate $\overrightarrow{BC} = (a - 40)i + (-52 - (-8))j = (a - 40)i - 44j$.
Setting $\overrightarrow{AB} = k(\overrightarrow{BC})$,we get $-20i - 11j = k((a - 40)i - 44j)$.
Comparing the coefficients of $j$,we have $-11 = -44k$,which gives $k = \frac{1}{4}$.
Comparing the coefficients of $i$,we have $-20 = k(a - 40)$.
Substituting $k = \frac{1}{4}$,we get $-20 = \frac{1}{4}(a - 40)$.
$-80 = a - 40$,which implies $a = -40$.

(A) Rutherford's scattering experiment is related to the size of the nucleus.
Rutherford observed that most of the $\alpha-$particles passed through the gold foil undeflected,indicating that the atom consists of a large empty space.
However,a very small fraction of $\alpha-$particles were deflected at large angles,which led to the conclusion that the positive charge and most of the mass of the atom are concentrated in a very small region called the nucleus.

(C) The Cannizzaro reaction is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking an $\alpha$-hydrogen atom.
Aldehydes that possess at least one $\alpha$-hydrogen atom undergo aldol condensation in the presence of a base,rather than the Cannizzaro reaction.
In $CH_3CHO$ (acetaldehyde),there are three $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,$CH_3CHO$ does not undergo the Cannizzaro reaction.

(C) Since there is no external torque acting on the system,the angular momentum $L$ is conserved.
Initial angular momentum $L_1 = I_1 \omega_1$,where $I_1 = MR^2$ is the moment of inertia of the ring.
Final angular momentum $L_2 = I_2 \omega_2$,where $I_2 = MR^2 + 2(mR^2) = (M + 2m)R^2$ is the moment of inertia of the ring with the two masses attached.
By the law of conservation of angular momentum,$I_1 \omega_1 = I_2 \omega_2$.
Substituting the values: $(MR^2) \omega = (M + 2m)R^2 \omega_2$.
Solving for $\omega_2$: $\omega_2 = \frac{MR^2 \omega}{(M + 2m)R^2} = \frac{M \omega}{M + 2m}$.

(B) The initial moment of inertia of the ring about its axis is $I_1 = Mr^2$.
The initial angular momentum of the system is $L_1 = I_1 \omega = Mr^2 \omega$.
When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia of the system becomes $I_2 = Mr^2 + mr^2 + mr^2 = (M + 2m)r^2$.
Since no external torque acts on the system,the angular momentum is conserved,so $L_1 = L_2$.
$Mr^2 \omega = (M + 2m)r^2 \omega'$
Solving for the new angular velocity $\omega'$,we get $\omega' = \frac{M \omega}{M + 2m}$.

(C) The initial moment of inertia of the ring about its axis is $I_1 = MR^2$.
The initial angular momentum of the system is $L_1 = I_1 \omega = MR^2 \omega$.
When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia of the system becomes $I_2 = MR^2 + mR^2 + mR^2 = (M + 2m)R^2$.
Since no external torque acts on the system,the angular momentum is conserved,so $L_1 = L_2$.
$MR^2 \omega = (M + 2m)R^2 \omega^{\prime}$.
Dividing both sides by $R^2$,we get $M \omega = (M + 2m) \omega^{\prime}$.
Therefore,the new angular velocity is $\omega^{\prime} = \frac{\omega M}{M + 2m}$.

(B) Faraday's first law of electrolysis states that the amount of a substance deposited or dissolved at an electrode is directly proportional to the charge $(Q)$ passing through the electrolyte.
Faraday's second law states that when the same quantity of electricity is passed through different electrolytes,the masses of different substances deposited at the electrodes are proportional to their chemical equivalent weights $(E)$.
Therefore,Faraday's laws are related to the equivalent weight of the electrolyte.

(B) $Cl_2 + H_2O \to 2HCl + [O]$ (Nascent oxygen)
$\text{Coloured matter} + [O] \xrightarrow{\text{Bleaching}} \text{Colourless matter}$ (Oxidation)
Chlorine acts as a bleaching agent only in the presence of moisture because it produces nascent oxygen,which is responsible for the bleaching action.

(B) According to Raoult's law for solutions containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P_1^o - P_1}{P_1^o} = x_2$,where $x_2$ is the mole fraction of the solute.
Thus,the relative lowering of vapour pressure is equal to the mole fraction of the solute.

(D) The specific rate constant $(k)$ of a reaction is independent of the concentration of reactants or products and the time of the reaction.
It is a characteristic constant for a given reaction at a specific temperature.
According to the Arrhenius equation,$k = A e^{-E_a / RT}$,the rate constant depends significantly on the temperature of the reaction.

(C) Faraday's first law of electrolysis states that the mass $(w)$ of the substance deposited or liberated at an electrode is directly proportional to the quantity of electricity $(Q)$ passed through the electrolyte.
Mathematically,$w = Z \times Q = Z \times i \times t$.
Here,$Z$ is the electrochemical equivalent,which is defined as $Z = \frac{E}{F}$,where $E$ is the equivalent weight of the substance and $F$ is Faraday's constant.
Therefore,$w = \frac{E \times i \times t}{F}$.
Since $F$ is constant,if the current $(i)$ and time $(t)$ are kept constant,then $w \propto E$.
Thus,Faraday's laws are related to the equivalent weight of the electrolyte.

(B) The formula for equivalent conductivity is $\Lambda_{eq} = \frac{\kappa \times 1000}{C}$.
Here,$\kappa$ is the specific conductivity $(ohm^{-1} \ cm^{-1})$ and $C$ is the concentration in $gm \ equivalent \ L^{-1}$ (or $gm \ equivalent \ cm^{-3} \times 1000$).
Substituting the units: $\Lambda_{eq} = \frac{ohm^{-1} \ cm^{-1}}{gm \ equivalent \ cm^{-3}} = ohm^{-1} \ cm^2 \ (gm \ equivalent)^{-1}$.
Therefore,the correct unit is $ohm^{-1} \ cm^2 \ (gm \ equivalent)^{-1}$.

(C) catalyst is a substance that increases the rate of a chemical reaction by providing an alternative pathway with a lower activation energy,thereby shortening the time required to reach equilibrium without affecting the equilibrium constant or the final concentration of the products.

(C) In the alumino-thermite process,aluminium $(Al)$ is used to reduce metal oxides (like $Fe_2O_3$) to their respective metals.
Since aluminium undergoes oxidation $(Al \rightarrow Al^{3+} + 3e^-)$ and reduces the metal oxide,it acts as a reducing agent.

(C) When diethyl ether $(C_2H_5-O-C_2H_5)$ is heated with excess concentrated $HI$,it undergoes cleavage to form $2$ moles of ethyl iodide $(C_2H_5I)$ and $1$ mole of water $(H_2O)$.
The reaction is: $C_2H_5-O-C_2H_5 + 2HI \to 2C_2H_5I + H_2O$.

(C) The Cannizzaro reaction is given by aldehydes that do not possess an $\alpha-H$ atom.
$CH_3CHO$ (acetaldehyde) contains $3$ $\alpha-H$ atoms,therefore,it does not undergo the Cannizzaro reaction.

(B) The reaction of acetamide with $NaOH + Br_2$ is known as the Hofmann bromamide degradation reaction.
In this reaction,an amide is converted into a primary amine with one carbon atom less than the original amide.
$CH_3CONH_2 + Br_2 + 4NaOH \to CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Thus,acetamide gives methyl amine.