AIPMT 2001 Chemistry Question Paper with Answer and Solution

(B) According to Wien's Displacement Law, the product of the wavelength corresponding to maximum spectral emissive power and the absolute temperature is constant.
${\lambda _m} T = \text{constant}$
Therefore, ${\lambda _{m1}} T_1 = {\lambda _{m2}} T_2$
Given:
$T_1 = 2000 \; K$
$T_2 = 3000 \; K$
${\lambda _{m1}} = {\lambda _m}$
Substituting the values:
${\lambda _m} \times 2000 = {\lambda _{m2}} \times 3000$
${\lambda _{m2}} = \frac{2000}{3000} \times {\lambda _m}$
${\lambda _{m2}} = \frac{2}{3}{\lambda _m}$

(C) The density of liquid $HCl$ is $1.17 \ g/cc$.
This means $1 \ cc$ of $HCl$ has a mass of $1.17 \ g$.
Therefore,$1000 \ cc$ $(1 \ L)$ of $HCl$ has a mass of $1170 \ g$.
The molar mass of $HCl$ is $1.008 + 35.45 = 36.458 \ g/mol$,which is approximately $36.5 \ g/mol$.
Molarity = $\frac{\text{mass of solute in grams}}{\text{molar mass} \times \text{volume of solution in litres}}$.
Molarity = $\frac{1170 \ g}{36.5 \ g/mol \times 1 \ L} = 32.05 \ mol/L$.

(A) The minimum molecular weight of an enzyme is calculated assuming that at least one atom of the element $(Se)$ is present in one molecule of the enzyme.
Given that $0.5 \ g$ of $Se$ is present in $100 \ g$ of the enzyme.
Therefore,$78.4 \ g$ (atomic weight of $Se$) of $Se$ will be present in $\frac{100 \times 78.4}{0.5} \ g$ of the enzyme.
Calculation: $\frac{7840}{0.5} = 15680 = 1.568 \times 10^4 \ g/mol$.
Thus,the minimum molecular weight is $1.568 \times 10^4$.

(A) The hardness of water is removed by reacting $Mg^{2+}$ ions with washing soda $(Na_2CO_3)$.
The reaction is: $Mg^{2+} + Na_2CO_3 \rightarrow MgCO_3 + 2Na^+$.
Number of milli-equivalents (meq) of $Mg^{2+} = \frac{\text{mass in mg}}{\text{equivalent weight}}$.
Equivalent weight of $Mg^{2+} = \frac{\text{Atomic mass}}{\text{Valency}} = \frac{24}{2} = 12$.
Number of meq of $Mg^{2+} = \frac{12.00 \ mg}{12} = 1 \text{ meq}$.
Since $1 \text{ meq}$ of $Mg^{2+}$ reacts with $1 \text{ meq}$ of $Na_2CO_3$,the required milli-equivalents of washing soda is $1$.

(B) The total number of electrons in the given configuration is $2 + 2 + 1 + 1 + 1 = 7$.
An element with $7$ electrons is Nitrogen $(N)$.
The electronic configuration of Nitrogen $(Z = 7)$ is $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$.

(B) $CH_4$ (Methane) has a regular tetrahedral geometry.
Due to the symmetrical arrangement of the four $C-H$ bonds,the individual bond dipoles cancel each other out,resulting in a net dipole moment of zero.

(B) The chemical equation for the reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.
The expression for the equilibrium constant $K_c$ is given by $K_c = \frac{[HI]^2}{[H_2][I_2]}$.
Given values are $[H_2] = 8.0 \ mol \ L^{-1}$,$[I_2] = 3.0 \ mol \ L^{-1}$,and $[HI] = 28.0 \ mol \ L^{-1}$.
Substituting these values into the expression:
$K_c = \frac{(28.0)^2}{8.0 \times 3.0} = \frac{784}{24} = 32.66$.

(D) process is spontaneous if the change in Gibbs free energy $\Delta G = \Delta H - T\Delta S$ is negative.
For endothermic processes,$\Delta H > 0$.
For the melting of ice $(H_2O(s) \rightarrow H_2O(l))$ and the evaporation of water $(H_2O(l) \rightarrow H_2O(g))$,the entropy change $\Delta S$ is positive.
At temperatures above the melting point or boiling point respectively,the term $T\Delta S$ becomes larger than $\Delta H$,making $\Delta G$ negative.
Thus,both melting of ice and evaporation of water are spontaneous endothermic processes under appropriate conditions.

(B) The stability of a compound is inversely proportional to its enthalpy of formation.
Since the enthalpy of formation of $Y$ $(-156 \ kJ)$ is more negative than that of $X$ $(-84 \ kJ)$,$Y$ has lower potential energy and is more stable than $X$.
Therefore,$X$ is less stable than $Y$.

(C) Lithium nitride is $Li_3N$.
In this compound,the nitride ion is $N^{3-}$.
The atomic number of nitrogen is $7$,which means it has $7$ protons.
Since the ion has a charge of $-3$,it has gained $3$ electrons.
Therefore,the number of electrons = $7 + 3 = 10$.
Thus,the nitride ion is composed of $7P + 10e$.

(D) The correct answer is $(D)$.
$He$ (Helium) has an atomic number of $2$,meaning its electronic configuration is $1s^2$.
It has only $2$ electrons in its outermost shell (duplet) and does not possess an octet ($8$ electrons) like other noble gases.

(C) To determine the $IUPAC$ name,we identify the longest carbon chain containing the functional group.
The structure is $CH_3-CH(CH_3)-CH_2-CH_2Br$.
The longest chain has $4$ carbon atoms (butane).
Numbering starts from the carbon attached to the bromine atom to give it the lowest possible locant.
At position $1$,there is a bromo group,and at position $3$,there is a methyl group.
Therefore,the $IUPAC$ name is $1-$bromo$-3-$methylbutane.

(C) The structure is $(CH_3)_2CH-CH_2-CH_2Br$.
First,identify the longest carbon chain containing the functional group,which is a $4-$carbon chain (butane).
Number the chain starting from the end closer to the substituent (bromine),so $C-1$ is attached to $Br$.
At $C-3$,there is a methyl group $(CH_3)$.
Therefore,the $IUPAC$ name is $1-$bromo$-3-$methylbutane.

(A) The molecular formula $C_3H_7Cl$ corresponds to a saturated alkyl halide.
Two structural isomers are possible for this formula:
$1$. $CH_3-CH_2-CH_2Cl$ ($1$-chloropropane)
$2$. $CH_3-CHCl-CH_3$ ($2$-chloropropane)
Thus,the total number of structural isomers is $2$.

(D) Propane is obtained from propene by catalytic hydrogenation.
In this reaction,propene reacts with hydrogen in the presence of a metal catalyst like $Ni$,$Pt$,or $Pd$ to form propane:
$CH_3-CH=CH_2 + H_2 \xrightarrow{Ni/Pt/Pd} CH_3-CH_2-CH_3$.

(A) The conversion of propene to propane involves the addition of hydrogen across the double bond,which is known as catalytic hydrogenation.
$CH_3-CH=CH_2 + H_2 \xrightarrow{Ni, 300 \ ^oC} CH_3-CH_2-CH_3$
Thus,the correct option is $(A)$.

(C) The reaction $CH_2 = CH_2 + H_2 \xrightarrow[250 - 300^{\circ}C]{Ni} CH_3 - CH_3$ is known as the Sabatier-Senderens reaction.
This process involves the catalytic hydrogenation of unsaturated hydrocarbons (alkenes or alkynes) in the presence of a metal catalyst like $Ni$,$Pt$,or $Pd$ at elevated temperatures to form alkanes.

(A) The spleen is primarily a large,bean-shaped lymphoid organ.
It acts as a filter for the blood by trapping blood-borne microorganisms.
It also contains a large reservoir of erythrocytes and acts as a site for the maturation of lymphocytes.
Therefore,it is classified as a secondary lymphoid organ.

(D) Long-day plants flower only when they are exposed to light periods longer than a critical day length.
Examples include Spinach [Spinacia oleracea] and Sugar beet.
Short-day plants flower only when they are exposed to light periods shorter than a critical day length.
Examples include Glycine max [Soybean] and Tobacco.

(A) Passive absorption of minerals is a process that does not require metabolic energy $(ATP)$.
Since this process occurs along the concentration gradient and does not involve active transport mechanisms,it is not affected by metabolic inhibitors.
However,temperature can influence the rate of passive diffusion by affecting the kinetic energy of the ions.
Therefore,passive absorption depends on temperature.

(A) Phytochrome is a proteinaceous pigment that acts as a photoreceptor in plants.
It exists in two interconvertible forms: $P_r$ (which absorbs red light) and $P_{fr}$ (which absorbs far-red light).
It plays a crucial role in regulating various light-dependent developmental processes such as seed germination,flowering,and stem elongation.

(C) The number of Barr bodies in an individual is calculated using the formula: $N = n - 1$,where $n$ is the total number of $X$ chromosomes.
In a $XXXX$ female,there are $4$ $X$ chromosomes $(n = 4)$.
Therefore,the number of Barr bodies = $4 - 1 = 3$.
Thus,the correct answer is $3$.

(B) According to the Oparin-Haldane hypothesis and the chemical evolution theory,the first life forms on Earth were simple,anaerobic,and chemoheterotrophic organisms. These organisms obtained their energy by consuming organic molecules present in the 'primordial soup' of the primitive ocean. They did not possess the ability to synthesize their own food (autotrophy) or perform photosynthesis,as these are more complex metabolic processes that evolved later.

(A) $1$. Reserpine is a drug derived from the plant $Rauwolfia$ $\text{serpentina}$ and is used as a tranquilizer to treat high blood pressure and mental disorders.
$2$. Cocaine is a stimulant derived from $Erythroxylum$ $\text{coca}$, not an opiate.
$3$. Morphine is an opiate analgesic (painkiller), not a hallucinogen.
$4$. Bhang is a preparation from the cannabis plant, which acts as a hallucinogen, not an analgesic.
Therefore, the correct pair is Reserpine - Tranquilizer.

(A) Phloem loading is the process of transporting sugars from the source (usually leaves) into the sieve tube elements of the phloem.
This process involves the active transport of sucrose into the sieve elements,which increases the solute concentration inside the phloem.
As a result,water enters the phloem from the adjacent xylem by osmosis,creating a high turgor pressure that drives the bulk flow of sap towards the sink.

(C) Glyptal is a polyester (alkyd resin) formed by the condensation of ethylene glycol and phthalic acid.
Nylon-$66$,Nylon-$6$,and Proteins are all polyamides containing the amide linkage $(-CONH-)$.

(B) Polyamides are polymers containing amide linkages $(-CONH-)$ in their backbone.
$1$. Nylon-$6$ is a polyamide formed by the polymerization of caprolactam.
$2$. Nylon-$66$ is a polyamide formed by the condensation of hexamethylenediamine and adipic acid.
$3$. Proteins are natural polyamides formed by the linkage of amino acids via peptide bonds $(-CONH-)$.
$4$. Glyptal is a polyester formed by the condensation of ethylene glycol and phthalic acid. It contains ester linkages $(-COO-)$,not amide linkages.

(B) The general trend for ionisation energy $(IE)$ increases across a period from left to right due to an increase in effective nuclear charge.
However,there are exceptions due to stable electronic configurations.
$Be$ $(1s^2 2s^2)$ has a fully filled $2s$ orbital,making its $IE$ higher than $B$ $(1s^2 2s^2 2p^1)$.
$N$ $(1s^2 2s^2 2p^3)$ has a stable half-filled $p$-orbital,making its $IE$ higher than $O$ $(1s^2 2s^2 2p^4)$.
Therefore,the correct order is $B < Be < C < O < N$.

(B) An electrophile is an electron-deficient species that can accept an electron pair.
$NO_2^+$,$H^+$,and $BF_3$ are all electron-deficient and act as electrophiles.
$Na^+$ is a stable cation with a complete octet configuration $(1s^2 2s^2 2p^6)$. It does not have a tendency to accept electron pairs in typical organic reactions,hence it is not considered an electrophile.

(A) The $Pan$ $troglodytes$ (Chimpanzee) is considered the closest living relative of humans ($Homo$ $sapiens$).
Genetic studies have shown that humans and chimpanzees share approximately $98-99\%$ of their $DNA$ sequences.
This high degree of genetic similarity indicates a very recent common ancestor in the evolutionary timeline compared to other primates like gorillas,orangutans,or gibbons.

(B) $1$. Analyze the circuit: The circuit consists of a $5 \, V$ battery and two parallel branches connected in series with a $20 \, \Omega$ resistor.
$2$. Check diode bias: In the upper branch,the diode is reverse-biased (the $p$-side is connected to the negative terminal relative to the $n$-side),so it acts as an open circuit (no current flows).
$3$. In the lower branch,the diode is forward-biased (the $p$-side is connected to the positive terminal),so it acts as a closed switch with negligible resistance.
$4$. Calculate total resistance: The total resistance in the active path is the sum of the $20 \, \Omega$ resistor in the main line and the $30 \, \Omega$ resistor in the lower branch: $R_{total} = 20 \, \Omega + 30 \, \Omega = 50 \, \Omega$.
$5$. Calculate current: Using Ohm's law,$i = \frac{V}{R_{total}} = \frac{5 \, V}{50 \, \Omega} = \frac{5}{50} \, A$.

(D) For a parallel plate capacitor,the capacitance is $C = \frac{A \varepsilon_{0}}{d}$.
The volume of the capacitor is $V_{vol} = Ad$.
The total energy stored in the capacitor is $U = \frac{1}{2} CV^{2}$.
The energy per unit volume (energy density) is given by $u = \frac{U}{V_{vol}} = \frac{\frac{1}{2} CV^{2}}{Ad}$.
Substituting $C = \frac{A \varepsilon_{0}}{d}$ into the expression:
$u = \frac{\frac{1}{2} (\frac{A \varepsilon_{0}}{d}) V^{2}}{Ad} = \frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$.

(B) $1$. Analyze the circuit: The circuit consists of two parallel branches connected to a $5 \text{ V}$ battery and a $20 \text{ } \Omega$ resistor in series with the battery.
$2$. Check the diodes: The top branch has a diode in reverse bias (the triangle points against the conventional current flow from the positive terminal),so it acts as an open circuit (no current flows).
$3$. The middle branch has a diode in forward bias (the triangle points in the direction of conventional current flow),so it acts as a closed circuit (current flows).
$4$. Calculate the total resistance: The effective resistance of the circuit is the sum of the $20 \text{ } \Omega$ resistor (in series with the battery) and the $30 \text{ } \Omega$ resistor (in the middle branch).
$5$. Total resistance $R = 20 \text{ } \Omega + 30 \text{ } \Omega = 50 \text{ } \Omega$.
$6$. Calculate the current: Using Ohm's law,$I = \frac{V}{R} = \frac{5 \text{ V}}{50 \text{ } \Omega} = \frac{5}{50} \text{ A}$.

(C) Among the given compounds,ferric chloride $(FeCl_3)$ is the most acidic in water.
This is due to the higher charge density and effective nuclear charge of the $Fe^{3+}$ ion compared to $Al^{3+}$ and $Be^{2+}$.
As the charge density of the central metal cation increases,its ability to polarize the $O-H$ bond in coordinated water molecules increases,leading to greater acidity.
All these compounds undergo hydrolysis in water to liberate $HCl$ gas:
$BeCl_2 + 2 H_2O \rightarrow Be(OH)_2 + 2 HCl \uparrow$
$AlCl_3 + 3 H_2O \rightarrow Al(OH)_3 + 3 HCl \uparrow$
$FeCl_3 + 3 H_2O \rightarrow Fe(OH)_3 + 3 HCl \uparrow$
$Fe^{3+}$ has a higher charge-to-size ratio,making it the strongest Lewis acid among the choices.

(C) Pure water can be obtained from sea water by the process of reverse osmosis. In this process,pressure greater than the osmotic pressure is applied to the sea water,forcing the water molecules to pass through a semi-permeable membrane,leaving behind the dissolved salts.

(B) In the contact process for the industrial preparation of sulphuric acid,$SO_2$ is oxidized to $SO_3$ in the presence of a catalyst,vanadium pentoxide $(V_2O_5)$.
The chemical reaction is: $2SO_2(g) + O_2(g) \xrightarrow{V_2O_5} 2SO_3(g)$.

(D) Fluorine $(F_2)$ is prepared by the electrolytic oxidation of fluoride ions. Specifically,it is obtained by the electrolysis of a mixture of potassium hydrogen fluoride $(KHF_2)$ and anhydrous hydrogen fluoride $(HF)$.
Other elements like $Ca$ are also prepared by electrolysis,but in the context of the $p$-block elements chapter,$F_2$ is the standard example of an element prepared by electrolysis.

(C) Given:
Percentage by weight of $HCl$ = $49\%$
Specific gravity of solution = $1.41 \ g/mL$
Density of solution $(d)$ = $1.41 \ g/mL$
Mass of $1000 \ mL$ of solution = $1.41 \times 1000 = 1410 \ g$
Mass of $HCl$ solute in $1000 \ mL$ solution = $49\% \text{ of } 1410 \ g = 0.49 \times 1410 = 690.9 \ g$
Molar mass of $HCl$ = $1 + 35.5 = 36.5 \ g/mol$
Molarity $(M)$ = $\frac{\text{Mass of solute}}{\text{Molar mass of solute} \times \text{Volume of solution in } L}$
$M = \frac{690.9}{36.5 \times 1} = 18.929 \ M \approx 18.93 \ M$
Therefore,the molarity is approximately $18.92 \ M$.

(C) According to Raoult's Law for non-volatile solutes: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$.
Given: $P^o = 195 \ mm \ Hg$,$P_s = 192.5 \ mm \ Hg$,$w_2 = 1 \ g$,$w_1 = 100 \ g$,$M_1 = 58 \ g/mol$.
Substituting the values: $\frac{195 - 192.5}{195} = \frac{1 \times 58}{M_2 \times 100}$.
$\frac{2.5}{195} = \frac{58}{100 \times M_2}$.
$M_2 = \frac{58 \times 195}{2.5 \times 100} = \frac{11310}{250} = 45.24 \ g/mol$.

(A) The relationship between half-life $(t_{1/2})$ and decay constant $(k)$ is given by:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 6.31 \times 10^{-4} \ yr^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{6.31 \times 10^{-4}} \ yr$
$t_{1/2} \approx 0.1098 \times 10^4 \ yr$
$t_{1/2} = 1098 \ yrs$.

(C) The reaction $2NO + O_2 \to 2NO_2$ follows the rate law expression: $Rate = k[NO]^2[O_2]^1$.
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law expression.
$\text{Order} = 2 + 1 = 3$.
Therefore,it is a third-order reaction.

(A) The standard $EMF$ of the cell is calculated using the formula: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
Here,$Cu^{2+}/Cu$ acts as the cathode $(E^o = +0.34\,V)$ and $Zn^{2+}/Zn$ acts as the anode $(E^o = -0.76\,V)$.
Substituting the values: $E^o_{cell} = 0.34\,V - (-0.76\,V) = 0.34\,V + 0.76\,V = 1.10\,V$.

(A) The cell reaction is $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$.
According to the Nernst equation,the cell potential is given by $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.
To increase the $E_{cell}$,the value of the logarithmic term $\frac{[Cu^{2+}]}{[Ag^{+}]^2}$ must decrease.
This can be achieved by increasing the concentration of the reactant $[Ag^{+}]$ or decreasing the concentration of the product $[Cu^{2+}]$.
Therefore,an increase in the concentration of $Ag^{+}$ ions will increase the voltage of the cell.

(D) The most convenient method to protect the bottom of a ship made of iron is by connecting it with a more reactive metal block,such as $Mg$ (magnesium).
This is known as cathodic protection or sacrificial protection.
Since $Mg$ is more reactive than $Fe$ (iron),it acts as the anode and undergoes oxidation,while the iron acts as the cathode and is protected from corrosion.

(D) Let the oxidation number of $Cr$ be $x$.
The oxidation number of $NH_3$ is $0$ and the oxidation number of $Cl$ is $-1$.
The sum of oxidation numbers in a neutral complex is $0$.
$x + 6 \times (0) + 3 \times (-1) = 0$
$x - 3 = 0$
$x = +3$
Therefore,the oxidation number of $Cr$ is $+3$.

(C) The given complexes $[Pt(NH_3)_4Cl_2]Br_2$ and $[Pt(NH_3)_4Br_2]Cl_2$ have the same molecular formula but produce different ions in an aqueous solution.
$[Pt(NH_3)_4Cl_2]Br_2$ dissociates to give $Br^-$ ions,whereas $[Pt(NH_3)_4Br_2]Cl_2$ dissociates to give $Cl^-$ ions.
This phenomenon is known as ionization isomerism.

(D) The reaction of ethyl alcohol with thionyl chloride in the presence of pyridine is known as the Darzens process.
The reaction is: $C_2H_5OH + SOCl_2 \xrightarrow{\text{Pyridine}} C_2H_5Cl + SO_2 + HCl$.
Pyridine is used to neutralize the $HCl$ produced in the reaction,which drives the reaction to completion.

(C) Freon $(CCl_2F_2)$ is an odourless,non-corrosive,and non-toxic gas which is stable even at high temperatures and pressures.
It has a low boiling point,low specific heat,and can be easily liquefied by applying pressure at room temperature.
It is,therefore,widely used as a refrigerant (cooling agent) in refrigerators and air conditioners.

(B) The correct answer is $(B)$.
Phenol $(C_6H_5OH)$ is acidic because the phenoxide ion formed after the loss of a proton is resonance-stabilized by the benzene ring.
Aliphatic alcohols like methanol,ethanol,and isopropyl alcohol are much less acidic than phenol because their corresponding alkoxide ions are not resonance-stabilized.

(D) Step $1$: Acetylene $(HC \equiv CH)$ undergoes hydration in the presence of $30\% \ H_2SO_4$ and $HgSO_4$ to form vinyl alcohol,which tautomerizes to acetaldehyde $(CH_3CHO)$. Thus,$A = CH_3CHO$.
Step $2$: Acetaldehyde contains $\alpha$-hydrogen atoms,so it undergoes aldol condensation in the presence of dilute $NaOH$ to form $3$-hydroxybutanal $(CH_3-CH(OH)-CH_2CHO)$. Thus,$B = CH_3-CH(OH)-CH_2CHO$.

(C) $FeCl_3$ is used to detect phenols by forming a violet-colored complex.
Fehling solution is used to detect reducing sugars like glucose.
$NaHSO_3$ (sodium bisulfite) is used to detect carbonyl compounds (aldehydes and ketones) by forming a crystalline addition product.
Tollen’s reagent is used to detect aldehydes,not unsaturation. Unsaturation is typically detected using bromine water or Baeyer's reagent.
Therefore,the statement that Tollen’s reagent is used in the detection of unsaturation is incorrect.

(C) Glyptal is a polyester resin formed by the condensation of ethylene glycol and phthalic acid. It is not a polyamide. Nylon-$6,6$,Nylon-$6$,and proteins all contain amide linkages $(-CONH-)$ in their structure.

(C) The correct statement is that starch is a polymer of $\alpha$-glucose.
Starch is a polysaccharide composed of two components: amylose and amylopectin,both of which are polymers of $\alpha$-$D$-glucose units.

(B) . Proteins are polymers of amino acids.
Amino acid $\to$ Dipeptide $\to$ Polypeptide $\to$ Protein.

(B) Antibodies are specialized proteins produced by the immune system to identify and neutralize foreign substances like bacteria and viruses. They are classified as $Globular \text{ } proteins$ and are also known as $Immunoglobulins$ $(Ig)$. Therefore,both options $B$ and $C$ are technically correct,but in the context of biochemical classification,they are primarily identified as $Globular \text{ } proteins$.

(A) The correct option is $(A)$.
Energy is stored in our body in the form of $ATP$ $(Adenosine \ Triphosphate)$,which acts as the primary energy currency of the cell.