AIPMT 1988 Chemistry Question Paper with Answer and Solution

(D) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \cdot l$ and mass $m = d \cdot V = d \cdot A \cdot l$ (where $d$ is density),we have $A = \frac{m}{d \cdot l}$.
Substituting $A$ into the resistance formula: $R = \rho \frac{l}{(m / (d \cdot l))} = \rho \cdot d \cdot \frac{l^2}{m}$.
Since $\rho$ and $d$ are constant for copper,$R \propto \frac{l^2}{m}$.
Given the ratios $m_1 : m_2 : m_3 = 1 : 3 : 5$ and $l_1 : l_2 : l_3 = 5 : 3 : 1$,the ratio of resistances is:
$R_1 : R_2 : R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$
$R_1 : R_2 : R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5} = 25 : 3 : 0.2$
Multiplying by $5$ to clear the fraction: $125 : 15 : 1$.

(A) Chlorine is used as a disinfectant for killing germs in water purification.

(C) In the titration of a strong acid and a strong base,the equivalence point occurs at $pH = 7$ (neutral).
Since both phenolphthalein (range $8.3 - 10.0$) and methyl orange (range $3.1 - 4.4$) show a sharp color change within the steep $pH$ change region of the titration curve (typically $pH \ 4$ to $10$),either indicator can be used.

(B) To name the compound $CH_2=CH-CH(CH_3)_2$ according to $IUPAC$ rules:
$1$. Identify the longest carbon chain containing the double bond. The longest chain has $4$ carbon atoms.
$2$. Number the chain starting from the end closer to the double bond. The double bond starts at carbon $1$.
$3$. The substituent (methyl group) is at position $3$.
$4$. Combining these,the name is $3-$methylbut$-1-$ene or $3-$methyl$-1-$butene.

(C) Isomers are defined as compounds that possess the same $Molecular \ formula$ but differ in the arrangement of atoms in space or their connectivity.
Since they share the same $Molecular \ formula$,they contain the same number and type of atoms.

(D) An enantiomorph (or enantiomer) exists if a molecule contains at least one chiral center (a carbon atom bonded to four different groups).
In $CH_3-CH_2-CH^*(NH_2)-CH_3$,the carbon atom marked with an asterisk $(*)$ is bonded to four different groups: $-H$,$-NH_2$,$-CH_3$,and $-CH_2CH_3$.
Since it possesses a chiral center,it can exist as a pair of non-superimposable mirror images,known as enantiomorphs.

(B) An optically active compound must contain at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $2-$chlorobutane $(CH_3-CH(Cl)-CH_2-CH_3)$,the second carbon atom is bonded to a hydrogen atom,a chlorine atom,a methyl group,and an ethyl group.
Since all four groups are different,$2-$chlorobutane is chiral and thus optically active.
The other options ($n-$propanol,$n-$butanol,and $4-$hydroxyheptane) do not contain a chiral carbon atom.

(A) The process of separating a racemic mixture into its individual $d$ and $l$ enantiomers is known as resolution.

(B) The stability of carbocations follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$.
$(CH_3)_3C^{+}$ is a $3^{\circ}$ carbocation,which is stabilized by the inductive effect of three methyl groups and hyperconjugation.
Therefore,$(CH_3)_3C^{+}$ is the most stable cation among the given options.

(C) The bond length between carbon atoms depends on the bond order. As the bond order increases, the bond length decreases.
The order of bond lengths is: $C-C > C=C > C \equiv C$.
$1.$ Ethane $(C_2H_6)$ has a $C-C$ single bond $(154 \ pm)$.
$2.$ Ethene $(C_2H_4)$ has a $C=C$ double bond $(134 \ pm)$.
$3.$ Ethyne $(C_2H_2)$ has a $C \equiv C$ triple bond $(120 \ pm)$.
Therefore, the carbon-carbon bond length is minimum in Ethyne.

(C) The chemical formula of Carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$.
It is a hydrated potassium magnesium chloride mineral,which serves as an important source of $Mg$.

(B) $Sporeine$ was the first bioinsecticide developed on a commercial scale in $Germany$.

(C) The most important function of cytokinin (Zeatin is a cytokinin found in $Zea\, mays$ or corn) is to promote cell division.
They are now proven to be essential components for cell division.

(B) Cytokinins are plant hormones that promote cell division and delay senescence (aging) in plant organs.
They help in maintaining the chlorophyll content in leaves for a longer period,thereby preventing yellowing and keeping the leaves green.
Therefore,applying cytokinin to the roots helps in delaying the senescence of leaves.

(D) In angiosperms,each Microspore Mother Cell $(MMC)$ undergoes meiosis to produce $4$ microspores (pollen grains).
Similarly,each Megaspore Mother Cell $(MMC)$ undergoes meiosis to produce $4$ megaspores,out of which only $1$ remains functional to form the embryo sac (ovule).
To produce $100$ seeds (grains) of wheat,we need $100$ functional megaspores and $100$ functional microspores.
For $100$ megaspores,$100$ meiotic divisions are required in $MMC$s.
For $100$ microspores,$100/4 = 25$ meiotic divisions are required in $MMC$s.
Total meiotic divisions = $100 + 25 = 125$.
However,the question specifically asks for the number of $MMC$s undergoing meiosis to produce $100$ grains. Since each grain requires one functional megaspore,$100$ meiotic divisions are required for the female gametophyte development. If the question implies the total number of divisions for both male and female gametes,it is $125$. Given standard textbook patterns for this specific question,the answer is $125$. Since $125$ is not an option,we evaluate the male contribution: $100$ grains require $100$ pollen grains,which is $25$ divisions. If the question refers only to the female side (ovule development),the answer is $100$.

(C) In photography,hypo $(Na_2S_2O_3)$ is used as a fixing agent.
It reacts with the undecomposed silver bromide $(AgBr)$ present on the photographic film to form a soluble complex,sodium dithiosulphatoargentate$(I)$.
The reaction is:
$AgBr + 2Na_2S_2O_3 \to Na_3[Ag(S_2O_3)_2] + NaBr$
This process removes the unexposed $AgBr$ from the film,making the image permanent.

(B) When a non-volatile solute like common salt $(NaCl)$ is dissolved in a solvent like water,the vapor pressure of the solution decreases.
To reach the atmospheric pressure,the solution must be heated to a higher temperature.
Therefore,the boiling point of the solution increases,which is known as the elevation of boiling point.

(A) Potassium ferrocyanide is a coordination compound containing the hexacyanidoferrate$(II)$ ion.
In this complex,the iron atom is in the $+2$ oxidation state.
According to the $IUPAC$ nomenclature rules,the formula is $K_4[Fe(CN)_6]$.

(D) $CCl_4$ is a covalent compound and does not ionize to provide $Cl^-$ ions in the solution.
Since $AgNO_3$ reacts with free $Cl^-$ ions to form a white precipitate of $AgCl$,no reaction occurs between $CCl_4$ and $AgNO_3$.
Therefore,the correct option is $D$.

(D) The reaction of $phenol$ with $phthalic \ anhydride$ in the presence of concentrated $H_2SO_4$ (acting as a dehydrating agent) is a condensation reaction.
Two molecules of $phenol$ react with one molecule of $phthalic \ anhydride$ to form $phenolphthalein$.
This is a classic synthesis of the acid-base indicator $phenolphthalein$.
Therefore,the correct option is $(D)$.

(A) The silver mirror test,also known as the $Tollens'$ test,is a chemical test used to distinguish between aldehydes and ketones.
Aldehydes are oxidized to carboxylic acids by $Tollens'$ reagent,which is an ammoniacal silver nitrate solution,while the silver ions are reduced to metallic silver,forming a silver mirror on the inner surface of the reaction vessel.
Ketones do not give this test.

(B) Glucose is a reducing sugar because it contains a free aldehyde group.
When glucose is heated with Fehling's solution,the aldehyde group is oxidized to a carboxylate group,and $Cu^{2+}$ ions are reduced to $Cu^+$ ions.
The reaction produces a red precipitate of cuprous oxide $(Cu_2O)$.
Reaction: $\text{Glucose} + \text{Fehling's solution} \to \text{Gluconic acid} + Cu_2O \text{ (Red precipitate)}$.