AIPMT 1990 Chemistry Question Paper with Answer and Solution

(B) The boiling point of water on the Centigrade scale is $100^{\circ}C$ and on the Fahrenheit scale is $212^{\circ}F$.
Initially,both thermometers are at the boiling point.
The final temperature on the Fahrenheit scale is $F = 140^{\circ}F$.
Using the conversion formula $\frac{C}{5} = \frac{F - 32}{9}$,we find the final temperature on the Centigrade scale:
$\frac{C}{5} = \frac{140 - 32}{9} = \frac{108}{9} = 12$.
$C = 12 \times 5 = 60^{\circ}C$.
The fall in temperature on the Centigrade scale is $\Delta C = 100^{\circ}C - 60^{\circ}C = 40^{\circ}C$.

(C) The magnetic flux linked with the coil is $\phi = NBA \cos(\omega t)$.
The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.
The amplitude of the induced $EMF$ is $V_0 = NBA\omega$.
The amplitude of the induced current is $i_0 = \frac{V_0}{R} = \frac{NBA\omega}{R}$.
Given: $B = 10^{-2} \, T$,$r = 0.3 \, m$,$R = \pi^2 \, \Omega$,$N = 1$,and frequency $f = \frac{200}{60} \, Hz = \frac{10}{3} \, Hz$.
Angular velocity $\omega = 2\pi f = 2\pi \times \frac{10}{3} = \frac{20\pi}{3} \, rad/s$.
Area $A = \pi r^2 = \pi (0.3)^2 = 0.09\pi \, m^2$.
Substituting the values: $i_0 = \frac{1 \times 10^{-2} \times 0.09\pi \times (20\pi/3)}{\pi^2}$.
$i_0 = \frac{10^{-2} \times 0.03 \times 20 \times \pi^2}{\pi^2} = 10^{-2} \times 0.6 = 0.006 \, A = 6 \, mA$.

(A) The average atomic mass is calculated by taking the weighted average of the isotopes:
Average atomic mass $= \frac{(10 \times 19) + (11 \times 81)}{100} = \frac{190 + 891}{100} = \frac{1081}{100} = 10.81$.
Thus,the atomic mass that should appear in the periodic table is approximately $10.8$.

(A) The molar mass of $CO_2$ is $44 \ g/mol$.
$4.4 \ g$ of $CO_2$ corresponds to $\frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$ of $CO_2$.
Each molecule of $CO_2$ contains $2$ oxygen atoms.
Therefore,$0.1 \ mol$ of $CO_2$ contains $0.1 \times 2 = 0.2 \ mol$ of oxygen atoms.
The number of oxygen atoms is $0.2 \times 6.022 \times 10^{23} \approx 1.2 \times 10^{23}$ atoms.

(B) Atomic number is defined as the number of protons in the nucleus.

(B) The principal quantum number is $n = 3$.
The azimuthal quantum number is $l = 1$,which corresponds to the $p$ subshell.
The number of orbitals in a subshell is given by the formula $(2l + 1)$.
For $l = 1$,the number of orbitals $= 2(1) + 1 = 3$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the total number of electrons $= 3 \times 2 = 6$ electrons.

(A) The atomic number of copper $(Cu)$ is $29$.
Electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$ or $2, 8, 18, 1$.
When $Cu$ loses one electron to form $Cu^{+}$,the configuration becomes $2, 8, 18$.
In $Cu^{+}$,the outermost shell is the third shell $(n=3)$,which contains $18$ electrons.
Thus,$Cu^{+}$ is the ion with $18$ electrons in its outermost shell.

(C) The electronic configuration is given as $1s^2 2s^2 2p^3$.
According to Hund's rule of maximum multiplicity,the $2p$ subshell has three orbitals $(2p_x, 2p_y, 2p_z)$ which are singly occupied by three electrons.
Therefore,there are $3$ unpaired electrons in the $2p$ subshell.

(C) $NH_4Cl$ (Ammonium chloride) contains both ionic and covalent bonds.
The ammonium ion $(NH_4^+)$ is formed by covalent bonds between the nitrogen atom and four hydrogen atoms.
The bond between the ammonium cation $(NH_4^+)$ and the chloride anion $(Cl^-)$ is ionic.

(A) $\sigma$ bond is formed by head-on overlapping of atomic orbitals, resulting in greater electron density between the nuclei, which makes it stronger than a $\pi$ bond formed by lateral overlapping.
Therefore, the statement that a $\sigma$ bond is weaker than a $\pi$ bond is incorrect.

(D) In the structure $H-C=C(O)-OH$,the carbon atom involved in the double bond with oxygen and the single bond with the other carbon and the hydroxyl group has a total of $5$ bonds ($1$ with $H$,$2$ with $C$,and $2$ with $O$). Since carbon can only form a maximum of $4$ covalent bonds,this representation is incorrect.

(C) The bond angles for the given molecules are as follows:
$BeF_2$: $180^o$ ($sp$ hybridization,linear geometry).
$CH_4$: $109^o 28'$ ($sp^3$ hybridization,tetrahedral geometry).
$NH_3$: $107^o$ ($sp^3$ hybridization,one lone pair,pyramidal geometry).
$H_2O$: $104.5^o$ ($sp^3$ hybridization,two lone pairs,bent geometry).
Due to the presence of two lone pairs on the oxygen atom in $H_2O$,the $lp-lp$ repulsion is greater than the $lp-bp$ repulsion in $NH_3$,resulting in the smallest bond angle among the given options. Thus,the correct option is $(C)$.

(B) $NH_3$ has a pyramidal structure because the nitrogen atom is $sp^3$ hybridized and contains one lone pair of electrons.
According to $VSEPR$ theory,the presence of one lone pair and three bond pairs results in a trigonal pyramidal geometry.

(C) The strength of a hydrogen bond depends on the electronegativity of the atom to which the hydrogen atom is bonded.
Since fluorine $(F)$ is the most electronegative element,the $H-F$ bond is highly polar,resulting in the strongest hydrogen bonding in $HF$ compared to $H_2O$,$NH_3$,or acetic acid.

(A) coordinate covalent bond (or dative bond) is formed when one atom donates a lone pair of electrons to another atom that needs them to complete its octet.
In the reaction $H_2O + H^{+} \to H_3O^{+}$,the oxygen atom in water has two lone pairs of electrons.
It donates one of these lone pairs to the $H^{+}$ ion,which has an empty $1s$ orbital.
This results in the formation of a hydronium ion $(H_3O^{+})$,where the bond between $O$ and $H^{+}$ is a coordinate covalent bond.
Therefore,the correct option is $A$.

(B) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature in Kelvin.
Since $M$ and $R$ are constants,$d \propto \frac{P}{T}$.
For option $A$ $(S.T.P.)$: $P = 1\,atm$,$T = 273\,K$,so $d \propto \frac{1}{273} \approx 0.0036$.
For option $B$: $P = 2\,atm$,$T = 273\,K$,so $d \propto \frac{2}{273} \approx 0.0073$.
For option $C$: $P = 1\,atm$,$T = 546\,K$,so $d \propto \frac{1}{546} \approx 0.0018$.
For option $D$: $P = 2\,atm$,$T = 546\,K$,so $d \propto \frac{2}{546} \approx 0.0036$.
Comparing the values,the density is highest at ${0\,^oC}$ and $2\,atm$.

(A) Absolute zero is the theoretical temperature of $0 \ K$ (or $-273.15 \ ^{\circ}C$) at which the enthalpy and entropy of a cooled ideal gas reach their minimum value,and all classical molecular motion ceases.

(C) The formula for root mean square velocity $(V_{rms})$ is given by:
$V_{rms} = \sqrt{\frac{3RT}{M}}$
where $R$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass (or $m$ as the molecular mass).
From the expression,it is clear that $V_{rms} \propto \frac{1}{\sqrt{m}}$,which can be written as $V_{rms} \propto m^{-1/2}$.

(C) The Van der Waals equation for a non-ideal gas is given by $\left( P + \frac{a}{V^2} \right)(V - b) = RT$.
In this equation,the term $\left( P + \frac{a}{V^2} \right)$ accounts for the pressure correction due to intermolecular forces of attraction between gas molecules.
The term $\frac{a}{V^2}$ represents the reduction in pressure caused by these attractive forces,where $a$ is the Van der Waals constant representing the magnitude of intermolecular attraction.

(D) The ideal gas equation is $PV = nRT$.
Real gases approach ideal behavior under conditions of high temperature and low pressure,where the intermolecular forces are negligible and the volume of the gas particles is insignificant compared to the total volume of the container.

(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
Initial moles: $n(N_2) = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$,$n(H_2) = \frac{6 \, g}{2 \, g/mol} = 3 \, mol$,$n(NH_3) = 0 \, mol$
Equilibrium moles of $NH_3 = \frac{27.54 \, g}{17 \, g/mol} = 1.62 \, mol$
According to stoichiometry,$2 \, mol$ of $NH_3$ are formed from $1 \, mol$ of $N_2$ and $3 \, mol$ of $H_2$.
So,$1.62 \, mol$ of $NH_3$ are formed from $0.81 \, mol$ of $N_2$ and $2.43 \, mol$ of $H_2$.
Equilibrium moles: $n(N_2) = 1 - 0.81 = 0.19 \, mol$,$n(H_2) = 3 - 2.43 = 0.57 \, mol$,$n(NH_3) = 1.62 \, mol$
Since the volume is $1 \, L$,concentration $[X] = n(X) \, mol/L$.
$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1.62)^2}{(0.19)(0.57)^3} \approx \frac{2.6244}{0.19 \times 0.185} \approx \frac{2.6244}{0.03515} \approx 74.66 \approx 75$.

(C) The equilibrium constant $K_c$ is a function of temperature only.
Since the temperature is kept constant at $25\,^oC$,changing the concentration of the reactants or products will shift the equilibrium position according to Le Chatelier's Principle,but it will not alter the value of the equilibrium constant $K_c$.
Therefore,$K_c$ remains the same.

(C) For the given reaction: ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$
The relationship between ${K_p}$ and ${K_c}$ is given by the formula: ${K_p} = {K_c}(RT)^{\Delta n}$
Here,$\Delta n$ is the change in the number of moles of gaseous products and reactants:
$\Delta n = (n_{products}) - (n_{reactants}) = 2 - (1 + 1) = 0$
Since $\Delta n = 0$,the expression becomes: ${K_p} = {K_c}(RT)^0 = {K_c} \times 1 = {K_c}$
Given that ${K_c} = 50$,therefore ${K_p} = 50$.

(D) The relationship between enthalpy change and internal energy change for a gaseous reaction is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous products and reactants,defined as $\Delta n_g = n_p - n_r$.
If the number of moles of products is less than the number of moles of reactants,then $\Delta n_g$ is negative $(\Delta n_g < 0)$.
In this case,$\Delta H = \Delta E - |\Delta n_g| RT$,which implies $\Delta H < \Delta E$.
Therefore,the correct option is $D$.

(A) The correct option is $(A)$.
Alkali metals belong to Group $1$ of the periodic table and have a general valence shell electronic configuration of $ns^1$.
Since their valence electron is present in the $s-$orbital,they lose this electron to form $M^+$ ions.

(A) Washing soda is the common name for sodium carbonate decahydrate.
Its chemical formula is $Na_2CO_3 \cdot 10H_2O$.

(D) $PbI_2$ is yellow and is known as 'Golden spangles'.
When a solution of $KI$ is added to a solution containing $Pb^{2+}$ ions,$PbI_2$ is formed,which is golden yellow in colour and is insoluble at room temperature.

(D) The functional group present is an aldehyde group,$-CHO$.
The compound $CH_3CHO$ contains two carbon atoms in its longest chain.
For a two-carbon aldehyde,the parent alkane is ethane,and the suffix for an aldehyde is $-al$.
Therefore,the $IUPAC$ name is $Ethanal$.

(A) For a molecule to exhibit geometrical isomerism,each carbon atom of the double bond must be attached to two different groups.
In $CH_3-CH=CH-CH_3$ (But$-2-$ene),the first carbon is attached to $-H$ and $-CH_3$,and the second carbon is also attached to $-H$ and $-CH_3$. Since both carbons have different substituents,it exhibits $cis$- and $trans$-isomerism.
In the other options,at least one carbon of the double bond is attached to two identical groups (e.g.,two $-H$ atoms),which prevents geometrical isomerism.

(B) The reaction of an alkene with cold,dilute,alkaline $KMnO_4$ solution (Baeyer's reagent) results in the formation of a vicinal diol.
During this reaction,the purple colour of $KMnO_4$ disappears,which serves as a qualitative test for the presence of a double or triple bond (unsaturation).
This specific test is known as $Baeyer's$ test.

(A) In $1928$,the scientist $Went$ proved through his experiments on $Avena$ coleoptiles that the highest concentration of $Auxin$ is found at the growing tips (apices).
$Auxin$ is synthesized in the apical regions and then transported downwards,where it promotes growth.

(B) The angular velocity $\omega$ is given by $\omega = 2 \pi f$,where $f$ is the frequency in $Hz$.
Given $f = 1200 \, r.p.m. = \frac{1200}{60} \, r.p.s. = 20 \, Hz$.
Thus,$\omega = 2 \pi \times 20 = 40 \pi \, rad/s$.
The centripetal acceleration $a$ is given by $a = \omega^2 r$.
Given $r = 30 \, cm = 0.3 \, m$.
$a = (40 \pi)^2 \times 0.3 = 1600 \pi^2 \times 0.3$.
Using $\pi^2 \approx 9.8696$,we get $a = 1600 \times 9.8696 \times 0.3 = 4737.4 \, m/s^2$.
Rounding to the nearest given option,the acceleration is approximately $4740 \, m/s^2$.

(B) The boiling point of water on the Fahrenheit scale is $212^{\circ}F$ and on the centigrade scale is $100^{\circ}C$.
The change in temperature on the Fahrenheit scale is $\Delta F = 212^{\circ}F - 140^{\circ}F = 72^{\circ}F$.
The relationship between the change in temperature on the Fahrenheit scale and the centigrade scale is given by $\frac{\Delta F}{9} = \frac{\Delta C}{5}$.
Substituting the value of $\Delta F$ into the equation: $\frac{72}{9} = \frac{\Delta C}{5}$.
$8 = \frac{\Delta C}{5}$.
$\Delta C = 8 \times 5 = 40^{\circ}C$.
Therefore,the fall in temperature registered by the centigrade thermometer is $40^{\circ}C$.

(B) The relationship between the Celsius scale $(T_C)$ and the Fahrenheit scale $(T_F)$ is given by: $\frac{T_C - 0}{100} = \frac{T_F - 32}{180}$.
Initially,both thermometers are in boiling water,so the initial temperature is $T_{C,i} = 100^{\circ}C$ and $T_{F,i} = 212^{\circ}F$.
When the Fahrenheit thermometer registers $T_{F,f} = 140^{\circ}F$,we find the corresponding Celsius temperature $T_{C,f}$:
$\frac{T_{C,f}}{100} = \frac{140 - 32}{180}$
$\frac{T_{C,f}}{100} = \frac{108}{180} = 0.6$
$T_{C,f} = 60^{\circ}C$.
The fall in temperature as registered by the centigrade thermometer is $\Delta T_C = T_{C,i} - T_{C,f} = 100^{\circ}C - 60^{\circ}C = 40^{\circ}C$.

(D) Colour blindness is an $X$-linked recessive disorder.
Let $X^C$ represent the allele for colour blindness and $X$ represent the normal allele.
The husband has normal vision,so his genotype is $XY$.
The wife has normal vision,but her father was colour blind. Therefore,she must be a carrier,with the genotype $X^CX$.
The cross is: $XY \times X^CX$.
The possible genotypes of the offspring are: $XX$ (normal daughter),$X^CX$ (carrier daughter),$XY$ (normal son),and $X^CY$ (colour blind son).
For a daughter to be colour blind,she must inherit the $X^C$ allele from both parents. Since the father is normal $(XY)$,he cannot pass the $X^C$ allele to his daughter.
Therefore,the probability of their daughter being colour blind is $0 \%$.

(B) Isoelectronic species are those that have the same number of electrons.
For $CO_2$: Total electrons = $6 + (8 \times 2) = 6 + 16 = 22$.
For $N_2O$: Total electrons = $(7 \times 2) + 8 = 14 + 8 = 22$.
Since both $CO_2$ and $N_2O$ have $22$ electrons,they are isoelectronic.
Therefore,the correct option is $(B)$.

(B) Some liquids,on mixing,form azeotropes,which are binary mixtures having the same composition in liquid and vapour phases and boil at a constant temperature.
In such cases,it is not possible to separate the components by fractional distillation.
This mixture is known as an Azeotropic mixture.

(C) The boiling point elevation is a colligative property,which depends on the van't Hoff factor $(i)$.
For $0.1 \ N \ Na_2SO_4$,$i = 3$ and molarity $= 0.05 \ M$,so $i \times M = 0.15$.
For $0.1 \ N \ MgSO_4$,$i = 2$ and molarity $= 0.05 \ M$,so $i \times M = 0.10$.
For $0.1 \ M \ Al_2(SO_4)_3$,$i = 5$ and molarity $= 0.1 \ M$,so $i \times M = 0.50$.
For $0.1 \ M \ BaSO_4$,it is sparingly soluble,so $i \times M$ is very low.
Since $Al_2(SO_4)_3$ has the highest value of $i \times M$,it will show the highest boiling point.

(C) For a first-order reaction,the rate constant $k$ is related to the half-life $t_{1/2}$ by the formula:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 69.35 \, \text{sec}$.
Substituting the value:
$k = \frac{0.693}{69.35} \approx 0.00999 \, \sec^{-1} \approx 0.01 \, \sec^{-1}$.
Therefore,the correct option is $(C)$.

(A) The formation of Prussian blue occurs when $Fe^{3+}$ ions react with potassium ferrocyanide,$K_4[Fe(CN)_6]$.
The chemical reaction is: $4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$.
This complex,$Fe_4[Fe(CN)_6]_3$,is known as ferric ferrocyanide or Prussian blue.

(A) To name the coordination compound $K_4[Fe(CN)_6]$ according to $IUPAC$ rules:
$1$. Name the cation first: $Potassium$.
$2$. Name the ligands in alphabetical order: $hexacyano$ (for $6$ $CN^-$ ions).
$3$. Name the central metal atom: Since the complex ion $[Fe(CN)_6]^{4-}$ is anionic,the metal $Fe$ is named as $ferrate$.
$4$. Determine the oxidation state of $Fe$: $x + 6(-1) = -4$,so $x = +2$. Thus,it is $ferrate(II)$.
Combining these,the name is $Potassium hexacyanoferrate(II)$.

(B) The formation of Prussian blue occurs when $Fe^{3+}$ ions react with potassium ferrocyanide,$K_4[Fe(CN)_6]$.
$Fe_2(SO_4)_3 \to 2Fe^{3+} + 3SO_4^{2-}$
$3K_4[Fe(CN)_6] + 4Fe^{3+} \to Fe_4[Fe(CN)_6]_3 + 12K^+$
$Fe_4[Fe(CN)_6]_3$ is the chemical formula for Prussian blue.

(B) $CHCl_3 + \frac{1}{2}O_2 \xrightarrow{\text{Sunlight}} COCl_2 + HCl$
$COCl_2$ is known as Phosgene or carbonyl chloride.

(B) The reaction described is the carbylamine test,which is characteristic of primary amines.
The reaction is: $C_6H_5NH_2 + 3 KOH + CHCl_3 \to C_6H_5NC + 3 KCl + 3 H_2O$.
Here,$(Y)$ is $CHCl_3$ (chloroform).
Chloroform $(Y)$ is prepared by the haloform reaction using ethanol $(Z)$ and chlorine in the presence of slaked lime $(Ca(OH)_2)$:
$C_2H_5OH + Cl_2$ $\xrightarrow{Ca(OH)_2} CH_3CHO$ $\xrightarrow{Cl_2} CCl_3CHO$ $\xrightarrow{Ca(OH)_2} CHCl_3$.
Thus,the compound $(Z)$ is $C_2H_5OH$ (ethanol).

(A) Lucas test is used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction with Lucas reagent (a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$).

(A) When ethanol $(C_2H_5OH)$ is reacted with ammonia $(NH_3)$ in the presence of alumina $(Al_2O_3)$ at high temperature,it undergoes an ammonolysis reaction to form ethylamine $(C_2H_5NH_2)$.
The chemical equation is: $C_2H_5OH + NH_3 \xrightarrow{Al_2O_3} C_2H_5NH_2 + H_2O$.

(D) Acetophenone $(C_6H_5COCH_3)$ contains a methyl ketone group $(-COCH_3)$,which gives a positive iodoform test with $I_2$ and $Na_2CO_3$ (or $NaOH$),resulting in a yellow precipitate of iodoform $(CHI_3)$.
Benzophenone $(C_6H_5COC_6H_5)$ does not contain a methyl ketone group and therefore does not give the iodoform test.
Thus,$I_2$ and $Na_2CO_3$ can be used to distinguish between them.