AIPMT 1993 Chemistry Question Paper with Answer and Solution

(B) The chemical reaction is: $NH_{3(g)} + HCl_{(g)} \to NH_4Cl_{(s)}$.
Given volumes are $V_{NH_3} = 20 \, mL$ and $V_{HCl} = 40 \, mL$.
Since the reaction stoichiometry is $1:1$,$NH_3$ acts as the limiting reagent.
$20 \, mL$ of $NH_3$ will react completely with $20 \, mL$ of $HCl$ to form solid $NH_4Cl$.
$NH_4Cl$ is a solid,so it does not contribute to the gaseous volume.
The remaining volume of $HCl$ is $40 \, mL - 20 \, mL = 20 \, mL$.
Thus,the final volume of the gas is $20 \, mL$.

(B) Ionic bonds are formed by the transfer of electrons from a metal to a non-metal.
$1$. The metal atom should have a $Low \ ionization \ potential$ to easily lose an electron to form a cation.
$2$. The non-metal atom should have a $High \ electron \ affinity$ to easily accept the electron to form an anion.
Therefore,the correct combination is $Low \ ionization \ potential$ and $High \ electron \ affinity$.

(C) The ideal gas equation is given by $PV = nRT$.
Rearranging for concentration $C = \frac{n}{V}$,we get $P = CRT$.
Given $P = 1 \, atm$,$C = 1 \, mol \, L^{-1}$,and $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$.
Substituting these values: $1 = 1 \times 0.082 \times T$.
Therefore,$T = \frac{1}{0.082} \approx 12.19 \, K$.
Thus,the condition is $T \approx 12 \, K$.

(A) For an ideal gas,the kinetic energy $(KE)$ is given by $KE = \frac{3}{2}PV$.
Since internal energy $(E)$ per unit volume is defined as $E = \frac{KE}{V}$,we have $E = \frac{3}{2}P$.
Rearranging this for pressure $(P)$,we get $P = \frac{2}{3}E$.

(A) The average kinetic energy of a gas molecule is given by the formula $KE_{avg} = \frac{3}{2}RT$.
Since the temperature $(T)$ is the same for both gases at $STP$,the average kinetic energy per mole of both gases will be equal.
Therefore,$0.50 \ mol$ of $H_2$ and $1.0 \ mol$ of $He$ will have equal average kinetic energies.

(D) The formulas for the three types of molecular velocities are:
Most probable velocity $(V_{mp})$ = $\sqrt{\frac{2RT}{M}}$
Mean velocity $(V_{avg})$ = $\sqrt{\frac{8RT}{\pi M}}$
Root mean square velocity $(V_{rms})$ = $\sqrt{\frac{3RT}{M}}$
Taking the ratio $V_{mp} : V_{avg} : V_{rms}$:
$= \sqrt{\frac{2RT}{M}} : \sqrt{\frac{8RT}{\pi M}} : \sqrt{\frac{3RT}{M}}$
Dividing by $\sqrt{\frac{RT}{M}}$,we get:
$= \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$
Thus,the correct option is $(D)$.

(B) An ideal gas is defined as a system in which there are no intermolecular or interatomic forces.
Any real system approaches ideal gas behaviour in the limit where the pressure is extremely low and the temperature is high enough to overcome attractive intermolecular forces.
$PV = nRT$
Here,$n$ is the number of moles of gas.
The ideal gas equation describes the state of an ideal gas.
At low temperature and high pressure,gases deviate more from ideal behaviour because as pressure increases,the force of attraction between molecules increases and the molecules come closer together,making the volume of the gas particles significant compared to the total volume.

(C) According to Dalton's Law of Partial Pressures,the total pressure of a gas collected over water is the sum of the partial pressure of the dry gas and the vapour pressure of water at that temperature.
$P_{\text{total}} = P_{O_2} + P_{H_2O}$
Given:
$P_{\text{total}} = 751\, mm\, Hg$
$P_{H_2O} = 21\, mm\, Hg$
Therefore,$P_{O_2} = P_{\text{total}} - P_{H_2O} = 751 - 21 = 730\, mm\, Hg$.
To convert this to atmospheres,we use the relation $1\, atm = 760\, mm\, Hg$:
$P_{O_2} = \frac{730}{760}\, atm \approx 0.96\, atm$.

(B) For pure water,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}]$.
Given $[H_3O^{+}] = 10^{-6}\,M$,therefore $[OH^{-}] = 10^{-6}\,M$.
The ionic product of water is defined as $K_w = [H_3O^{+}][OH^{-}]$.
Substituting the values: $K_w = (10^{-6}) \times (10^{-6}) = 10^{-12}$.

(C) The thermal decomposition of ammonium dichromate is a classic reaction used in the 'volcano' experiment.
When heated,it decomposes to produce chromium$(III)$ oxide,nitrogen gas,and water vapor.
The balanced chemical equation is:
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4H_2O$
Therefore,the products are chromium oxide and nitrogen.

(A) To find the percentage of nitrogen,we calculate the mass percentage for each compound:
$1$. Urea $(NH_2CONH_2)$: Molar mass = $60 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/60) \times 100 = 46.66 \%$.
$2$. Ammonium sulphate $(NH_4)_2SO_4$: Molar mass = $132 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/132) \times 100 = 21.21 \%$.
$3$. Ammonium nitrate $(NH_4NO_3)$: Molar mass = $80 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/80) \times 100 = 35 \%$.
$4$. Calcium nitrate $Ca(NO_3)_2$: Molar mass = $164 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/164) \times 100 = 17.07 \%$.
Thus,Urea has the highest percentage of nitrogen.

(A) In the process of urine formation,the glomerular filtrate contains substances like glucose,amino acids,and electrolytes that are essential for the body.
These substances are reabsorbed from the renal tubules back into the blood.
Glucose is reabsorbed from the proximal convoluted tubule $(PCT)$ through the process of active transport,which requires energy $(ATP)$ to move molecules against their concentration gradient.

(B) Vitamin $K$ is essential for the synthesis of several clotting factors in the liver,including prothrombin (Factor $II$),Factor $VII$,Factor $IX$,and Factor $X$.
Specifically,it acts as a cofactor for the enzyme gamma-glutamyl carboxylase,which carboxylates glutamic acid residues on these clotting factors,enabling them to bind calcium ions and participate in the blood coagulation cascade.
Therefore,the synthesis of prothrombin in the liver is dependent on Vitamin $K$.

(C) $o$-Nitrophenol and $p$-nitrophenol can be separated by steam distillation.
$o$-Nitrophenol is steam volatile due to the presence of intramolecular hydrogen bonding,which reduces its intermolecular attraction.
In contrast,$p$-nitrophenol exhibits intermolecular hydrogen bonding,leading to the association of molecules,which makes it less volatile.
Therefore,$o$-nitrophenol distills over with steam,while $p$-nitrophenol remains in the distillation flask.

(A) Pulses are seeds of leguminous plants belonging to the family $Fabaceae$. This family was earlier known as $Papilionoideae$,a subfamily of the family $Leguminosae$. Plants in this family are a major source of protein in the human diet.

(D) The moment of inertia $I$ of a body about an axis is given by $I = \int r^2 dm$,where $r$ is the perpendicular distance of the mass element $dm$ from the axis. To minimize the moment of inertia,the mass distribution should be as close to the axis as possible.
For a rectangle of sides $a$ and $b$ (where $b = 2a$),the axes $HF$ and $EG$ pass through the center. Axis $HF$ is parallel to the sides $AB$ and $CD$ (length $a$),and axis $EG$ is parallel to the sides $AD$ and $BC$ (length $2a$).
The moment of inertia about an axis passing through the center and parallel to a side of length $L$ is given by $I = \frac{1}{12} M L^2$ (where $L$ is the side perpendicular to the axis).
For axis $HF$ (parallel to $AB$),the perpendicular side is $BC = 2a$,so $I_{HF} = \frac{1}{12} M (2a)^2 = \frac{4}{12} M a^2 = \frac{1}{3} M a^2$.
For axis $EG$ (parallel to $BC$),the perpendicular side is $AB = a$,so $I_{EG} = \frac{1}{12} M a^2$.
Comparing the two,$I_{EG} < I_{HF}$. Thus,the moment of inertia is minimum along the axis $EG$.

(D) Most probable velocity,$U_{MP} = \sqrt{\frac{2RT}{M}}$
Average velocity,$U_{avg} = \sqrt{\frac{8RT}{\pi M}}$
Root mean square velocity,$U_{RMS} = \sqrt{\frac{3RT}{M}}$
Taking the ratio $U_{MP} : U_{avg} : U_{RMS}$:
$= \sqrt{\frac{2RT}{M}} : \sqrt{\frac{8RT}{\pi M}} : \sqrt{\frac{3RT}{M}}$
$= \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$
Hence,option $D$ is correct.

(D) According to the photoelectric effect,the number of photo-electrons emitted per second is directly proportional to the number of incident photons per second.
Since the intensity of light is defined as the energy incident per unit area per unit time,it is directly proportional to the number of incident photons.
Therefore,increasing the intensity of the incident light increases the number of incident photons,which in turn increases the number of photo-electrons emitted per second.

(D) Let the length $AB = CD = a$ and the width $BC = AD = 2a$. The mass of the rectangle is $M$.
For a rectangular lamina of sides $a$ and $b$,the moment of inertia about an axis passing through the center and parallel to side $b$ is $I = \frac{Ma^2}{12}$.
$1$. For axis $EG$ (parallel to $BC$): The distance of mass distribution is along $AB$. Here,$I_{EG} = \frac{M(AB)^2}{12} = \frac{Ma^2}{12}$.
$2$. For axis $HF$ (parallel to $AB$): The distance of mass distribution is along $BC$. Here,$I_{HF} = \frac{M(BC)^2}{12} = \frac{M(2a)^2}{12} = \frac{4Ma^2}{12} = \frac{Ma^2}{3}$.
Comparing the two,$I_{EG} < I_{HF}$.
Since the moment of inertia is proportional to the square of the distance of the mass from the axis,the axis that passes through the center and is parallel to the longer side will have the minimum moment of inertia.
Therefore,the moment of inertia is minimum about the axis $EG$.

(C) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $\mu_0$ is the permeability of free space,$N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
From this expression,it is clear that the self-inductance $L$ is directly proportional to the square of the number of turns,i.e.,$L \propto N^2$.

(B) The given reactions are oxidation half-reactions. The standard reduction potentials $(E^o_{red})$ are the negative of the standard oxidation potentials $(E^o_{ox})$.
$E^o_{red}(Zn^{2+}/Zn) = -0.76 \ V$
$E^o_{red}(Fe^{2+}/Fe) = -0.41 \ V$
In the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$,$Zn$ is oxidized (anode) and $Fe^{2+}$ is reduced (cathode).
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{red}(Fe^{2+}/Fe) - E^o_{red}(Zn^{2+}/Zn)$
$E^o_{cell} = -0.41 \ V - (-0.76 \ V)$
$E^o_{cell} = -0.41 \ V + 0.76 \ V = +0.35 \ V$.

(A) Argentite is an ore of silver with the chemical formula $Ag_2S$.
Stibnite is an ore of antimony with the chemical formula $Sb_2S_3$.
Haematite is an ore of iron with the chemical formula $Fe_2O_3$.
Bauxite is an ore of aluminium with the chemical formula $Al_2O_3 \cdot 2H_2O$.