AIIMS 1998 Chemistry Question Paper with Answer and Solution

(C) The angular displacement is given by $\theta(t) = 2t^3 + 0.5$.
The angular velocity $\omega$ is the rate of change of angular displacement with respect to time,given by $\omega = \frac{d\theta}{dt}$.
Differentiating $\theta(t)$ with respect to $t$: $\omega = \frac{d}{dt}(2t^3 + 0.5) = 6t^2$.
To find the angular velocity at $t = 2 \, s$,substitute $t = 2$ into the expression for $\omega$:
$\omega = 6 \times (2)^2 = 6 \times 4 = 24 \, rad/s$.

(B) According to the law of conservation of mechanical energy,the potential energy of the water at the top of the dam is converted into kinetic energy at the turbine.
$PE_{top} = KE_{bottom}$
$mgh = \frac{1}{2}mv^2$
By canceling the mass $m$ from both sides,we get:
$v^2 = 2gh$
$v = \sqrt{2gh}$
Substituting the given values $g = 9.8\, m/s^2$ and $h = 19.6\, m$:
$v = \sqrt{2 \times 9.8 \times 19.6}$
$v = \sqrt{19.6 \times 19.6}$
$v = 19.6\, m/s$
Therefore,the velocity of the water at the turbine is $19.6\, m/s$.

(C) This is because chargeless particles do not undergo any deflection in an electric or magnetic field,making them difficult to detect using traditional particle detection methods.

(C) The Aufbau principle states that in the ground state of an atom,the orbitals are filled in the increasing order of their energies,starting from the lowest energy orbital.

(B) The bond angle depends on the hybridisation state of the carbon atom.
For $sp^3$ hybridisation,the bond angle is $109.5^\circ$.
For $sp^2$ hybridisation,the bond angle is $120^\circ$.
For $sp$ hybridisation,the bond angle is $180^\circ$.
Therefore,as the hybridisation state changes from $sp^3$ $\rightarrow sp^2$ $\rightarrow sp$,the bond angle increases gradually.

(A) The correct answer is $(A)$.
Ethanol $(C_2H_5OH)$ contains an $-OH$ group,which allows for intermolecular hydrogen bonding.
Dimethyl ether $(CH_3OCH_3)$ lacks an $-OH$ group and cannot form hydrogen bonds.
Due to the presence of strong intermolecular hydrogen bonding in ethanol,more energy is required to overcome these forces,resulting in a higher boiling point compared to dimethyl ether.

(C) The given reaction is $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$.
Since the volume of the vessel is $1 L$,the molar concentrations are equal to the number of moles:
$[H_2S] = 0.5 mol L^{-1}$
$[H_2] = 0.10 mol L^{-1}$
$[S_2] = 0.4 mol L^{-1}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[H_2]^2 [S_2]}{[H_2S]^2}$
Substituting the values:
$K_c = \frac{(0.10)^2 \times (0.4)}{(0.5)^2} = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016 mol L^{-1}$.

(D) According to Le Chatelier's principle,whenever a constraint (such as change in concentration,pressure,or temperature) is applied to a system in equilibrium,the system tends to readjust so as to nullify the effect of the constraint by bringing a shift in the equilibrium position.
Therefore,it helps in predicting the direction of the shift in equilibrium.

(B) The solubility of $BaSO_4$ in $g/L$ is given as $2.33 \times 10^{-3} \ g/L$.
First,convert the solubility into $mol/L$ $(S)$:
$S = \frac{\text{solubility in } g/L}{\text{molar mass}} = \frac{2.33 \times 10^{-3}}{233} = 1 \times 10^{-5} \ mol/L$.
Since $BaSO_4$ is a $1:1$ electrolyte,it dissociates as: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = S \times S = S^2$.
Substituting the value of $S$:
$K_{sp} = (1 \times 10^{-5})^2 = 1 \times 10^{-10}$.

(D) Hess's law of constant heat summation states that the total enthalpy change for a chemical reaction is the same,whether the reaction occurs in one step or several steps.
This principle allows for the calculation of the heat of reaction,heat of formation,heat of transition,and other enthalpy changes by using known thermochemical data.
Therefore,it is applicable to all the given processes.

(B) The combustion reaction for $C_2H_{4(g)}$ is: $C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)}$
$\Delta H_{combustion} = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$
$\Delta H_{combustion} = [2 \times \Delta H_f^o(CO_2) + 2 \times \Delta H_f^o(H_2O)] - [\Delta H_f^o(C_2H_4) + 3 \times \Delta H_f^o(O_2)]$
Given: $\Delta H_f^o(C_2H_4) = 52 \ kJ \ mol^{-1}$,$\Delta H_f^o(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta H_f^o(H_2O) = -286 \ kJ \ mol^{-1}$,and $\Delta H_f^o(O_2) = 0 \ kJ \ mol^{-1}$.
$\Delta H_{combustion} = [2(-394) + 2(-286)] - [52 + 3(0)]$
$\Delta H_{combustion} = [-788 - 572] - 52 = -1360 - 52 = -1412 \ kJ \ mol^{-1}$.

(B) In an endothermic reaction,heat is absorbed from the surroundings.
Therefore,the enthalpy change $\Delta H$ is always positive $(+ve)$.

(B) The given reaction is $4Fe + 3O_2 \to 4Fe^{3+} + 6O^{2-}$.
In this reaction,the oxidation state of $Fe$ increases from $0$ to $+3$,which means $Fe$ is oxidised.
Since $Fe$ is oxidised,it acts as a reducing agent.
Conversely,the oxidation state of $O$ decreases from $0$ to $-2$,meaning $O_2$ is reduced.
Statement $B$ is incorrect because metallic iron is oxidised to $Fe^{3+}$,not reduced.

(D) For the molecule $Na_2S_4O_6$,let the oxidation number of sulphur be $x$.
Assigning oxidation states: $Na = +1$,$O = -2$.
The sum of oxidation states in a neutral molecule is $0$.
$2(+1) + 4(x) + 6(-2) = 0$
$2 + 4x - 12 = 0$
$4x - 10 = 0$
$4x = 10$
$x = 10/4 = 5/2$
Therefore,the average oxidation number of sulphur is $5/2$.

(B) The electropositive character (metallic character) of elements decreases as we move from left to right across a period in the periodic table.
Among the given elements,$Mg$ (Magnesium) is in Group $2$,$Al$ (Aluminium) is in Group $13$,$P$ (Phosphorus) is in Group $15$,and $S$ (Sulphur) is in Group $16$.
Since $Mg$ is the leftmost element among these in the $3^{rd}$ period,it has the highest metallic character and is the most electropositive.
Therefore,the correct option is $(B)$.

(D) $Anhydrous \ CaCl_2$ (Anhydrous calcium chloride) is a common drying agent used in the laboratory for neutral gases.
It absorbs moisture to form a hydrate,$CaCl_2 \cdot 6H_2O$.
It is effective for drying neutral gases like $N_2$,$O_2$,and $CO_2$.

(C) The equivalent mass of an acid is defined as the mass of the acid that provides one mole of $H^+$ ions in a reaction.
It is calculated using the formula:
$\text{Equivalent mass of acid} = \frac{\text{Molecular mass of the acid}}{\text{Basicity of the acid}}$
Here,the basicity of an acid is the number of replaceable $H^+$ ions per molecule of the acid.

(B) The empirical formula is $CH_2O$.
The empirical formula mass is calculated as: $12 + (2 \times 1) + 16 = 30 \ g/mol$.
The molecular mass is calculated using the relation: $\text{Molecular Mass} = 2 \times \text{Vapour Density} = 2 \times 30 = 60 \ g/mol$.
Calculate the value of $n$: $n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{60}{30} = 2$.
The molecular formula is given by: $(Empirical \ Formula)_n = (CH_2O)_2 = C_2H_4O_2$.

(A) To find the empirical formula,we calculate the molar ratio of each element:

Element	Moles (Percentage / Atomic Mass)	Simple Ratio
$C$	$18.5 / 12 = 1.54$	$1.54 / 1.54 = 1$
$H$	$1.55 / 1 = 1.55$	$1.55 / 1.54 \approx 1$
$Cl$	$55.04 / 35.5 = 1.55$	$1.55 / 1.54 \approx 1$
$O$	$24.81 / 16 = 1.55$	$1.55 / 1.54 \approx 1$

The simple ratio of $C:H:Cl:O$ is $1:1:1:1$.
Therefore,the empirical formula is $CHClO$.

(A) $1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has the highest priority,so the parent chain is named as an alkanoic acid.
$2$. Number the chain: Start numbering from the carbon of the $-COOH$ group as $C-1$. The chain is $C_1(OOH)-C_2H_2-C_3H=C_4(OH)-C_5H_3$.
$3$. Identify substituents and unsaturation: There is a hydroxyl group $(-OH)$ at $C-4$ and a double bond starting at $C-3$.
$4$. Combine the parts: The parent chain has $5$ carbons (pent),the double bond is at position $3$ (pent-$3$-ene),and the $-OH$ group is at position $4$. Combining these,we get $4-$hydroxypent-$3-$enoic acid.

(A) Baeyer's reagent is an alkaline solution of cold $KMnO_4$.
It is primarily used for the detection of unsaturation (double or triple bonds) in a molecule,as it causes the purple color of $KMnO_4$ to disappear.

(A) The stability of an alkene is directly proportional to the number of alkyl groups attached to the double-bonded carbon atoms due to the hyperconjugation effect and inductive effect.
$A$ represents a tetra-substituted alkene $(R_2C=CR_2)$,which has the maximum number of alkyl groups.
$B$ represents a di-substituted alkene.
$C$ is chemically incorrect as written.
$D$ represents ethene,which is unsubstituted.
Therefore,$R_2C=CR_2$ is the most stable alkene.

(A) The given reaction $C_6H_6 + CH_3Cl \xrightarrow{anhydrous \ AlCl_3} C_6H_5CH_3 + HCl$ involves the alkylation of benzene in the presence of a Lewis acid catalyst $(AlCl_3)$.
This is a classic example of a Friedel-Crafts alkylation reaction.

(A) Electrophilic aromatic substitution is facilitated by electron-donating groups on the benzene ring,which increase the electron density of the ring.
$1$. The $-CH_3$ group in Toluene is an electron-donating group due to the $+I$ effect and hyperconjugation.
$2$. Benzene has no substituents.
$3$. The $-COOH$ group in Benzoic acid and the $-NO_2$ group in Nitrobenzene are electron-withdrawing groups,which decrease the electron density of the ring.
Therefore,Toluene is the most reactive towards electrophilic nitration among the given options.

(C) $Arbor$ $vitae$ is a branched,tree-like structure found in the cerebellum of the brain. It is primarily composed of myelinated axons,which constitute the white matter,surrounded by the outer grey matter (cerebellar cortex).

(A) The stability of alkenes increases with the number of alkyl groups attached to the double-bonded carbon atoms due to the hyperconjugation effect.
$R_2C = CR_2$ is a tetra-substituted alkene,which is the most stable among the given options.

(C) Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom such as $F, O,$ or $N$.
In $H_2O$ (water),$HF$ (hydrogen fluoride),and glycerine $(C_3H_8O_3)$,hydrogen is bonded to $O$ or $F$,allowing for hydrogen bonding.
In $H_2S$ (hydrogen sulphide),the electronegativity of sulfur is not high enough to facilitate significant hydrogen bonding.
Therefore,the correct option is $C$.

(D) The spring constant $k$ is given by Hooke's Law: $k = \frac{F}{x}$.
Given $F = 10\,N$ and $x = 1\,mm = 1 \times 10^{-3}\,m$.
So,$k = \frac{10}{1 \times 10^{-3}} = 10^{4}\,N/m$.
The work done $W$ in stretching the spring by a displacement $x = 40\,mm = 40 \times 10^{-3}\,m$ is given by $W = \frac{1}{2}kx^{2}$.
Substituting the values: $W = \frac{1}{2} \times 10^{4} \times (40 \times 10^{-3})^{2}$.
$W = \frac{1}{2} \times 10^{4} \times (1600 \times 10^{-6}) = \frac{1}{2} \times 16 = 8\,J$.

(D) The energy of a photon is given by the formula $E = h\nu = \frac{hc}{\lambda}$.
From this relation,it is clear that the energy of a photon is inversely proportional to its wavelength $(\lambda)$.
Therefore,photons of different wavelengths possess different amounts of energy.
Since both the Assertion and the Reason are false,the correct option is $D$.

(B) The Van der Waal's equation is a modified version of the ideal gas equation that accounts for the finite volume of gas molecules and the intermolecular forces of attraction.
It is specifically designed to describe the behavior of real (non-ideal) gases.

(D) According to Graham's law of diffusion,the rate of diffusion $(r)$ of a gas is directly proportional to its pressure $(P)$ and inversely proportional to the square root of its molar mass $(M)$.
Mathematically,this is expressed as $r \propto \frac{P}{\sqrt{M}}$.

(D) $FeCl_3$ is a salt of a weak base $(Fe(OH)_3)$ and a strong acid $(HCl)$.
When dissolved in water,it undergoes hydrolysis to form $Fe(OH)_3$ and $HCl$.
The presence of $HCl$ makes the solution acidic,not basic.
Therefore,the Assertion is incorrect,but the Reason is correct because $FeCl_3$ does undergo hydrolysis in water.
The reaction is: $FeCl_3 + 3H_2O \to Fe(OH)_3 + 3HCl$.

(C) Baeyer's process is used for the purification of red bauxite,which contains iron oxide $(Fe_2O_3)$ as the main impurity.
Serpeck's process is used for the purification of white bauxite,which contains silica $(SiO_2)$ as the main impurity.

(A) The Van't Hoff factor $i$ represents the number of particles a solute dissociates into in a solution.
For $K_4[Fe(CN)_6]$,the dissociation is: $K_4[Fe(CN)_6] \rightarrow 4K^{+} + [Fe(CN)_6]^{4-}$.
Total number of particles $i = 4 + 1 = 5$.
Now,checking the options:
$A$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 2 + 3 = 5$.
$B$. $NaCl \rightarrow Na^{+} + Cl^{-}$,so $i = 1 + 1 = 2$.
$C$. $Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$,so $i = 2 + 1 = 3$.
$D$. $Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^{-}$,so $i = 1 + 3 = 4$.
Therefore,$Al_2(SO_4)_3$ has the same Van't Hoff factor as $K_4[Fe(CN)_6]$.

(C) Given:
Initial boiling point of water = $100\,^{\circ}C$
Final boiling point of water = $100.52\,^{\circ}C$
Mass of solute $(w)$ = $3\,g$
Mass of solvent $(W)$ = $200\,g$ (assuming density of water is $1\,g/mL$)
${K_b}$ for water = $0.6\,K\,kg\,mol^{-1}$
Elevation in boiling point $(\Delta {T_b})$ = $100.52 - 100 = 0.52\,^{\circ}C$
Using the formula:
$M = \frac{{K_b \times w \times 1000}}{{\Delta {T_b} \times W}}$
$M = \frac{{0.6 \times 3 \times 1000}}{{0.52 \times 200}}$
$M = \frac{{1800}}{{104}} \approx 17.3\,g\,mol^{-1}$

(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given $t = 30 \ min$ and the reaction is $30\%$ complete,so $[A]_t = 100 - 30 = 70$.
$k = \frac{2.303}{30} \log \frac{100}{70} = \frac{2.303}{30} \log(1.428) \approx \frac{2.303 \times 0.1547}{30} \approx 0.01188 \ min^{-1}$.
The half-life period $T_{1/2}$ is given by $T_{1/2} = \frac{0.693}{k}$.
$T_{1/2} = \frac{0.693}{0.01188} \approx 58.3 \ min$.
Rounding to the nearest option,the correct answer is $58.2 \ min$.

(B) The unit of the rate constant $(min^{-1})$ indicates that the reaction is of the first order.
For a first-order reaction,the half-life period $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given $k = 0.69 \times 10^{-1} \ min^{-1}$.
$t_{1/2} = \frac{0.693}{0.69 \times 10^{-1}} \ min \approx 10 \ min$.
To convert this into seconds: $10 \ min \times 60 \ \sec/min = 600 \ \sec$.

(B) According to Faraday's second law of electrolysis,the mass of different substances liberated by the same quantity of electricity is proportional to their equivalent weights.
$\frac{\text{Weight of } Cu}{\text{Weight of } H_2} = \frac{\text{Equivalent weight of } Cu}{\text{Equivalent weight of } H}$
Given:
Weight of $H_2 = 0.504 \ g$
Equivalent weight of $H = 1$
Equivalent weight of $Cu = \frac{63.5}{2} = 31.75$
$\frac{\text{Weight of } Cu}{0.504} = \frac{31.75}{1}$
$\text{Weight of } Cu = 0.504 \times 31.75 = 15.999 \approx 16.0 \ g$
Rounding to the nearest provided option,the correct answer is $15.9 \ g$.

(D) The $Standard \ Hydrogen \ Electrode$ $(SHE)$ is universally accepted as the primary reference electrode.
By convention,the potential of the $SHE$ is assigned a value of $0.00 \ V$ at all temperatures.
The half-cell reaction is represented as:
$H_2(g) \rightarrow 2H^+(aq) + 2e^-$ (as an anode)
$2H^+(aq) + 2e^- \rightarrow H_2(g)$ (as a cathode)

(D) In physisorption,the enthalpy of adsorption is very low because the accumulation of substances on the surface is due to $Van \ der \ Waal's$ forces,which are weak in nature.

(A) $(I) [Fe(CN)_6]^{3-}$
Here,the oxidation state of $Fe$ is $+3$.
The electronic configuration of $Fe$ (ground state) is $[Ar] 3d^6 4s^2$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5 4s^0$.
Number of unpaired electrons $= 5$.
$(II) [Fe(CN)_6]^{4-}$
Here,the oxidation state of $Fe$ is $+2$.
The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6 4s^0$.
Due to the strong field ligand $CN^-$,the electrons pair up,resulting in $0$ unpaired electrons.
$(III) [Cr(H_2O)_6]^{3+}$
Here,the oxidation state of $Cr$ is $+3$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3 4s^0$.
Number of unpaired electrons $= 3$.
$(IV) [Cu(H_2O)_6]^{2+}$
Here,the oxidation state of $Cu$ is $+2$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9 4s^0$.
Number of unpaired electrons $= 1$.
The paramagnetic character is directly proportional to the number of unpaired electrons.
Therefore,$[Fe(CN)_6]^{3-}$ has the maximum paramagnetic character.

(A) The reaction of a carboxylic acid with $PCl_5$ produces an acyl chloride (acid chloride).
The chemical equation is: $CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$
Therefore,the correct option is $A$.

(C) The reaction between acetic acid $(CH_3COOH)$ and phosphorus trichloride $(PCl_3)$ is a standard method for the preparation of acid chlorides.
The balanced chemical equation is:
$3CH_3COOH + PCl_3 \to 3CH_3COCl + H_3PO_3$
Thus,the product obtained is acetyl chloride $(CH_3COCl)$.

(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.

(D) The reaction described is the Liebermann's nitroso reaction.
In this reaction,a secondary amine reacts with nitrous acid $(HNO_2)$ to form a nitroso amine $(R_2N-N=O)$.
When this nitroso amine is heated with concentrated $H_2SO_4$,it undergoes hydrolysis to regenerate the secondary amine.

(D) Teflon (Polytetrafluoroethylene) has great chemical inertness and high thermal stability,hence it is used for making non-stick utensils.
For this purpose,a thin layer of Teflon is coated on the inner side of the vessel.

(A) The $f$-block elements are divided into two series: the lanthanides ($4f$-series) and the actinides ($5f$-series).
Because both series involve the filling of $f$-orbitals and exhibit similar chemical properties,the actinides are considered to be analogous or homologous to the lanthanides.

(B) The reaction of a primary amine with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction.
This reaction produces an isocyanide (carbylamine),which has a foul smell.
The general chemical equation is: $R-NH_2 + CHCl_3 + 3KOH_{(alc)} \to R-NC + 3KCl + 3H_2O$.

(A) Colligative properties depend on the number of particles in the solution.
Benzoic acid $(C_6H_5COOH)$ undergoes dimerisation in non-polar solvents like benzene due to hydrogen bonding.
This process reduces the total number of particles in the solution.
Since the colligative property is inversely proportional to the molar mass,a decrease in the number of particles leads to an abnormally high observed molecular mass compared to the theoretical value.

(D) The Assertion is incorrect because the electrical conductivity of metals like copper decreases with an increase in temperature. This happens because an increase in temperature causes increased vibrations of metal ions,which increases the resistance to the flow of electrons.
The Reason is correct because the electrical conductivity of metals is indeed due to the motion of free electrons.
Therefore,the Assertion is incorrect,but the Reason is correct.

(A) catalyst works by providing an alternative reaction pathway with lower activation energy. The reaction occurs on the surface of the catalyst. By converting the catalyst into a finely divided form,the total surface area available for the reactant molecules to adsorb increases significantly. This leads to a higher rate of reaction,making the catalyst more effective.

(C) Deep sea divers use a mixture of $He$ and $O_2$ for respiration because $He$ is much less soluble in blood than $N_2$.
This prevents the formation of bubbles of $N_2$ in the blood when the diver ascends to the surface,a condition known as decompression sickness or 'bends'.
Therefore,the Assertion is correct,but the Reason is incorrect because $He$ is actually sparingly soluble in blood.

(C) Pure iron is soft and stretches easily when hot. It is not suitable for making tools and machines because it lacks the necessary hardness and strength.
To make it useful,iron is mixed with small amounts of carbon to form steel,which is much harder and more durable.
Therefore,the Assertion is correct,but the Reason is incorrect because pure iron is soft,not hard.

(A) $AgCl$ is a sparingly soluble salt in water.
When $NH_4OH$ is added,it reacts with $AgCl$ to form a soluble coordination complex,diamminesilver$(I)$ chloride,which causes the precipitate to dissolve.
The chemical reaction is: $AgCl(s) + 2NH_4OH(aq) \to [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(B) Proteins are polymers of $\alpha$-amino acids linked by peptide bonds. On hydrolysis,they break down into their constituent amino acids.
Amino acids are bifunctional molecules containing both an amino group $(-NH_2)$ and a carboxylic acid group $(-COOH)$.
While both statements are scientifically correct,the presence of these functional groups in amino acids is not the direct reason why proteins produce amino acids upon hydrolysis; the hydrolysis is due to the cleavage of peptide bonds. Therefore,the Reason is not the correct explanation of the Assertion.