ChemistryQ1–29 of 29 questions
Page 1 of 1 · English
1
ChemistryMediumMCQAIIMS · 1992
$A$ sample of pure carbon dioxide,irrespective of its source,contains $27.27\%$ carbon and $72.73\%$ oxygen. This data supports:
A
Law of constant composition
B
Law of conservation of mass
C
Law of reciprocal proportions
D
Law of multiple proportions
Solution
(A) The law of constant composition (also known as the law of definite proportions) states that a given chemical compound always contains its component elements in a fixed ratio by mass,regardless of its source or method of preparation.
Since carbon dioxide $(CO_2)$ always contains $27.27\%$ carbon and $72.73\%$ oxygen by mass regardless of its source,this data supports the Law of constant composition.
Since carbon dioxide $(CO_2)$ always contains $27.27\%$ carbon and $72.73\%$ oxygen by mass regardless of its source,this data supports the Law of constant composition.
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2
ChemistryEasyMCQAIIMS · 1992
The normality of $2 \ M$ sulphuric acid is: (in $N$)
A
$2$
B
$4$
C
$0.5$
D
$0.25$
Solution
(B) The relationship between normality $(N)$ and molarity $(M)$ is given by the formula: $N = M \times \text{n-factor}$.
For sulphuric acid $(H_2SO_4)$,the basicity (n-factor) is $2$ because it can provide $2$ protons ($H^+$ ions) per molecule.
Given molarity $(M)$ = $2 \ M$.
Therefore,$N = 2 \times 2 = 4 \ N$.
For sulphuric acid $(H_2SO_4)$,the basicity (n-factor) is $2$ because it can provide $2$ protons ($H^+$ ions) per molecule.
Given molarity $(M)$ = $2 \ M$.
Therefore,$N = 2 \times 2 = 4 \ N$.
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3
ChemistryMediumMCQAIIMS · 1992
How many $g$ of a dibasic acid (Mol. wt. $= 200$) should be present in $100 \ mL$ of its aqueous solution to give decinormal strength?
A
$1$
B
$2$
C
$10$
D
$20$
Solution
(A) For a dibasic acid,the equivalent weight $E$ is calculated as $E = \frac{\text{Molecular weight}}{2} = \frac{200}{2} = 100 \ g/eq$.
Decinormal strength means the normality $N = \frac{1}{10} = 0.1 \ N$.
The formula for normality is $N = \frac{W \times 1000}{E \times V(mL)}$,where $W$ is the weight in grams.
Substituting the values: $0.1 = \frac{W \times 1000}{100 \times 100}$.
$0.1 = \frac{W \times 1000}{10000} \implies 0.1 = \frac{W}{10}$.
$W = 0.1 \times 10 = 1 \ g$.
Decinormal strength means the normality $N = \frac{1}{10} = 0.1 \ N$.
The formula for normality is $N = \frac{W \times 1000}{E \times V(mL)}$,where $W$ is the weight in grams.
Substituting the values: $0.1 = \frac{W \times 1000}{100 \times 100}$.
$0.1 = \frac{W \times 1000}{10000} \implies 0.1 = \frac{W}{10}$.
$W = 0.1 \times 10 = 1 \ g$.
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4
ChemistryMediumMCQAIIMS · 1992
If pressure becomes double at the same absolute temperature on $2 \ L$ $CO_2$,then the volume of $CO_2$ becomes ............ $L$.
A
$2$
B
$4$
C
$0.5$
D
$1$
Solution
(D) According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$.
Given: $V_1 = 2 \ L$,$P_2 = 2P_1$.
Substituting the values: $P_1 \times 2 \ L = (2P_1) \times V_2$.
$V_2 = \frac{P_1 \times 2 \ L}{2P_1} = 1 \ L$.
Therefore,the volume becomes $1 \ L$.
Given: $V_1 = 2 \ L$,$P_2 = 2P_1$.
Substituting the values: $P_1 \times 2 \ L = (2P_1) \times V_2$.
$V_2 = \frac{P_1 \times 2 \ L}{2P_1} = 1 \ L$.
Therefore,the volume becomes $1 \ L$.
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5
ChemistryEasyMCQAIIMS · 1992
$A$ well-stoppered thermos flask contains some ice cubes. This is an example of a
A
Closed system
B
Open system
C
Isolated system
D
Non-thermodynamic system
Solution
(C) In an isolated system,neither the exchange of matter nor the exchange of energy is possible with the surroundings. Since a well-stoppered thermos flask prevents both heat transfer and matter exchange,it acts as an isolated system.
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6
ChemistryMediumMCQAIIMS · 1992
The chloride of an element $A$ gives a neutral solution in water. In the periodic table,the element $A$ belongs to
A
First group
B
Third group
C
Fifth group
D
First transition series
Solution
(A) The chloride of an element $A$ forms a neutral solution in water if it is a salt of a strong acid and a strong base.
Elements of the $1^{st}$ group (alkali metals) form chlorides like $NaCl$ or $KCl$,which are salts of a strong base $(NaOH/KOH)$ and a strong acid $(HCl)$.
These salts do not undergo hydrolysis,resulting in a neutral solution with $pH = 7$.
Therefore,the element $A$ belongs to the $1^{st}$ group.
Elements of the $1^{st}$ group (alkali metals) form chlorides like $NaCl$ or $KCl$,which are salts of a strong base $(NaOH/KOH)$ and a strong acid $(HCl)$.
These salts do not undergo hydrolysis,resulting in a neutral solution with $pH = 7$.
Therefore,the element $A$ belongs to the $1^{st}$ group.
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7
ChemistryDifficultMCQAIIMS · 1992
Which of the following transitions involves the maximum amount of energy?
A
$M^{-}_{(g)} \to M_{(g)}$
B
$M_{(g)} \to M^{+}_{(g)}$
C
$M^{+}_{(g)} \to M^{2+}_{(g)}$
D
$M^{2+}_{(g)} \to M^{3+}_{(g)}$
Solution
(D) The correct option is $D$.
In the given transitions,the energy required for the removal of an electron is known as the ionization energy.
As the positive charge on the ion increases,the effective nuclear charge experienced by the remaining electrons increases.
This leads to a stronger electrostatic attraction between the nucleus and the valence electrons.
Therefore,the energy required to remove an electron from $M^{2+}_{(g)}$ to form $M^{3+}_{(g)}$ is significantly higher than the energy required for the previous ionization steps.
In the given transitions,the energy required for the removal of an electron is known as the ionization energy.
As the positive charge on the ion increases,the effective nuclear charge experienced by the remaining electrons increases.
This leads to a stronger electrostatic attraction between the nucleus and the valence electrons.
Therefore,the energy required to remove an electron from $M^{2+}_{(g)}$ to form $M^{3+}_{(g)}$ is significantly higher than the energy required for the previous ionization steps.
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8
ChemistryEasyMCQAIIMS · 1992
The type of bonding in $HCl$ molecule is
A
Pure covalent
B
Polar covalent
C
Highly covalent
D
$H$-bonding
Solution
(B) In an $HCl$ molecule,the electronegativity difference between $H$ $(2.1)$ and $Cl$ $(3.0)$ is approximately $0.9$.
Because the electronegativity difference is greater than $0$ but less than $1.7$,the bond formed is a polar covalent bond,where the shared pair of electrons is shifted towards the more electronegative chlorine atom.
Because the electronegativity difference is greater than $0$ but less than $1.7$,the bond formed is a polar covalent bond,where the shared pair of electrons is shifted towards the more electronegative chlorine atom.
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9
ChemistryEasyMCQAIIMS · 1992
Which of the following pairs represents types of stereoisomerism?
A
Geometrical isomerism,position isomerism
B
Geometrical isomerism,conformational isomerism
C
Optical isomerism,geometrical isomerism
D
Optical isomerism,metamerism
Solution
(C) Stereoisomerism is a type of isomerism in which molecules have the same molecular formula and sequence of bonded atoms (constitution),but differ in the three-dimensional orientations of their atoms in space.
Stereoisomerism is broadly classified into two main types:
$1$. Geometrical isomerism (cis-trans isomerism).
$2$. Optical isomerism (enantiomers and diastereomers).
Therefore,the pair representing stereoisomerism is $Optical \ isomerism$ and $geometrical \ isomerism$.
Stereoisomerism is broadly classified into two main types:
$1$. Geometrical isomerism (cis-trans isomerism).
$2$. Optical isomerism (enantiomers and diastereomers).
Therefore,the pair representing stereoisomerism is $Optical \ isomerism$ and $geometrical \ isomerism$.
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10
ChemistryEasyMCQAIIMS · 1992
How many grams of dibasic acid (mol. wt. $200$) should be present in $100 \ mL$ of its aqueous solution to give decinormal strength?
A
$1$
B
$2$
C
$10$
D
$20$
Solution
(A) Given: Molecular weight $(M_w)$ = $200 \ g/mol$.
Since the acid is dibasic,its basicity is $2$.
Equivalent weight $(E)$ = $\frac{M_w}{\text{basicity}} = \frac{200}{2} = 100 \ g/eq$.
Strength $(N)$ = $0.1 \ N$ (decinormal).
Volume $(V)$ = $100 \ mL = 0.1 \ L$.
Using the formula: $N = \frac{w}{E \times V(L)}$,where $w$ is the weight in grams.
$0.1 = \frac{w}{100 \times 0.1}$.
$0.1 = \frac{w}{10}$.
$w = 0.1 \times 10 = 1 \ g$.
Since the acid is dibasic,its basicity is $2$.
Equivalent weight $(E)$ = $\frac{M_w}{\text{basicity}} = \frac{200}{2} = 100 \ g/eq$.
Strength $(N)$ = $0.1 \ N$ (decinormal).
Volume $(V)$ = $100 \ mL = 0.1 \ L$.
Using the formula: $N = \frac{w}{E \times V(L)}$,where $w$ is the weight in grams.
$0.1 = \frac{w}{100 \times 0.1}$.
$0.1 = \frac{w}{10}$.
$w = 0.1 \times 10 = 1 \ g$.
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11
ChemistryEasyMCQAIIMS · 1992
Normality of $2 \, M$ sulphuric acid is
A
$2 \, N$
B
$4 \, N$
C
$N / 2$
D
$N / 4$
Solution
(B) The relationship between Normality $(N)$ and Molarity $(M)$ is given by the formula: $N = M \times \text{basicity}$.
For sulphuric acid $(H_2SO_4)$,the basicity is $2$ because it can provide $2$ replaceable $H^+$ ions.
Given $M = 2 \, M$.
Therefore,$N = 2 \times 2 = 4 \, N$.
For sulphuric acid $(H_2SO_4)$,the basicity is $2$ because it can provide $2$ replaceable $H^+$ ions.
Given $M = 2 \, M$.
Therefore,$N = 2 \times 2 = 4 \, N$.
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12
ChemistryMCQAIIMS · 1992
Arsenic drugs are mainly used in the treatment of
A
Jaundice
B
Typhoid
C
Syphilis
D
Cholera
Solution
(C) Arsenic-based drugs,such as Salvarsan,were historically developed by Paul Ehrlich and were primarily used for the treatment of $Syphilis$.
Therefore,the correct option is $(C)$.
Therefore,the correct option is $(C)$.
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13
ChemistryMCQAIIMS · 1992
The formula of sodium nitroprusside is:
A
$Na_4[Fe(CN)_5NOS]$
B
$Na_2[Fe(CN)_5NO]$
C
$NaFe[Fe(CN)_6]$
D
$Na_2[Fe(CN)_6NO_2]$
Solution
(B) Sodium nitroprusside is a well-known coordination compound used in laboratory tests for the detection of sulfide ions.
Its chemical formula is $Na_2[Fe(CN)_5NO]$.
In this complex,the iron is in the $+II$ oxidation state,and the ligand $NO^+$ (nitrosonium) is coordinated to the central metal atom.
Its chemical formula is $Na_2[Fe(CN)_5NO]$.
In this complex,the iron is in the $+II$ oxidation state,and the ligand $NO^+$ (nitrosonium) is coordinated to the central metal atom.
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14
ChemistryMediumMCQAIIMS · 1992
The $IUPAC$ name for $CH_3CH(OH)CH_2-C(CH_3)_2OH$ is
A
$1,1-$dimethyl$-1,3-$butanediol
B
$2-$methyl$-2,4-$pentanediol
C
$4-$methyl$-2,4-$pentanediol
D
$1,3,3-$trimethyl$-1,3-$propanediol
Solution
(B) The given structure is $CH_3-CH(OH)-CH_2-C(CH_3)_2OH$.
To determine the $IUPAC$ name,we select the longest carbon chain containing the functional groups (hydroxyl groups).
The chain has $5$ carbons,so the parent alkane is pentane.
Numbering the chain from the end that gives the lowest locants to the substituents and functional groups:
Starting from the right,the hydroxyl groups are at positions $2$ and $4$,and the methyl group is at position $2$.
Thus,the name is $2-$methyl$-2,4-$pentanediol.
To determine the $IUPAC$ name,we select the longest carbon chain containing the functional groups (hydroxyl groups).
The chain has $5$ carbons,so the parent alkane is pentane.
Numbering the chain from the end that gives the lowest locants to the substituents and functional groups:
Starting from the right,the hydroxyl groups are at positions $2$ and $4$,and the methyl group is at position $2$.
Thus,the name is $2-$methyl$-2,4-$pentanediol.
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15
ChemistryMediumMCQAIIMS · 1992
Which of the following exhibits optical isomerism?
A
Butan$-1-$ol
B
Butan$-2-$ol
C
But$-1-$ene
D
But$-2-$ene
Solution
(B) The correct answer is $(b)$.
Butan$-2-$ol,represented as $CH_3-CH(OH)-CH_2-CH_3$,exhibits optical isomerism.
This is because the carbon atom at the $C-2$ position is a chiral center,meaning it is bonded to four different groups: a hydrogen atom $(-H)$,a hydroxyl group $(-OH)$,a methyl group $(-CH_3)$,and an ethyl group $(-CH_2CH_3)$.
Butan$-2-$ol,represented as $CH_3-CH(OH)-CH_2-CH_3$,exhibits optical isomerism.
This is because the carbon atom at the $C-2$ position is a chiral center,meaning it is bonded to four different groups: a hydrogen atom $(-H)$,a hydroxyl group $(-OH)$,a methyl group $(-CH_3)$,and an ethyl group $(-CH_2CH_3)$.
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16
ChemistryMCQAIIMS · 1992
The formula of sodium nitroprusside is .........
A
$Na_4[Fe(CN)_5NOS]$
B
$Na_2[Fe(CN)_5NO]$
C
$NaFe[Fe(CN)_6]$
D
$Na_2[Fe(CN)_6NO_2]$
Solution
(B) Sodium nitroprusside is a coordination compound with the chemical formula $Na_2[Fe(CN)_5NO]$.
In this complex,the iron is in the $+II$ oxidation state,and the ligand $NO^+$ (nitrosonium ion) is coordinated to the central metal atom.
In this complex,the iron is in the $+II$ oxidation state,and the ligand $NO^+$ (nitrosonium ion) is coordinated to the central metal atom.
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17
ChemistryMediumMCQAIIMS · 1992
Identify the incorrect statement with respect to ozone.
A
Ozone is formed in the upper atmosphere by a photochemical reaction involving dioxygen.
B
Ozone is more reactive than oxygen.
C
Ozone is diamagnetic whereas dioxygen is paramagnetic.
D
Ozone protects the earth's inhabitants by absorbing $\gamma$ radiations.
Solution
(D) Ozone protects the earth's inhabitants by absorbing $UV$ radiations,not $\gamma$ radiations.
Ozone $(O_3)$ is diamagnetic,whereas dioxygen $(O_2)$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals.
Ozone is thermodynamically unstable and decomposes to give atomic oxygen,making it more reactive than stable dioxygen.
Ozone is formed in the upper atmosphere by the action of $UV$ radiation on dioxygen.
Therefore,the statement in option $D$ is incorrect.
Ozone $(O_3)$ is diamagnetic,whereas dioxygen $(O_2)$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals.
Ozone is thermodynamically unstable and decomposes to give atomic oxygen,making it more reactive than stable dioxygen.
Ozone is formed in the upper atmosphere by the action of $UV$ radiation on dioxygen.
Therefore,the statement in option $D$ is incorrect.
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18
ChemistryDifficultMCQAIIMS · 1992
$Cu^{2+}$ ions will be reduced to $Cu^{+}$ ions by the addition of an aqueous solution of
A
$KF$
B
$KCl$
C
$KI$
D
$KOH$
Solution
(C) The reduction of $Cu^{2+}$ to $Cu^{+}$ occurs when $Cu^{2+}$ reacts with iodide ions $(I^-)$ because $CuI$ is insoluble and the reaction is thermodynamically favorable.
The balanced chemical equation is: $2Cu^{2+} + 4I^- \to 2CuI(s) + I_2(s)$.
The balanced chemical equation is: $2Cu^{2+} + 4I^- \to 2CuI(s) + I_2(s)$.
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19
ChemistryEasyMCQAIIMS · 1992
An aqueous solution of glucose is $10\%$ in strength. The volume in which $1 \, \text{g-mole}$ of it is dissolved will be ........... $L$.
A
$18$
B
$9$
C
$0.9$
D
$1.8$
Solution
(D) $10\%$ aqueous solution of glucose means $10 \, g$ of glucose is present in $100 \, mL$ of solution.
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \, g/mol$.
Number of moles in $100 \, mL$ solution = $\frac{10 \, g}{180 \, g/mol} = \frac{1}{18} \, mol$.
This means $\frac{1}{18} \, mol$ is present in $100 \, mL$ $(0.1 \, L)$.
Therefore,$1 \, mol$ of glucose will be present in volume = $\frac{0.1 \, L \times 1 \, mol}{1/18 \, mol} = 0.1 \times 18 = 1.8 \, L$.
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \, g/mol$.
Number of moles in $100 \, mL$ solution = $\frac{10 \, g}{180 \, g/mol} = \frac{1}{18} \, mol$.
This means $\frac{1}{18} \, mol$ is present in $100 \, mL$ $(0.1 \, L)$.
Therefore,$1 \, mol$ of glucose will be present in volume = $\frac{0.1 \, L \times 1 \, mol}{1/18 \, mol} = 0.1 \times 18 = 1.8 \, L$.
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20
ChemistryMediumMCQAIIMS · 1992
The number of electrons required to deposit $1 \ g$ atom of aluminium (at. wt. $= 27$) from a solution of aluminium chloride will be ............ $N$ (where $N$ is Avogadro's number).
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(C) The reduction reaction for the deposition of aluminium from aluminium chloride $(AlCl_3)$ is:
$Al^{3+} + 3e^- \rightarrow Al(s)$
From the stoichiometry of the reaction,$1 \ mole$ of $Al^{3+}$ ions requires $3 \ moles$ of electrons to be reduced to $1 \ mole$ of $Al$ metal.
Since $1 \ g$ atom of $Al$ is equivalent to $1 \ mole$ of $Al$ atoms,the number of electrons required is $3 \ moles$.
Since $1 \ mole$ contains $N$ electrons (where $N$ is Avogadro's number),the total number of electrons required is $3N$.
$Al^{3+} + 3e^- \rightarrow Al(s)$
From the stoichiometry of the reaction,$1 \ mole$ of $Al^{3+}$ ions requires $3 \ moles$ of electrons to be reduced to $1 \ mole$ of $Al$ metal.
Since $1 \ g$ atom of $Al$ is equivalent to $1 \ mole$ of $Al$ atoms,the number of electrons required is $3 \ moles$.
Since $1 \ mole$ contains $N$ electrons (where $N$ is Avogadro's number),the total number of electrons required is $3N$.
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21
ChemistryEasyMCQAIIMS · 1992
Charge required to liberate $11.5 \ g$ of sodium is:
A
$0.5 \ F$
B
$0.1 \ F$
C
$1.5 \ F$
D
$96500 \ C$
Solution
(A) The reduction reaction for sodium is: $Na^{+} + e^{-} \to Na$
From the stoichiometry of the reaction,$1 \ mol$ of $e^{-}$ (which is $1 \ F$ of charge) is required to deposit $1 \ mol$ of $Na$.
Molar mass of $Na = 23 \ g/mol$.
Moles of $Na = \frac{11.5 \ g}{23 \ g/mol} = 0.5 \ mol$.
Therefore,the charge required $= 0.5 \ mol \times 1 \ F/mol = 0.5 \ F$.
From the stoichiometry of the reaction,$1 \ mol$ of $e^{-}$ (which is $1 \ F$ of charge) is required to deposit $1 \ mol$ of $Na$.
Molar mass of $Na = 23 \ g/mol$.
Moles of $Na = \frac{11.5 \ g}{23 \ g/mol} = 0.5 \ mol$.
Therefore,the charge required $= 0.5 \ mol \times 1 \ F/mol = 0.5 \ F$.
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22
ChemistryMediumMCQAIIMS · 1992
The formula of sodium nitroprusside is
A
$Na_4[Fe(CN)_5NOS]$
B
$Na_2[Fe(CN)_5NO]$
C
$NaFe[Fe(CN)_6]$
D
$Na_2[Fe(CN)_6NO_2]$
Solution
(B) The chemical formula of sodium nitroprusside is $Na_2[Fe(CN)_5NO]$.
It is a coordination compound where the central metal ion is $Fe^{2+}$ and the ligands are five $CN^-$ ions and one $NO^+$ (nitrosonium) ion.
The correct option is $B$.
It is a coordination compound where the central metal ion is $Fe^{2+}$ and the ligands are five $CN^-$ ions and one $NO^+$ (nitrosonium) ion.
The correct option is $B$.
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23
ChemistryAdvancedMCQAIIMS · 1992
Treatment of ammonia with excess of ethyl chloride will yield
A
Diethyl amine
B
Ethane
C
Tetraethyl ammonium chloride
D
Methyl amine
Solution
(C) The reaction of ammonia with an alkyl halide is a nucleophilic substitution reaction that proceeds until the quaternary ammonium salt is formed if the alkyl halide is in excess.
$NH_3 + C_2H_5Cl \rightarrow C_2H_5NH_2 + HCl$
$C_2H_5NH_2 + C_2H_5Cl \rightarrow (C_2H_5)_2NH + HCl$
$(C_2H_5)_2NH + C_2H_5Cl \rightarrow (C_2H_5)_3N + HCl$
$(C_2H_5)_3N + C_2H_5Cl \rightarrow [(C_2H_5)_4N]^+ Cl^-$
Since ethyl chloride is in excess,the final product is the quaternary ammonium salt,which is $Tetraethyl \ ammonium \ chloride$.
$NH_3 + C_2H_5Cl \rightarrow C_2H_5NH_2 + HCl$
$C_2H_5NH_2 + C_2H_5Cl \rightarrow (C_2H_5)_2NH + HCl$
$(C_2H_5)_2NH + C_2H_5Cl \rightarrow (C_2H_5)_3N + HCl$
$(C_2H_5)_3N + C_2H_5Cl \rightarrow [(C_2H_5)_4N]^+ Cl^-$
Since ethyl chloride is in excess,the final product is the quaternary ammonium salt,which is $Tetraethyl \ ammonium \ chloride$.
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24
ChemistryDifficultMCQAIIMS · 1992
In the presence of $NaOH$,phenol reacts with $CHCl_3$ to form $o$-hydroxybenzaldehyde. This reaction is called:
A
Reimer-Tiemann reaction
B
Sandmeyer's reaction
C
Hoffmann's degradation reaction
D
Gattermann's aldehyde synthesis
Solution
(A) When phenol is treated with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ at $340 \ K$,it forms $2$-hydroxybenzaldehyde,also known as salicylaldehyde.
This specific chemical transformation is known as the Reimer-Tiemann reaction.
This specific chemical transformation is known as the Reimer-Tiemann reaction.
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25
ChemistryMediumMCQAIIMS · 1992
On boiling with concentrated hydrobromic acid,phenyl ethyl ether will yield:
A
Phenol and ethyl bromide
B
Phenol and ethane
C
Bromobenzene and ethanol
D
Bromobenzene and ethane
Solution
(A) The reaction of phenyl ethyl ether $(C_6H_5-O-C_2H_5)$ with concentrated hydrobromic acid $(HBr)$ involves the cleavage of the $C-O$ bond.
Since the $C-O$ bond between the phenyl group and oxygen has partial double bond character due to resonance,it is stronger and does not break easily.
Therefore,the cleavage occurs at the $O-C_2H_5$ bond,resulting in the formation of phenol $(C_6H_5OH)$ and ethyl bromide $(C_2H_5Br)$.
The reaction is: $C_6H_5OC_2H_5 + HBr \rightarrow C_6H_5OH + C_2H_5Br$.
Since the $C-O$ bond between the phenyl group and oxygen has partial double bond character due to resonance,it is stronger and does not break easily.
Therefore,the cleavage occurs at the $O-C_2H_5$ bond,resulting in the formation of phenol $(C_6H_5OH)$ and ethyl bromide $(C_2H_5Br)$.
The reaction is: $C_6H_5OC_2H_5 + HBr \rightarrow C_6H_5OH + C_2H_5Br$.
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26
ChemistryMediumMCQAIIMS · 1992
Which of the following compounds does not give the iodoform test?
A
$CH_3CH_2OH$
B
$CH_3OH$
C
$CH_3CHO$
D
$PhCOCH_3$
Solution
(B) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$CH_3CH_2OH$ contains the $CH_3CH(OH)-$ group and gives a positive test.
$CH_3CHO$ contains the $CH_3CO-$ group and gives a positive test.
$PhCOCH_3$ (acetophenone) contains the $CH_3CO-$ group attached to a phenyl ring and gives a positive test.
$CH_3OH$ (methanol) does not contain either of these groups,so it does not give the iodoform test.
$CH_3CH_2OH$ contains the $CH_3CH(OH)-$ group and gives a positive test.
$CH_3CHO$ contains the $CH_3CO-$ group and gives a positive test.
$PhCOCH_3$ (acetophenone) contains the $CH_3CO-$ group attached to a phenyl ring and gives a positive test.
$CH_3OH$ (methanol) does not contain either of these groups,so it does not give the iodoform test.
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27
ChemistryDifficultMCQAIIMS · 1992
Aniline on treatment with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ yields:
A
$o-$ and $p-$nitroanilines
B
$m-$nitroaniline
C
$A$ black tarry matter
D
No reaction
Solution
(C) Direct nitration of aniline with conc. $HNO_3$ and conc. $H_2SO_4$ is not possible because the amino group $(-NH_2)$ is highly sensitive to oxidation.
$HNO_3$ acts as a strong oxidizing agent and oxidizes the aniline ring to form a complex mixture of oxidation products,which appears as a black tarry mass.
$HNO_3$ acts as a strong oxidizing agent and oxidizes the aniline ring to form a complex mixture of oxidation products,which appears as a black tarry mass.
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28
ChemistryMediumMCQAIIMS · 1992
$A$ $500 \ g$ toothpaste sample has $0.2 \ g$ fluoride concentration. What is the concentration of $F^-$ in terms of $ppm$ level?
A
$250$
B
$200$
C
$400$
D
$1000$
Solution
(C) The concentration in $ppm$ is calculated using the formula:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
Given:
Mass of solute $(F^-)$ = $0.2 \ g$
Mass of solution = $500 \ g$
$ppm = \frac{0.2}{500} \times 10^6 = 0.0004 \times 10^6 = 400 \ ppm$
Therefore,the correct option is $C$.
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
Given:
Mass of solute $(F^-)$ = $0.2 \ g$
Mass of solution = $500 \ g$
$ppm = \frac{0.2}{500} \times 10^6 = 0.0004 \times 10^6 = 400 \ ppm$
Therefore,the correct option is $C$.
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29
ChemistryMediumMCQAIIMS · 1992
Arsenic drugs are mainly used in the treatment of
A
jaundice
B
typhoid
C
syphilis
D
cholera
Solution
(C) The first magic bullet was fired at syphilis on this day in $1909$. Although specific diseases responded better to some drugs than to others,before the early $1900s$ development of $Salvarsan$,an arsenic-based drug to treat syphilis,drugs were not developed to target a specific disease.
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