ChemistryQ51–66 of 103 questions
Page 2 of 2 · English
51
ChemistryMCQAIPMT · 1989
Maximum contribution of $O_2$ is from
A
Phytoplankton
B
Grasslands
C
Herbs and shrubs
D
Dense forests
Solution
(A) The correct answer is $A$.
Phytoplankton are microscopic aquatic organisms that perform photosynthesis on a massive scale.
They are responsible for approximately $50\%$ to $80\%$ of the Earth's oxygen production.
Because they are present in vast numbers across the world's oceans and perform primary production efficiently,they contribute more to the global $O_2$ supply than terrestrial forests or grasslands.
Phytoplankton are microscopic aquatic organisms that perform photosynthesis on a massive scale.
They are responsible for approximately $50\%$ to $80\%$ of the Earth's oxygen production.
Because they are present in vast numbers across the world's oceans and perform primary production efficiently,they contribute more to the global $O_2$ supply than terrestrial forests or grasslands.
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52
ChemistryMCQAIPMT · 1989
Which of the following has the lowest boiling point?
A
$NH_3$
B
$PH_3$
C
$AsH_3$
D
$SbH_3$
Solution
(B) The boiling point of hydrides of group $15$ elements generally increases down the group due to an increase in molecular mass and the resulting increase in van der Waals forces.
However,$NH_3$ has an anomalously high boiling point due to the presence of intermolecular hydrogen bonding.
Comparing the remaining hydrides $(PH_3, AsH_3, SbH_3)$,$PH_3$ has the lowest molecular mass,resulting in the weakest van der Waals forces.
Therefore,$PH_3$ has the lowest boiling point among the given options.
However,$NH_3$ has an anomalously high boiling point due to the presence of intermolecular hydrogen bonding.
Comparing the remaining hydrides $(PH_3, AsH_3, SbH_3)$,$PH_3$ has the lowest molecular mass,resulting in the weakest van der Waals forces.
Therefore,$PH_3$ has the lowest boiling point among the given options.
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53
ChemistryEasyMCQAIPMT · 1989
The modern periodic table is based on the atomic number of the elements. The experiment which proved the significance of the atomic number was:
A
Millikan's oil drop experiment
B
Moseley's work on $X$-ray spectra
C
Bragg's work on $X$-ray diffraction
D
Discovery of $X$-rays by Rontgen
Solution
(B) Moseley's work on $X$-ray spectra established that the frequency of $X$-rays emitted by an element is related to its atomic number $(Z)$ by the equation $\sqrt{\nu} = a(Z - b)$,which proved the significance of the atomic number over atomic mass.
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54
ChemistryDifficultMCQAIPMT · 1989
Which of the following structures corresponds to the product expected when an excess of $C_6H_6$ reacts with $CH_2Cl_2$ in the presence of anhydrous $AlCl_3$?
A
B
C
D
Solution
(D) The reaction of benzene $(C_6H_6)$ with dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction. Since benzene is in excess,both chlorine atoms of $CH_2Cl_2$ are replaced by phenyl groups,resulting in the formation of diphenylmethane $(C_6H_5-CH_2-C_6H_5)$ and $2HCl$. The reaction is: $2C_6H_6 + CH_2Cl_2 \xrightarrow{AlCl_3} C_6H_5-CH_2-C_6H_5 + 2HCl$.
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55
ChemistryMCQAIPMT · 1989
Which of the following structures corresponds to the product expected when an excess of $C_6H_6$ reacts with $CH_2Cl_2$ in the presence of anhydrous $AlCl_3$?
A
B
C
D
Solution
(D) The reaction between benzene $(C_6H_6)$ and dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
Since $C_6H_6$ is in excess,both chlorine atoms of $CH_2Cl_2$ will be replaced by phenyl groups.
Step $1$: $C_6H_6 + CH_2Cl_2 \xrightarrow{AlCl_3} C_6H_5-CH_2Cl + HCl$
Step $2$: $C_6H_5-CH_2Cl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5-CH_2-C_6H_5 + HCl$
The final product is diphenylmethane $(C_6H_5-CH_2-C_6H_5)$.
Since $C_6H_6$ is in excess,both chlorine atoms of $CH_2Cl_2$ will be replaced by phenyl groups.
Step $1$: $C_6H_6 + CH_2Cl_2 \xrightarrow{AlCl_3} C_6H_5-CH_2Cl + HCl$
Step $2$: $C_6H_5-CH_2Cl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5-CH_2-C_6H_5 + HCl$
The final product is diphenylmethane $(C_6H_5-CH_2-C_6H_5)$.
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56
ChemistryMCQAIPMT · 1989
Earthworms are ........
A
Useful
B
Harmful
C
More useful than harmful
D
More harmful
Solution
(C) Earthworms are known as the 'friends of farmers' because they improve soil fertility by burrowing,which aerates the soil,and by producing vermicompost through the decomposition of organic matter. While they may occasionally damage delicate seedlings,their overall contribution to agriculture and soil health is significantly beneficial. Therefore,they are considered more useful than harmful.
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57
ChemistryMCQAIPMT · 1989
Which of the following hormones can be used to replace vernalization?
A
Auxin
B
Cytokinins
C
Gibberellin
D
Ethylene
Solution
(C) Vernalization is the process of inducing flowering by exposure to low temperatures. $Gibberellins$ are plant hormones that can promote flowering in many plants, effectively substituting for the cold treatment required for vernalization. Therefore, $Gibberellin$ is the correct answer.
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58
ChemistryMCQAIPMT · 1989
Which of the following statements does not form part of Bohr's model of the hydrogen atom?
A
Energy of the electrons in the orbit is quantized
B
The electron in the orbit nearest to the nucleus has the lowest energy
C
Electrons revolve in different orbits around the nucleus
D
The position and velocity of the electrons in the orbit cannot be determined simultaneously
Solution
(D) The statement $(D)$ does not form a part of Bohr's model of the hydrogen atom.
Bohr assumed that an electron in an atom is located at a definite distance from the nucleus and is revolving with a definite velocity around it.
This assumption contradicts Heisenberg's uncertainty principle,which states that the position and velocity of an electron cannot be determined simultaneously with absolute precision.
Bohr assumed that an electron in an atom is located at a definite distance from the nucleus and is revolving with a definite velocity around it.
This assumption contradicts Heisenberg's uncertainty principle,which states that the position and velocity of an electron cannot be determined simultaneously with absolute precision.
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59
ChemistryMCQAIPMT · 1989
The ratio of $C_p$ and $C_v$ of a gas $'X'$ is $1.4$. The number of atoms of the gas $'X'$ present in $11.2 \ L$ of it at $NTP$ will be:
A
$6.02 \times 10^{23}$
B
$1.2 \times 10^{23}$
C
$3.01 \times 10^{23}$
D
$2.01 \times 10^{23}$
Solution
(A) The ratio $\frac{C_p}{C_v} = 1.4$ indicates that the gas is diatomic (e.g.,$O_2, N_2$).
Since the gas is diatomic,the number of atoms per molecule is $2$.
At $NTP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,moles of gas $'X' = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.
Number of molecules $= 0.5 \times N_A$.
Number of atoms $= 0.5 \times N_A \times 2 = N_A = 6.02 \times 10^{23}$.
Since the gas is diatomic,the number of atoms per molecule is $2$.
At $NTP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,moles of gas $'X' = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.
Number of molecules $= 0.5 \times N_A$.
Number of atoms $= 0.5 \times N_A \times 2 = N_A = 6.02 \times 10^{23}$.
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60
ChemistryMCQAIPMT · 1989
Two gaseous equilibria $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ and $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ have equilibrium constants $K_{1}$ and $K_{2}$ respectively at $298 \ K$. Which of the following relationships between $K_{1}$ and $K_{2}$ is correct?
A
$K_{1} = K_{2}$
B
$K_{2} = (K_{1})^{2}$
C
$K_{2} = (\frac{1}{K_{1}})^{2}$
D
$K_{2} = \frac{1}{K_{1}}$
Solution
(C) For the first reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_{1} = \frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$.
For the second reaction: $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_{2} = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$.
Comparing the two expressions,we can see that $K_{2} = \left( \frac{[SO_{2}][O_{2}]^{1/2}}{[SO_{3}]} \right)^{2}$.
Since $\frac{[SO_{2}][O_{2}]^{1/2}}{[SO_{3}]} = \frac{1}{K_{1}}$,it follows that $K_{2} = (\frac{1}{K_{1}})^{2}$.
For the second reaction: $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_{2} = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$.
Comparing the two expressions,we can see that $K_{2} = \left( \frac{[SO_{2}][O_{2}]^{1/2}}{[SO_{3}]} \right)^{2}$.
Since $\frac{[SO_{2}][O_{2}]^{1/2}}{[SO_{3}]} = \frac{1}{K_{1}}$,it follows that $K_{2} = (\frac{1}{K_{1}})^{2}$.
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61
ChemistryMCQAIPMT · 1989
If we use pyrene $(CCl_4)$ in the Riemer-Tiemann reaction in place of chloroform,the product formed is
A
Salicylaldehyde
B
Phenolphthalein
C
Salicylic acid
D
Cyclohexanol
Solution
(C) In the standard Riemer-Tiemann reaction,phenol reacts with chloroform $(CHCl_3)$ and aqueous $NaOH$ to form salicylaldehyde.
When carbon tetrachloride $(CCl_4)$ is used instead of chloroform in the presence of a base,the reaction proceeds via the formation of a dichlorocarbene intermediate (or similar electrophilic species) that leads to the introduction of a carboxyl group $(-COOH)$ at the ortho position of the phenol.
Therefore,the product formed is salicylic acid ($2$-hydroxybenzoic acid).
When carbon tetrachloride $(CCl_4)$ is used instead of chloroform in the presence of a base,the reaction proceeds via the formation of a dichlorocarbene intermediate (or similar electrophilic species) that leads to the introduction of a carboxyl group $(-COOH)$ at the ortho position of the phenol.
Therefore,the product formed is salicylic acid ($2$-hydroxybenzoic acid).
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62
ChemistryMCQAIPMT · 1989
Which of the following has the lowest boiling point?
A
$NH_3$
B
$PH_3$
C
$AsH_3$
D
$SbH_3$
Solution
(B) As we move down the group,the boiling point of hydrides generally increases due to an increase in molecular mass and van der Waals forces.
However,$NH_3$ has an abnormally high boiling point due to intermolecular $H$-bonding.
The order of boiling points for these hydrides is $PH_3 < AsH_3 < NH_3 < SbH_3$.
Therefore,$PH_3$ has the lowest boiling point.
However,$NH_3$ has an abnormally high boiling point due to intermolecular $H$-bonding.
The order of boiling points for these hydrides is $PH_3 < AsH_3 < NH_3 < SbH_3$.
Therefore,$PH_3$ has the lowest boiling point.
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63
ChemistryMCQAIPMT · 1989
Which of the following has the lowest boiling point?
A
$NH_3$
B
$PH_3$
C
$AsH_3$
D
$SbH_3$
Solution
(B) The boiling point of hydrides of group $15$ elements generally increases down the group due to an increase in molecular mass and the resulting increase in van der Waals forces.
However,$NH_3$ has an anomalously high boiling point due to intermolecular hydrogen bonding.
Comparing the remaining hydrides $(PH_3, AsH_3, SbH_3)$,$PH_3$ has the lowest molecular mass and therefore the weakest van der Waals forces.
Thus,$PH_3$ has the lowest boiling point among the given options.
However,$NH_3$ has an anomalously high boiling point due to intermolecular hydrogen bonding.
Comparing the remaining hydrides $(PH_3, AsH_3, SbH_3)$,$PH_3$ has the lowest molecular mass and therefore the weakest van der Waals forces.
Thus,$PH_3$ has the lowest boiling point among the given options.
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64
ChemistryDifficultMCQAIPMT · 1989
The most important method of preparation of hydrocarbons of lower carbon number is
A
Pyrolysis of higher carbon number of hydrocarbons
B
Electrolysis of salts of fatty acids
C
Sabatier and Senderens reaction
D
Direct synthesis
Solution
(A) Pyrolysis (or cracking) is the process of breaking down higher alkanes into lower alkanes and alkenes by heating them in the absence of air.
For example: $C_6H_{14} \xrightarrow[\Delta]{\text{Pyrolysis}} C_2H_4 + C_4H_{10}$
This is the most effective industrial method for obtaining hydrocarbons with a lower carbon number from higher ones.
For example: $C_6H_{14} \xrightarrow[\Delta]{\text{Pyrolysis}} C_2H_4 + C_4H_{10}$
This is the most effective industrial method for obtaining hydrocarbons with a lower carbon number from higher ones.
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65
ChemistryMCQAIPMT · 1989
Lichens indicate $SO_2$ pollution because they
A
show association between algae and fungi.
B
grow faster than others.
C
are sensitive to $SO_2$.
D
flourish in $SO_2$ rich environment.
Solution
(C) Lichens are formed by a symbiotic relationship between algae or cyanobacteria and fungi.
Lichens typically grow in various environments,but most lichens,especially epiphytic fruticose species and those containing cyanobacteria,are highly sensitive to atmospheric pollutants.
Specifically,they are unable to survive in areas with high concentrations of sulfur dioxide $(SO_2)$.
Therefore,they are widely used as bio-indicators to monitor $SO_2$ pollution levels in the environment.
Lichens typically grow in various environments,but most lichens,especially epiphytic fruticose species and those containing cyanobacteria,are highly sensitive to atmospheric pollutants.
Specifically,they are unable to survive in areas with high concentrations of sulfur dioxide $(SO_2)$.
Therefore,they are widely used as bio-indicators to monitor $SO_2$ pollution levels in the environment.
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66
ChemistryMCQAIPMT · 1989
Which of the following molecules does not have a linear arrangement of atoms?
A
$H_2S$
B
$C_2H_2$
C
$BeCl_2$
D
$CO_2$
Solution
(A) The central atom in $H_2S$ is $S$,which has $6$ valence electrons. It forms $2$ single bonds with $H$ atoms and has $2$ lone pairs of electrons. According to $VSEPR$ theory,the presence of $2$ lone pairs on the central atom causes repulsion,resulting in a bent (angular) geometry.
In contrast,$C_2H_2$ $(HC \equiv CH)$,$BeCl_2$ $(Cl-Be-Cl)$,and $CO_2$ $(O=C=O)$ all have a linear geometry.
In contrast,$C_2H_2$ $(HC \equiv CH)$,$BeCl_2$ $(Cl-Be-Cl)$,and $CO_2$ $(O=C=O)$ all have a linear geometry.
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