ChemistryQ1–59 of 59 questions
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1
ChemistryDifficultMCQAIPMT · 2000
Assuming complete decomposition,the volume of $CO_2$ released at $STP$ on heating $9.85 \ g$ of $BaCO_3$ (Atomic mass of $Ba = 137$) will be ................ $L$.
A
$0.84$
B
$2.24$
C
$4.06$
D
$1.12$
Solution
(D) The balanced chemical equation for the decomposition of $BaCO_3$ is:
$BaCO_3(s) \to BaO(s) + CO_2(g)$
The molar mass of $BaCO_3 = 137 + 12 + (3 \times 16) = 197 \ g/mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ $(197 \ g)$ produces $1 \ mol$ of $CO_2$ gas,which occupies $22.4 \ L$ at $STP$.
Therefore,the volume of $CO_2$ produced from $9.85 \ g$ of $BaCO_3$ is:
$V = \frac{22.4 \ L}{197 \ g} \times 9.85 \ g = 1.12 \ L$.
$BaCO_3(s) \to BaO(s) + CO_2(g)$
The molar mass of $BaCO_3 = 137 + 12 + (3 \times 16) = 197 \ g/mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ $(197 \ g)$ produces $1 \ mol$ of $CO_2$ gas,which occupies $22.4 \ L$ at $STP$.
Therefore,the volume of $CO_2$ produced from $9.85 \ g$ of $BaCO_3$ is:
$V = \frac{22.4 \ L}{197 \ g} \times 9.85 \ g = 1.12 \ L$.
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2
ChemistryMediumMCQAIPMT · 2000
$A$ compound contains atoms of three elements $A$,$B$ and $C$. If the oxidation number of $A$ is $+2$,$B$ is $+5$ and that of $C$ is $-2$,the possible formula of the compound is:
A
$A_3(BC_4)_2$
B
$A_3(B_4C)_2$
C
$ABC_2$
D
$A_2(BC_3)_2$
Solution
(A) For a compound to be neutral,the sum of the oxidation numbers of all atoms must be $0$.
Check option $A$: $A_3(BC_4)_2$
Sum $= 3 \times (+2) + 2 \times [1 \times (+5) + 4 \times (-2)]$
Sum $= 6 + 2 \times [5 - 8]$
Sum $= 6 + 2 \times (-3) = 6 - 6 = 0$.
Since the sum is $0$,the formula $A_3(BC_4)_2$ is correct.
Check option $A$: $A_3(BC_4)_2$
Sum $= 3 \times (+2) + 2 \times [1 \times (+5) + 4 \times (-2)]$
Sum $= 6 + 2 \times [5 - 8]$
Sum $= 6 + 2 \times (-3) = 6 - 6 = 0$.
Since the sum is $0$,the formula $A_3(BC_4)_2$ is correct.
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3
ChemistryDifficultMCQAIPMT · 2000
If change in energy $(\Delta E) = 3 \times 10^{-8} \ J$,$h = 6.64 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$,then the wavelength of the light is:
A
$6.36 \times 10^3 \ \mathring{A}$
B
$6.36 \times 10^5 \ \mathring{A}$
C
$6.64 \times 10^{-8} \ \mathring{A}$
D
$6.36 \times 10^{18} \ \mathring{A}$
Solution
(C) The relationship between energy and wavelength is given by the formula: $\Delta E = \frac{hc}{\lambda}$.
Rearranging for wavelength: $\lambda = \frac{hc}{\Delta E}$.
Substituting the given values: $\lambda = \frac{6.64 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{3 \times 10^{-8} \ J}$.
$\lambda = 6.64 \times 10^{-18} \ m$.
Since $1 \ m = 10^{10} \ \mathring{A}$,we have $\lambda = 6.64 \times 10^{-18} \times 10^{10} \ \mathring{A} = 6.64 \times 10^{-8} \ \mathring{A}$.
Rearranging for wavelength: $\lambda = \frac{hc}{\Delta E}$.
Substituting the given values: $\lambda = \frac{6.64 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{3 \times 10^{-8} \ J}$.
$\lambda = 6.64 \times 10^{-18} \ m$.
Since $1 \ m = 10^{10} \ \mathring{A}$,we have $\lambda = 6.64 \times 10^{-18} \times 10^{10} \ \mathring{A} = 6.64 \times 10^{-8} \ \mathring{A}$.
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4
ChemistryEasyMCQAIPMT · 2000
The bond formed in a crystal by an anion and a cation is:
A
Ionic
B
Metallic
C
Covalent
D
Dipole
Solution
(A) The electrostatic force of attraction between a positively charged cation and a negatively charged anion results in the formation of an $Ionic$ bond. Therefore,the correct option is $(A)$.
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5
ChemistryEasyMCQAIPMT · 2000
At $STP$,$0.50 \ mol$ $H_2$ gas and $1.0 \ mol$ $He$ gas:
A
Have equal average kinetic energies
B
Have equal molecular speeds
C
Occupy equal volumes
D
Have equal effusion rates
Solution
(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.
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.
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6
ChemistryMediumMCQAIPMT · 2000
Which of the following expressions correctly represents the relationship between the average molar kinetic energy,$\overline{K.E.}$,of $CO$ and $N_2$ molecules at the same temperature?
A
$\overline{KE}_{CO} = \overline{KE}_{N_2}$
B
$\overline{KE}_{CO} > \overline{KE}_{N_2}$
C
$\overline{KE}_{CO} < \overline{KE}_{N_2}$
D
Cannot be predicted unless the volumes of the gases are given
Solution
(A) The average molar kinetic energy of an ideal gas is given by the expression $\overline{K.E.} = \frac{3}{2}RT$.
Since the average molar kinetic energy depends only on the temperature $(T)$ and the universal gas constant $(R)$,it is independent of the nature of the gas or its molecular mass.
Therefore,at the same temperature,the average molar kinetic energy of $CO$ and $N_2$ molecules will be equal.
Thus,$\overline{KE}_{CO} = \overline{KE}_{N_2}$.
Since the average molar kinetic energy depends only on the temperature $(T)$ and the universal gas constant $(R)$,it is independent of the nature of the gas or its molecular mass.
Therefore,at the same temperature,the average molar kinetic energy of $CO$ and $N_2$ molecules will be equal.
Thus,$\overline{KE}_{CO} = \overline{KE}_{N_2}$.
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7
ChemistryMediumMCQAIPMT · 2000
Value of $K_p$ in the reaction $MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2(g)}$ is
A
$K_p = P_{CO_2}$
B
$K_p = P_{CO_2} \times \frac{P_{CO_2} \times P_{MgO}}{P_{MgCO_3}}$
C
$K_p = \frac{P_{CO_2} \times P_{MgO}}{P_{MgCO_3}}$
D
$K_p = \frac{P_{MgCO_3}}{P_{CO_2} \times P_{MgO}}$
Solution
(A) For the given heterogeneous equilibrium reaction: $MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2(g)}$
The equilibrium constant $K_p$ is defined as the product of the partial pressures of the gaseous products divided by the partial pressures of the gaseous reactants,each raised to the power of their stoichiometric coefficients.
Pure solids and liquids are assigned an activity of $1$ and are not included in the expression for $K_p$.
Therefore,$K_p = P_{CO_2}$.
The equilibrium constant $K_p$ is defined as the product of the partial pressures of the gaseous products divided by the partial pressures of the gaseous reactants,each raised to the power of their stoichiometric coefficients.
Pure solids and liquids are assigned an activity of $1$ and are not included in the expression for $K_p$.
Therefore,$K_p = P_{CO_2}$.
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8
ChemistryMediumMCQAIPMT · 2000
The conjugate acid of $NH_2^-$ is
A
$NH_3$
B
$NH_4^+$
C
$NH_2OH$
D
$N_2H_4$
Solution
(A) The conjugate acid of a base is formed by the addition of a proton $(H^+)$ to the base.
For the base $NH_2^-$,the conjugate acid is formed as follows:
$NH_2^- + H^+ \rightarrow NH_3$
Therefore,the correct option is $(A)$.
For the base $NH_2^-$,the conjugate acid is formed as follows:
$NH_2^- + H^+ \rightarrow NH_3$
Therefore,the correct option is $(A)$.
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9
ChemistryMediumMCQAIPMT · 2000
$A$ base dissolved in water yields a solution with a hydroxyl ion concentration of $0.05 \ mol \ L^{-1}$. The solution is
A
Basic
B
Acid
C
Neutral
D
Either $(b)$ or $(c)$
Solution
(A) Given the concentration of hydroxyl ions: $[OH^-] = 0.05 \ mol \ L^{-1} = 5 \times 10^{-2} \ M$.
Calculate $pOH$: $pOH = -\log[OH^-] = -\log(5 \times 10^{-2}) = 2 - \log 5 = 2 - 0.699 = 1.301$.
Calculate $pH$: $pH = 14 - pOH = 14 - 1.301 = 12.699$.
Since the $pH$ is greater than $7$,the solution is basic.
Calculate $pOH$: $pOH = -\log[OH^-] = -\log(5 \times 10^{-2}) = 2 - \log 5 = 2 - 0.699 = 1.301$.
Calculate $pH$: $pH = 14 - pOH = 14 - 1.301 = 12.699$.
Since the $pH$ is greater than $7$,the solution is basic.
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10
ChemistryEasyMCQAIPMT · 2000
Which of the following expressions represents the first law of thermodynamics?
A
$ \Delta E = -q + W $
B
$ \Delta E = q - W $
C
$ \Delta E = q + W $
D
$ \Delta E = -q - W $
Solution
(C) The first law of thermodynamics states that the change in internal energy ($\Delta E$ or $\Delta U$) of a system is equal to the sum of heat added to the system $(q)$ and the work done on the system $(W)$.
Therefore,the mathematical expression is $\Delta E = q + W$.
Therefore,the mathematical expression is $\Delta E = q + W$.
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11
ChemistryEasyMCQAIPMT · 2000
When enthalpy and entropy change for a chemical reaction are $-2.5 \times 10^3 \ cal$ and $7.4 \ cal \ K^{-1}$ respectively,predict if the reaction at $298 \ K$ is:
A
Spontaneous
B
Reversible
C
Irreversible
D
Non-spontaneous
Solution
(A) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T\Delta S$.
Given: $\Delta H = -2.5 \times 10^3 \ cal$,$\Delta S = 7.4 \ cal \ K^{-1}$,and $T = 298 \ K$.
Calculating $\Delta G$: $\Delta G = (-2500) - (298 \times 7.4) = -2500 - 2205.2 = -4705.2 \ cal$.
Since $\Delta G < 0$,the reaction is spontaneous.
Given: $\Delta H = -2.5 \times 10^3 \ cal$,$\Delta S = 7.4 \ cal \ K^{-1}$,and $T = 298 \ K$.
Calculating $\Delta G$: $\Delta G = (-2500) - (298 \times 7.4) = -2500 - 2205.2 = -4705.2 \ cal$.
Since $\Delta G < 0$,the reaction is spontaneous.
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12
ChemistryMediumMCQAIPMT · 2000
The values of heat of formation of $SO_2$ and $SO_3$ are $-298.2 \ kJ$ and $-395.2 \ kJ$ respectively. The heat of reaction for the following reaction will be $... \ kJ$
$SO_2 + \frac{1}{2}O_2 \to SO_3$
$SO_2 + \frac{1}{2}O_2 \to SO_3$
A
$-97.0$
B
$-356.2$
C
$+200.0$
D
$-396.2$
Solution
(A) The heat of reaction $\Delta H_r$ is calculated using the enthalpies of formation of products and reactants:
$\Delta H_r = \Delta H_f(SO_3) - [\Delta H_f(SO_2) + \frac{1}{2} \Delta H_f(O_2)]$
Given:
$\Delta H_f(SO_2) = -298.2 \ kJ/mol$
$\Delta H_f(SO_3) = -395.2 \ kJ/mol$
$\Delta H_f(O_2) = 0 \ kJ/mol$ (standard state)
Substituting the values:
$\Delta H_r = -395.2 - (-298.2) = -395.2 + 298.2 = -97.0 \ kJ$
$\Delta H_r = \Delta H_f(SO_3) - [\Delta H_f(SO_2) + \frac{1}{2} \Delta H_f(O_2)]$
Given:
$\Delta H_f(SO_2) = -298.2 \ kJ/mol$
$\Delta H_f(SO_3) = -395.2 \ kJ/mol$
$\Delta H_f(O_2) = 0 \ kJ/mol$ (standard state)
Substituting the values:
$\Delta H_r = -395.2 - (-298.2) = -395.2 + 298.2 = -97.0 \ kJ$
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13
ChemistryDifficultMCQAIPMT · 2000
When $KMnO_4$ is reduced with oxalic acid in acidic solution,the oxidation number of $Mn$ changes from
A
$7$ to $4$
B
$6$ to $4$
C
$7$ to $2$
D
$4$ to $2$
Solution
(C) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in an acidic medium is:
$5(COOH)_2 + 2KMnO_4 + 3H_2SO_4 \to K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$
In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1 + 4(-2) = 0$,which gives $x = +7$.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$ because the sulfate ion $(SO_4^{2-})$ has a charge of $-2$.
Therefore,the oxidation number of $Mn$ changes from $+7$ to $+2$.
$5(COOH)_2 + 2KMnO_4 + 3H_2SO_4 \to K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$
In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1 + 4(-2) = 0$,which gives $x = +7$.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$ because the sulfate ion $(SO_4^{2-})$ has a charge of $-2$.
Therefore,the oxidation number of $Mn$ changes from $+7$ to $+2$.
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14
ChemistryMediumMCQAIPMT · 2000
$Hydrogen$ $peroxide$ is reduced by:
A
$Ozone$
B
$Barium$ $peroxide$
C
$Acidic$ $solution$ of $KMnO_4$
D
$Lead$ $sulphide$ suspension
Solution
(D) $Hydrogen$ $peroxide$ $(H_2O_2)$ acts as a reducing agent when it reacts with strong oxidizing agents like $O_3$ or $KMnO_4$ in acidic medium.
However,$H_2O_2$ acts as an oxidizing agent when it reacts with $PbS$ (lead sulphide).
The reaction is:
$PbS(s) + 4H_2O_2(aq) \to PbSO_4(s) + 4H_2O(l)$
In this reaction,$PbS$ is oxidized to $PbSO_4$ and $H_2O_2$ is reduced to $H_2O$.
However,$H_2O_2$ acts as an oxidizing agent when it reacts with $PbS$ (lead sulphide).
The reaction is:
$PbS(s) + 4H_2O_2(aq) \to PbSO_4(s) + 4H_2O(l)$
In this reaction,$PbS$ is oxidized to $PbSO_4$ and $H_2O_2$ is reduced to $H_2O$.
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15
ChemistryMediumMCQAIPMT · 2000
In which of the following processes,fused sodium hydroxide is electrolysed at a $330 \ ^oC$ temperature for the extraction of sodium?
A
Castner’s process
B
Down’s process
C
Cyanide process
D
Both $(b)$ and $(c)$
Solution
(A) The correct answer is $A$.
In Castner’s process,fused sodium hydroxide $(NaOH)$ is electrolysed at a temperature of $330 \ ^oC$ to extract metallic sodium $(Na)$.
The reactions are:
At cathode: $Na^+ + e^- \rightarrow Na$
At anode: $4OH^- \rightarrow 2H_2O + O_2 + 4e^-$
In Castner’s process,fused sodium hydroxide $(NaOH)$ is electrolysed at a temperature of $330 \ ^oC$ to extract metallic sodium $(Na)$.
The reactions are:
At cathode: $Na^+ + e^- \rightarrow Na$
At anode: $4OH^- \rightarrow 2H_2O + O_2 + 4e^-$
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16
ChemistryMediumMCQAIPMT · 2000
Glass reacts with $HF$ to produce
A
$SiF_4$
B
$H_2SiF_6$
C
$H_2SiO_3$
D
$Na_3AlF_6$
Solution
(A) Glass is primarily composed of silica $(SiO_2)$. When it reacts with hydrofluoric acid $(HF)$,it forms silicon tetrafluoride $(SiF_4)$ and water.
The reaction is: $SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O$.
Further,$SiF_4$ can react with excess $HF$ to form hydrofluorosilicic acid $(H_2SiF_6)$,but the primary product of the reaction with glass is $SiF_4$.
The reaction is: $SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O$.
Further,$SiF_4$ can react with excess $HF$ to form hydrofluorosilicic acid $(H_2SiF_6)$,but the primary product of the reaction with glass is $SiF_4$.
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17
ChemistryDifficultMCQAIPMT · 2000
Empirical formula of a compound is $CH_2O$ and its vapour density is $30$. The molecular formula of the compound is:
A
$C_3H_6O_3$
B
$C_2H_4O_2$
C
$C_2H_4O$
D
$CH_2O$
Solution
(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: $(Empirical \ Formula)_n = (CH_2O)_2 = C_2H_4O_2$.
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: $(Empirical \ Formula)_n = (CH_2O)_2 = C_2H_4O_2$.
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18
ChemistryMediumMCQAIPMT · 2000
Which of the following $C-H$ bonds has the lowest bond dissociation energy?
A
Primary $(1^o)$ $C-H$ bond
B
Secondary $(2^o)$ $C-H$ bond
C
Tertiary $(3^o)$ $C-H$ bond
D
All of these have equal bond dissociation energy
Solution
(C) The bond dissociation energy of a $C-H$ bond depends on the stability of the resulting alkyl radical formed after homolytic cleavage.
Stability of alkyl radicals follows the order: $3^o > 2^o > 1^o > \text{methyl}$.
Since the $3^o$ radical is the most stable,the energy required to break the $3^o$ $C-H$ bond is the lowest.
Therefore,the tertiary $(3^o)$ $C-H$ bond has the lowest bond dissociation energy.
Stability of alkyl radicals follows the order: $3^o > 2^o > 1^o > \text{methyl}$.
Since the $3^o$ radical is the most stable,the energy required to break the $3^o$ $C-H$ bond is the lowest.
Therefore,the tertiary $(3^o)$ $C-H$ bond has the lowest bond dissociation energy.
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19
ChemistryMediumMCQAIPMT · 2000
The reason for geometrical isomerism in $2-butene$ is:
A
Chiral carbon
B
Free rotation about single bond
C
Free rotation about double bond
D
Restricted rotation about double bond
Solution
(D) Restricted rotation about the $C=C$ double bond is an essential condition for the existence of geometrical isomerism in alkenes like $2-butene$.
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20
ChemistryMediumMCQAIPMT · 2000
$C_6H_6 + CH_3Cl \xrightarrow{anhydrous \ AlCl_3} C_6H_5CH_3 + HCl$ is an example of
A
Friedel-Crafts reaction
B
Kolbe's synthesis
C
Wurtz reaction
D
Grignard reaction
Solution
(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.
This is a classic example of a Friedel-Crafts alkylation reaction.
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21
ChemistryEasyMCQAIPMT · 2000
The strongest reducing agent among the alkali metals is
A
$Li$
B
$Na$
C
$K$
D
$Cs$
Solution
(A) The reducing power of an element is determined by its standard electrode potential $(E^\circ)$.
$Lithium$ $(Li)$ has the most negative standard reduction potential $(E^\circ = -3.04 \ V)$ among all alkali metals.
This high negative value indicates that $Li$ has the greatest tendency to lose electrons in an aqueous medium,making it the strongest reducing agent.
$Lithium$ $(Li)$ has the most negative standard reduction potential $(E^\circ = -3.04 \ V)$ among all alkali metals.
This high negative value indicates that $Li$ has the greatest tendency to lose electrons in an aqueous medium,making it the strongest reducing agent.
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22
ChemistryMCQAIPMT · 2000
Which of the following reagents convert the propene to $1-$propanol?
A
$H_2O, H_2SO_4$
B
$B_2H_6, H_2O_2, OH^{-}$
C
$MgSO_4, NaBH_4/H_2O$
D
Aqueous $KOH$
Solution
(B) The conversion of propene $(CH_3-CH=CH_2)$ to $1-$propanol $(CH_3-CH_2-CH_2OH)$ is an anti-Markovnikov hydration reaction.
This is achieved via the hydroboration-oxidation process using $B_2H_6$ followed by $H_2O_2$ in the presence of $OH^-$.
The reaction is: $CH_3-CH=CH_2 + B_2H_6/H_2O_2, OH^{-} \to CH_3-CH_2-CH_2OH$.
This is achieved via the hydroboration-oxidation process using $B_2H_6$ followed by $H_2O_2$ in the presence of $OH^-$.
The reaction is: $CH_3-CH=CH_2 + B_2H_6/H_2O_2, OH^{-} \to CH_3-CH_2-CH_2OH$.
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23
ChemistryMCQAIPMT · 2000
Which of the following is responsible for mechanical support, enzyme transport and protein synthesis?
A
Dictyosomes
B
Cell membrane
C
Mitochondria
D
Endoplasmic reticulum
Solution
(D) The $Endoplasmic \text{ } reticulum$ $(ER)$ is a network of membranous tubules that extends from the plasma membrane to the nuclear membrane, providing mechanical support to the cell.
$Rough \text{ } Endoplasmic \text{ } Reticulum$ $(RER)$ has ribosomes attached to its surface, which are the sites of protein synthesis.
Furthermore, the $ER$ system acts as a transport network for various enzymes and proteins within the cell.
$Rough \text{ } Endoplasmic \text{ } Reticulum$ $(RER)$ has ribosomes attached to its surface, which are the sites of protein synthesis.
Furthermore, the $ER$ system acts as a transport network for various enzymes and proteins within the cell.
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24
ChemistryMCQAIPMT · 2000
What is the function of Copper-$T$?
A
Checks mutation
B
Stops fertilization
C
Stops zygote formation
D
Stops obliteration of blastocoel
Solution
(B) The Copper-$T$ is a type of intra-uterine device $(IUD)$.
It is placed in the uterus to act as a contraceptive.
It releases copper ions $(Cu^{2+})$ which suppress the motility of sperm and their fertilizing capacity,thereby preventing fertilization.
Additionally,it also prevents implantation of the embryo in the uterus.
It is placed in the uterus to act as a contraceptive.
It releases copper ions $(Cu^{2+})$ which suppress the motility of sperm and their fertilizing capacity,thereby preventing fertilization.
Additionally,it also prevents implantation of the embryo in the uterus.
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25
ChemistryMCQAIPMT · 2000
Which of the following is a geocarpic fruit?
A
Onion
B
Watermelon
C
Groundnut
D
Carrot
Solution
(C) Geocarpic fruits are those that develop underground.
Groundnut $(Arachis hypogaea)$ is a classic example of a geocarpic fruit.
After fertilization, the flower stalk (gynophore or peg) elongates and pushes the ovary into the soil, where the fruit matures.
Onion is a modified stem (bulb), and carrot is a modified root; they are not fruits.
Groundnut $(Arachis hypogaea)$ is a classic example of a geocarpic fruit.
After fertilization, the flower stalk (gynophore or peg) elongates and pushes the ovary into the soil, where the fruit matures.
Onion is a modified stem (bulb), and carrot is a modified root; they are not fruits.
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26
ChemistryMCQAIPMT · 2000
Which of the following is an essential amino acid?
A
Serine
B
Aspartic acid
C
Glycine
D
Phenylalanine
Solution
(D) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
Among the given options,$Phenylalanine$ is an essential amino acid.
$Serine$,$Aspartic$ $acid$,and $Glycine$ are non-essential amino acids,meaning they can be synthesized by the human body.
Among the given options,$Phenylalanine$ is an essential amino acid.
$Serine$,$Aspartic$ $acid$,and $Glycine$ are non-essential amino acids,meaning they can be synthesized by the human body.
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27
ChemistryMCQAIPMT · 2000
What is a conjugated protein called if it contains a carbohydrate as a prosthetic group?
A
Chromoprotein
B
Glycoprotein
C
Lipoprotein
D
Nucleoprotein
Solution
(B) Conjugated proteins are proteins that are covalently bonded to non-protein components called prosthetic groups.
When the prosthetic group is a carbohydrate,the protein is known as a $Glycoprotein$.
$Chromoproteins$ contain a pigment as a prosthetic group.
$Lipoproteins$ contain lipids as a prosthetic group.
$Nucleoproteins$ contain nucleic acids as a prosthetic group.
Therefore,the correct option is $B$.
When the prosthetic group is a carbohydrate,the protein is known as a $Glycoprotein$.
$Chromoproteins$ contain a pigment as a prosthetic group.
$Lipoproteins$ contain lipids as a prosthetic group.
$Nucleoproteins$ contain nucleic acids as a prosthetic group.
Therefore,the correct option is $B$.
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28
ChemistryMCQAIPMT · 2000
Enzymatic activity is inhibited by which of the following?
A
End product
B
Substrate
C
Enzyme
D
Increase in temperature
Solution
(A) Enzymatic activity is often regulated by a process called feedback inhibition. In this process,the end product of a metabolic pathway binds to an allosteric site of an enzyme that catalyzes an earlier step in the pathway. This binding changes the conformation of the enzyme,thereby inhibiting its activity and preventing the overproduction of the end product. Therefore,the end product acts as an inhibitor.
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29
ChemistryMCQAIPMT · 2000
How is the rate of an enzymatic reaction increased?
A
By forming an enzyme-substrate complex.
B
By changing the equilibrium point of the reaction.
C
By binding and immediately forming the product.
D
By lowering the activation energy of the reaction.
Solution
(D) Enzymes increase the rate of a reaction by lowering the activation energy $(E_a)$ required for the substrate to be converted into the product.
They provide an alternative pathway for the reaction to proceed,which requires less energy than the uncatalyzed reaction.
Enzymes do not change the equilibrium point of the reaction,nor do they change the energy levels of the substrate or the product.
They provide an alternative pathway for the reaction to proceed,which requires less energy than the uncatalyzed reaction.
Enzymes do not change the equilibrium point of the reaction,nor do they change the energy levels of the substrate or the product.
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30
ChemistryMCQAIPMT · 2000
In the study of a transistor as an amplifier,if $\alpha = \frac{I_C}{I_E}$ and $\beta = \frac{I_C}{I_B}$,where $I_C, I_B$,and $I_E$ are the collector,base,and emitter currents,respectively,then:
A
$\beta = \frac{\alpha}{1 + \alpha}$
B
$\beta = \frac{\alpha}{1 - \alpha}$
C
$\beta = \frac{1 + \alpha}{\alpha}$
D
$\beta = \frac{1 - \alpha}{\alpha}$
Solution
(B) We know that the relationship between the currents in a transistor is given by $I_E = I_C + I_B$.
Dividing both sides by $I_C$,we get $\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$.
Since $\alpha = \frac{I_C}{I_E}$,it follows that $\frac{1}{\alpha} = \frac{I_E}{I_C}$.
Similarly,since $\beta = \frac{I_C}{I_B}$,it follows that $\frac{1}{\beta} = \frac{I_B}{I_C}$.
Substituting these into the equation,we get $\frac{1}{\alpha} = 1 + \frac{1}{\beta}$.
Rearranging for $\frac{1}{\beta}$,we have $\frac{1}{\beta} = \frac{1}{\alpha} - 1 = \frac{1 - \alpha}{\alpha}$.
Therefore,$\beta = \frac{\alpha}{1 - \alpha}$.
Dividing both sides by $I_C$,we get $\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$.
Since $\alpha = \frac{I_C}{I_E}$,it follows that $\frac{1}{\alpha} = \frac{I_E}{I_C}$.
Similarly,since $\beta = \frac{I_C}{I_B}$,it follows that $\frac{1}{\beta} = \frac{I_B}{I_C}$.
Substituting these into the equation,we get $\frac{1}{\alpha} = 1 + \frac{1}{\beta}$.
Rearranging for $\frac{1}{\beta}$,we have $\frac{1}{\beta} = \frac{1}{\alpha} - 1 = \frac{1 - \alpha}{\alpha}$.
Therefore,$\beta = \frac{\alpha}{1 - \alpha}$.
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31
ChemistryMediumMCQAIPMT · 2000
Which of the following reagents convert propene to $1-$propanol?
A
$B_2H_6, H_2O_2, OH^{-}$
B
Aqueous $KOH$
C
$MgSO_4, NaBH_4/H_2O$
D
$H_2O, H_2SO_4$
Solution
(A) The conversion of propene $(CH_3-CH=CH_2)$ to $1-$propanol $(CH_3-CH_2-CH_2-OH)$ is achieved via hydroboration-oxidation.
This process uses diborane $(B_2H_6)$ followed by alkaline hydrogen peroxide $(H_2O_2/OH^{-})$.
This reaction follows the anti-Markovnikov addition of water across the double bond.
This process uses diborane $(B_2H_6)$ followed by alkaline hydrogen peroxide $(H_2O_2/OH^{-})$.
This reaction follows the anti-Markovnikov addition of water across the double bond.
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32
ChemistryMCQAIPMT · 2000
The main organelle involved in the modification and routing of newly synthesized proteins to their destinations is
A
chloroplast
B
mitochondria
C
lysosome
D
Golgi apparatus
Solution
(D) The $Golgi$ apparatus is the primary organelle responsible for the modification,sorting,and packaging of proteins synthesized in the $ER$ (Endoplasmic Reticulum). These proteins are modified by the addition of carbohydrates (glycosylation) or other molecules and are then routed to their specific destinations,such as lysosomes,the plasma membrane,or for secretion outside the cell.
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33
ChemistryMCQAIPMT · 2000
The correct relation between the degree of freedom $f$ and the ratio of specific heat $\gamma$ is -
A
$f=\frac{1}{\gamma+1}$
B
$f=\frac{\gamma+1}{2}$
C
$f=\frac{2}{\gamma+1}$
D
$f=\frac{2}{\gamma-1}$
Solution
(D) The ratio of specific heats $\gamma$ is defined as the ratio of molar specific heat at constant pressure $C_p$ to molar specific heat at constant volume $C_v$,i.e.,$\gamma = \frac{C_p}{C_v}$.
For an ideal gas,$C_v = \frac{f}{2}R$ and $C_p = C_v + R = (\frac{f}{2} + 1)R$.
Thus,$\gamma = \frac{(\frac{f}{2} + 1)R}{\frac{f}{2}R} = \frac{\frac{f+2}{2}}{\frac{f}{2}} = \frac{f+2}{f} = 1 + \frac{2}{f}$.
Rearranging the equation to solve for $f$:
$\gamma - 1 = \frac{2}{f}$
$f = \frac{2}{\gamma - 1}$.
For an ideal gas,$C_v = \frac{f}{2}R$ and $C_p = C_v + R = (\frac{f}{2} + 1)R$.
Thus,$\gamma = \frac{(\frac{f}{2} + 1)R}{\frac{f}{2}R} = \frac{\frac{f+2}{2}}{\frac{f}{2}} = \frac{f+2}{f} = 1 + \frac{2}{f}$.
Rearranging the equation to solve for $f$:
$\gamma - 1 = \frac{2}{f}$
$f = \frac{2}{\gamma - 1}$.
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34
ChemistryMCQAIPMT · 2000
Gravitational force is required for
A
Stirring of liquid
B
Convection
C
Conduction
D
Radiation
Solution
(B) The correct answer is $B$.
Convection is the process of heat transfer in fluids (liquids and gases) where the actual movement of particles occurs.
This movement is driven by density differences caused by temperature variations.
In a gravitational field,the warmer,less dense fluid rises while the cooler,denser fluid sinks,creating convection currents.
Without gravity,there is no buoyancy to drive this circulation,making convection impossible.
Convection is the process of heat transfer in fluids (liquids and gases) where the actual movement of particles occurs.
This movement is driven by density differences caused by temperature variations.
In a gravitational field,the warmer,less dense fluid rises while the cooler,denser fluid sinks,creating convection currents.
Without gravity,there is no buoyancy to drive this circulation,making convection impossible.
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35
ChemistryMediumMCQAIPMT · 2000
For a hypothetical reaction $A \to B$,the activation energies for forward and backward reactions are $19 \ kJ/mol$ and $9 \ kJ/mol$ respectively. The heat of reaction is $.... \ kJ$.
A
$28$
B
$19$
C
$10$
D
$9$
Solution
(C) The heat of reaction $(\Delta H)$ is defined as the difference between the activation energy of the forward reaction $(E_{a,f})$ and the activation energy of the backward reaction $(E_{a,b})$.
$\Delta H = E_{a,f} - E_{a,b}$
Given $E_{a,f} = 19 \ kJ/mol$ and $E_{a,b} = 9 \ kJ/mol$.
$\Delta H = 19 \ kJ/mol - 9 \ kJ/mol = 10 \ kJ/mol$.
Therefore,the correct option is $C$.
$\Delta H = E_{a,f} - E_{a,b}$
Given $E_{a,f} = 19 \ kJ/mol$ and $E_{a,b} = 9 \ kJ/mol$.
$\Delta H = 19 \ kJ/mol - 9 \ kJ/mol = 10 \ kJ/mol$.
Therefore,the correct option is $C$.
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36
ChemistryMediumMCQAIPMT · 2000
Which one of the following combines with $Fe(II)$ ions to form a brown complex?
A
$N_2O$
B
$NO$
C
$N_2O_3$
D
$N_2O_5$
Solution
(B) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
In this test,$Fe(II)$ ions react with $NO$ (nitric oxide) to form a brown-colored coordination complex.
The reaction is: $[Fe(H_2O)_6]^{2+} + NO \to [Fe(H_2O)_5(NO)]^{2+} + H_2O$.
This complex,$[Fe(H_2O)_5(NO)]^{2+}$,is responsible for the brown ring observed at the junction of the two layers.
In this test,$Fe(II)$ ions react with $NO$ (nitric oxide) to form a brown-colored coordination complex.
The reaction is: $[Fe(H_2O)_6]^{2+} + NO \to [Fe(H_2O)_5(NO)]^{2+} + H_2O$.
This complex,$[Fe(H_2O)_5(NO)]^{2+}$,is responsible for the brown ring observed at the junction of the two layers.
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37
ChemistryMediumMCQAIPMT · 2000
Copper turnings when heated with concentrated sulphuric acid will give:
A
$SO_2$
B
$SO_3$
C
$H_2S$
D
$O_2$
Solution
(A) When copper $(Cu)$ is heated with concentrated sulphuric acid $(H_2SO_4)$,it acts as an oxidizing agent and oxidizes copper to copper$(II)$ sulphate,while itself being reduced to sulphur dioxide $(SO_2)$.
The balanced chemical equation is:
$Cu(s) + 2H_2SO_4(conc.) \to CuSO_4(aq) + 2H_2O(l) + SO_2(g)$
The balanced chemical equation is:
$Cu(s) + 2H_2SO_4(conc.) \to CuSO_4(aq) + 2H_2O(l) + SO_2(g)$
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38
ChemistryMediumMCQAIPMT · 2000
Which of the following combines with $Fe(II)$ ions to form a brown complex?
A
$N_2O$
B
$NO$
C
$N_2O_3$
D
$N_2O_5$
Solution
(B) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
When a freshly prepared ferrous sulfate solution is added to the nitrate solution followed by the careful addition of concentrated sulfuric acid along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The reaction involves the reduction of nitrate to nitric oxide $(NO)$,which then reacts with the $Fe(II)$ ions to form the brown complex $[Fe(H_2O)_5(NO)]SO_4$.
When a freshly prepared ferrous sulfate solution is added to the nitrate solution followed by the careful addition of concentrated sulfuric acid along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The reaction involves the reduction of nitrate to nitric oxide $(NO)$,which then reacts with the $Fe(II)$ ions to form the brown complex $[Fe(H_2O)_5(NO)]SO_4$.
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39
ChemistryEasyMCQAIPMT · 2000
What will be the molality of a solution having $18 \ g$ of glucose (mol. wt. $= 180$) dissolved in $500 \ g$ of water?
A
$1$
B
$0.5$
C
$0.2$
D
$2$
Solution
(C) The formula for molality $(m)$ is given by: $m = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}}$.
Given:
Mass of glucose $= 18 \ g$
Molar mass of glucose $= 180 \ g/mol$
Mass of water (solvent) $= 500 \ g = 0.5 \ kg$.
Substituting the values:
$m = \frac{18}{180 \times 0.5} = \frac{0.1}{0.5} = 0.2 \ m$.
Therefore,the correct option is $(C)$.
Given:
Mass of glucose $= 18 \ g$
Molar mass of glucose $= 180 \ g/mol$
Mass of water (solvent) $= 500 \ g = 0.5 \ kg$.
Substituting the values:
$m = \frac{18}{180 \times 0.5} = \frac{0.1}{0.5} = 0.2 \ m$.
Therefore,the correct option is $(C)$.
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40
ChemistryMediumMCQAIPMT · 2000
An aqueous solution containing $1 \ g$ of urea boils at $100.25 \ ^oC$. The aqueous solution containing $3 \ g$ of glucose in the same volume will boil at ........ $^oC$. (Molecular weight of urea and glucose are $60$ and $180$ respectively)
A
$100.75$
B
$100.5$
C
$100.25$
D
$100$
Solution
(C) The elevation in boiling point is given by $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
For urea: $m_1 = \frac{1 \ g / 60 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.
For glucose: $m_2 = \frac{3 \ g / 180 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.
Since the molality $(m)$ is the same for both solutions,the elevation in boiling point $(\Delta T_b)$ will be identical.
Given $\Delta T_b = 100.25 \ ^oC - 100 \ ^oC = 0.25 \ ^oC$ for urea.
Therefore,the boiling point of the glucose solution will be $100 \ ^oC + 0.25 \ ^oC = 100.25 \ ^oC$.
For urea: $m_1 = \frac{1 \ g / 60 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.
For glucose: $m_2 = \frac{3 \ g / 180 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.
Since the molality $(m)$ is the same for both solutions,the elevation in boiling point $(\Delta T_b)$ will be identical.
Given $\Delta T_b = 100.25 \ ^oC - 100 \ ^oC = 0.25 \ ^oC$ for urea.
Therefore,the boiling point of the glucose solution will be $100 \ ^oC + 0.25 \ ^oC = 100.25 \ ^oC$.
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41
ChemistryMediumMCQAIPMT · 2000
$A$ compound is formed by elements $A$ and $B$. This crystallizes in a cubic structure where atoms $A$ are at the corners of the cube and atoms $B$ are at the body center. The simplest formula of the compound is:
A
$AB_3$
B
$AB_2$
C
$AB$
D
$A_2B$
Solution
(C) Atoms $A$ are present at the $8$ corners of the cube. The contribution of each corner atom to the unit cell is $\frac{1}{8}$.
Number of $A$ atoms per unit cell $= 8 \times \frac{1}{8} = 1$.
Atoms $B$ are present at the body center of the cube. The contribution of a body-centered atom is $1$.
Number of $B$ atoms per unit cell $= 1 \times 1 = 1$.
Therefore,the ratio of $A:B$ is $1:1$,and the simplest formula of the compound is $AB$.
Number of $A$ atoms per unit cell $= 8 \times \frac{1}{8} = 1$.
Atoms $B$ are present at the body center of the cube. The contribution of a body-centered atom is $1$.
Number of $B$ atoms per unit cell $= 1 \times 1 = 1$.
Therefore,the ratio of $A:B$ is $1:1$,and the simplest formula of the compound is $AB$.
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42
ChemistryEasyMCQAIPMT · 2000
In the reaction $2A + B \to A_2B$,if the concentration of $A$ is doubled and the concentration of $B$ is halved,then the rate of the reaction will:
A
Increase by four times
B
Decrease by two times
C
Increase by two times
D
Remain the same
Solution
(C) The rate law for the reaction is given by $Rate = k[A]^2[B]^1$.
Let the initial rate be $R_1 = k[A]^2[B]$.
When the concentration of $A$ is doubled $(2[A])$ and the concentration of $B$ is halved $(0.5[B])$,the new rate $R_2$ is:
$R_2 = k(2[A])^2(0.5[B])$
$R_2 = k(4[A]^2)(0.5[B])$
$R_2 = 2 \times k[A]^2[B]$
$R_2 = 2 \times R_1$.
Therefore,the rate of the reaction increases by two times.
Let the initial rate be $R_1 = k[A]^2[B]$.
When the concentration of $A$ is doubled $(2[A])$ and the concentration of $B$ is halved $(0.5[B])$,the new rate $R_2$ is:
$R_2 = k(2[A])^2(0.5[B])$
$R_2 = k(4[A]^2)(0.5[B])$
$R_2 = 2 \times k[A]^2[B]$
$R_2 = 2 \times R_1$.
Therefore,the rate of the reaction increases by two times.
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43
ChemistryMediumMCQAIPMT · 2000
After how many seconds will the concentration of the reactants in a first order reaction be halved,if the decay constant is $1.155 \times 10^{-3} \, \sec^{-1}$?
A
$100$
B
$200$
C
$400$
D
$600$
Solution
(D) 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 the decay constant $k = 1.155 \times 10^{-3} \, \sec^{-1}$.
Substituting the value of $k$:
$t_{1/2} = \frac{0.693}{1.155 \times 10^{-3}} = 600 \, \sec$
Therefore,the concentration of the reactants will be halved after $600 \, \sec$.
$t_{1/2} = \frac{0.693}{k}$
Given the decay constant $k = 1.155 \times 10^{-3} \, \sec^{-1}$.
Substituting the value of $k$:
$t_{1/2} = \frac{0.693}{1.155 \times 10^{-3}} = 600 \, \sec$
Therefore,the concentration of the reactants will be halved after $600 \, \sec$.
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44
ChemistryEasyMCQAIPMT · 2000
Faraday's laws of electrolysis state that the mass deposited on an electrode is proportional to:
A
$m \propto I^2$
B
$m \propto Q$
C
$m \propto Q^2$
D
None of these
Solution
(B) According to Faraday's first law of electrolysis,the mass $(m)$ of the substance deposited or liberated at an electrode is directly proportional to the quantity of electricity $(Q)$ passed through the electrolyte.
Mathematically,$m \propto Q$ or $m = ZQ$,where $Z$ is the electrochemical equivalent.
Mathematically,$m \propto Q$ or $m = ZQ$,where $Z$ is the electrochemical equivalent.
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45
ChemistryMediumMCQAIPMT · 2000
The specific conductance of a $0.1 \ N \ KCl$ solution at $23 \ ^oC$ is $0.012 \ \Omega^{-1} \ cm^{-1}$. The resistance of a cell containing the solution at the same temperature was found to be $55 \ \Omega$. The cell constant will be .............. $cm^{-1}$.
A
$0.142$
B
$0.66$
C
$0.918$
D
$1.12$
Solution
(B) The relationship between specific conductance $(K)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula:
$K = \frac{1}{R} \times G^*$
Rearranging to solve for the cell constant:
$G^* = K \times R$
Given:
$K = 0.012 \ \Omega^{-1} \ cm^{-1}$
$R = 55 \ \Omega$
Calculation:
$G^* = 0.012 \ \Omega^{-1} \ cm^{-1} \times 55 \ \Omega = 0.66 \ cm^{-1}$
Therefore,the cell constant is $0.66 \ cm^{-1}$.
$K = \frac{1}{R} \times G^*$
Rearranging to solve for the cell constant:
$G^* = K \times R$
Given:
$K = 0.012 \ \Omega^{-1} \ cm^{-1}$
$R = 55 \ \Omega$
Calculation:
$G^* = 0.012 \ \Omega^{-1} \ cm^{-1} \times 55 \ \Omega = 0.66 \ cm^{-1}$
Therefore,the cell constant is $0.66 \ cm^{-1}$.
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46
ChemistryEasyMCQAIPMT · 2000
Which of the following is used for the destruction of colloids?
A
Dialysis
B
Condensation
C
By ultrafiltration
D
By adding electrolyte
Solution
(D) The stability of colloids is maintained by the charge on the particles. By adding an electrolyte,the charge on the colloidal particles is neutralized. This leads to the coagulation or precipitation of the colloidal particles,effectively destroying the colloid.
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47
ChemistryEasyMCQAIPMT · 2000
An example of an associated colloid is
A
Milk
B
Soap solution
C
Rubber latex
D
Vegetable oil
Solution
(B) Associated colloids are substances that behave as normal strong electrolytes at low concentrations but exhibit colloidal behavior at higher concentrations due to the formation of aggregates called micelles.
Soap solution is a classic example of an associated colloid because soap molecules (like sodium stearate) form micelles in water above the critical micelle concentration $(CMC)$.
Soap solution is a classic example of an associated colloid because soap molecules (like sodium stearate) form micelles in water above the critical micelle concentration $(CMC)$.
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48
ChemistryEasyMCQAIPMT · 2000
The most abundant metal in the earth's crust is
A
$Na$
B
$Mg$
C
$Al$
D
$Fe$
Solution
(C) The most abundant metal in the earth's crust is $Al$ (Aluminum),which constitutes approximately $8.1\%$ by weight of the earth's crust.
Therefore,the correct option is $(C)$.
Therefore,the correct option is $(C)$.
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49
ChemistryMediumMCQAIPMT · 2000
The aqueous solution containing which one of the following ions will be colourless?
A
$Sc^{3+}$
B
$Fe^{2+}$
C
$Ti^{3+}$
D
$Mn^{2+}$ (Atomic number $Sc = 21, Fe = 26, Ti = 22, Mn = 25$)
Solution
(A) An ion is colourless if it does not have any unpaired electrons in its $d$-orbitals.
$1. _{21}Sc^{3+}: [Ar] 3d^0$. It has no unpaired electrons,so it is colourless.
$2. _{26}Fe^{2+}: [Ar] 3d^6$. It has $4$ unpaired electrons,so it is coloured.
$3. _{22}Ti^{3+}: [Ar] 3d^1$. It has $1$ unpaired electron,so it is coloured.
$4. _{25}Mn^{2+}: [Ar] 3d^5$. It has $5$ unpaired electrons,so it is coloured.
Therefore,the correct option is $A$.
$1. _{21}Sc^{3+}: [Ar] 3d^0$. It has no unpaired electrons,so it is colourless.
$2. _{26}Fe^{2+}: [Ar] 3d^6$. It has $4$ unpaired electrons,so it is coloured.
$3. _{22}Ti^{3+}: [Ar] 3d^1$. It has $1$ unpaired electron,so it is coloured.
$4. _{25}Mn^{2+}: [Ar] 3d^5$. It has $5$ unpaired electrons,so it is coloured.
Therefore,the correct option is $A$.
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50
ChemistryEasyMCQAIPMT · 2000
Percentage of silver in German silver is (in $\%$)
A
$2$
B
$1$
C
$5$
D
$0$
Solution
(D) German silver is an alloy of copper,zinc,and nickel. It does not contain any silver. Its composition is approximately $Cu = 56.0\%$,$Zn = 24.0\%$,and $Ni = 20.0\%$.
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51
ChemistryEasyMCQAIPMT · 2000
German silver is an alloy of
A
Copper,zinc and nickel
B
Copper and silver
C
Copper,zinc and tin
D
Copper,zinc and silver
Solution
(A) German silver is a well-known alloy consisting of copper,zinc,and nickel.
Its typical composition is $Cu$ $(60\%)$,$Zn$ $(20\%)$,and $Ni$ $(20\%)$.
Despite its name,it does not contain any silver.
Its typical composition is $Cu$ $(60\%)$,$Zn$ $(20\%)$,and $Ni$ $(20\%)$.
Despite its name,it does not contain any silver.
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52
ChemistryMediumMCQAIPMT · 2000
What is the shape of $Fe(CO)_5$?
A
Linear
B
Tetrahedral
C
Square planar
D
Trigonal bipyramidal
Solution
(D) . $Fe(CO)_5$ undergoes $dsp^3$ hybridization,which results in a trigonal bipyramidal geometry.
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53
ChemistryMediumMCQAIPMT · 2000
In the following reaction,product $P$ is: $RCOCl \xrightarrow{H_2, Pd-BaSO_4} P$
A
$RCH_2OH$
B
$RCOOH$
C
$RCHO$
D
$RCH_3$
Solution
(C) The reaction of an acid chloride with $H_2$ in the presence of $Pd-BaSO_4$ is known as the Rosenmund reduction.
This reaction specifically reduces the acid chloride group $(-COCl)$ to an aldehyde group $(-CHO)$.
Therefore,the product $P$ is $RCHO$.
This reaction specifically reduces the acid chloride group $(-COCl)$ to an aldehyde group $(-CHO)$.
Therefore,the product $P$ is $RCHO$.
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54
ChemistryMediumMCQAIPMT · 2000
Which reagent will bring about the conversion of carboxylic acids into esters?
A
$C_2H_5OH$
B
$Dry \ HCl + C_2H_5OH$
C
$LiAlH_4$
D
$Al(OC_2H_5)_3$
Solution
(B) The reaction of carboxylic acids with alcohols in the presence of an acid catalyst like $Dry \ HCl$ is known as Fischer esterification.
The reaction is: $RCOOH + C_2H_5OH \xrightarrow{Dry \ HCl} RCOOC_2H_5 + H_2O$.
Thus,$Dry \ HCl + C_2H_5OH$ is the correct reagent.
The reaction is: $RCOOH + C_2H_5OH \xrightarrow{Dry \ HCl} RCOOC_2H_5 + H_2O$.
Thus,$Dry \ HCl + C_2H_5OH$ is the correct reagent.
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55
ChemistryMediumMCQAIPMT · 2000
What will happen if $LiAlH_4$ is added to an ester?
A
Two units of alcohol are obtained
B
One unit of alcohol and one unit of acid is obtained
C
Two units of acids are obtained
D
None of these
Solution
(A) $LiAlH_4$ is a strong reducing agent that reduces esters to alcohols. The reaction is: $R-COO-R' \xrightarrow{LiAlH_4} R-CH_2OH + R'-OH$. Thus,two units of alcohols are obtained.
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56
ChemistryDifficultMCQAIPMT · 2000
What is formed when nitrobenzene is reduced using zinc and alkali?
A
Phenol
B
Aniline
C
Nitrosobenzene
D
Hydrazobenzene
Solution
(D) The reduction of nitrobenzene with zinc and alkali (like $NaOH$) is a bimolecular reduction.
$2C_{6}H_{5}NO_{2} \xrightarrow[Zn/NaOH]{10[H]} C_{6}H_{5}NH-NHC_{6}H_{5} + 4H_{2}O$
The product formed is $Hydrazobenzene$.
$2C_{6}H_{5}NO_{2} \xrightarrow[Zn/NaOH]{10[H]} C_{6}H_{5}NH-NHC_{6}H_{5} + 4H_{2}O$
The product formed is $Hydrazobenzene$.
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57
ChemistryEasyMCQAIPMT · 2000
$F_2C = CF_2$ is the monomer of
A
Nylon-$6$
B
Buna-$S$
C
Glyptal
D
Teflon
Solution
(D) The monomer $F_2C = CF_2$ is known as tetrafluoroethene.
Polymerization of tetrafluoroethene under high pressure in the presence of a peroxide or persulphate catalyst yields polytetrafluoroethene,commonly known as Teflon.
Therefore,$F_2C = CF_2$ is the monomer of Teflon.
Polymerization of tetrafluoroethene under high pressure in the presence of a peroxide or persulphate catalyst yields polytetrafluoroethene,commonly known as Teflon.
Therefore,$F_2C = CF_2$ is the monomer of Teflon.
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58
ChemistryDifficultMCQAIPMT · 2000
Ebonite is
A
Polypropene
B
Natural rubber
C
Synthetic rubber
D
Highly vulcanized rubber
Solution
(D) Ebonite is a hard and highly $(20-30\%)$ vulcanized rubber.
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59
ChemistryMediumMCQAIPMT · 2000
The number of essential amino acids in man is
A
$8$
B
$10$
C
$18$
D
$20$
Solution
(B) The amino acids which cannot be synthesized by the human body are called essential amino acids,as they must be obtained through the diet.
There are $10$ such essential amino acids.
There are $10$ such essential amino acids.
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