ChemistryQ1–15 of 15 questions
Page 1 of 1 · English
1
ChemistryEasyMCQGUJCET · 2011
In the given reaction sequence,identify the processes $X$ and $Y$ respectively to obtain salicylaldehyde from chlorobenzene:
$Chlorobenzene$ $\xrightarrow{X} Phenol$ $\xrightarrow{Y} Salicylaldehyde$
$Chlorobenzene$ $\xrightarrow{X} Phenol$ $\xrightarrow{Y} Salicylaldehyde$
A
Fries rearrangement and Kolbe-Schmidt
B
Cumene and Reimer-Tiemann
C
Dow and Reimer-Tiemann
D
Dow and Friedel-Crafts
Solution
(C) The conversion of chlorobenzene to phenol is known as the $Dow$ process,where chlorobenzene reacts with $NaOH$ at high temperature and pressure.
The conversion of phenol to salicylaldehyde is known as the $Reimer-Tiemann$ reaction,which involves treatment with $CHCl_3$ and $NaOH$ followed by hydrolysis.
Therefore,$X$ is the $Dow$ process and $Y$ is the $Reimer-Tiemann$ reaction.
The conversion of phenol to salicylaldehyde is known as the $Reimer-Tiemann$ reaction,which involves treatment with $CHCl_3$ and $NaOH$ followed by hydrolysis.
Therefore,$X$ is the $Dow$ process and $Y$ is the $Reimer-Tiemann$ reaction.
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2
ChemistryEasyMCQGUJCET · 2011
Hybridisation shown by Carbon and Oxygen of $-OH$ group in phenol are respectively . . . . . . .
A
$sp^2, sp^2$
B
$sp^3, sp^3$
C
$sp, sp^2$
D
$sp^2, sp^3$
Solution
(D) In phenol $(C_6H_5OH)$,the carbon atom attached to the $-OH$ group is part of the benzene ring and is bonded to two other carbons and one oxygen atom via double and single bonds,giving it $sp^2$ hybridisation.
The oxygen atom in the $-OH$ group is bonded to one carbon atom and one hydrogen atom,and it possesses two lone pairs of electrons. Therefore,the steric number of oxygen is $4$ ($2$ bond pairs + $2$ lone pairs),which corresponds to $sp^3$ hybridisation.
Thus,the hybridisation of carbon and oxygen is $sp^2$ and $sp^3$ respectively.
The oxygen atom in the $-OH$ group is bonded to one carbon atom and one hydrogen atom,and it possesses two lone pairs of electrons. Therefore,the steric number of oxygen is $4$ ($2$ bond pairs + $2$ lone pairs),which corresponds to $sp^3$ hybridisation.
Thus,the hybridisation of carbon and oxygen is $sp^2$ and $sp^3$ respectively.
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3
ChemistryEasyMCQGUJCET · 2011
Which of the following is a nucleophilic addition reaction?
A
Hydrolysis of Ethyl Chloride by $NaOH$
B
Purification of Acetaldehyde by $NaHSO_3$
C
Alkylation of Anisole
D
Decarboxylation of Acetic acid
Solution
(B) The reaction of acetaldehyde with sodium bisulphite $(NaHSO_3)$ is a classic example of a nucleophilic addition reaction.
In this reaction,the bisulphite ion $(HSO_3^-)$ acts as a nucleophile and attacks the electrophilic carbonyl carbon of the acetaldehyde to form a crystalline bisulphite addition compound.
This reaction is reversible and is commonly used for the purification and separation of aldehydes and ketones.
In this reaction,the bisulphite ion $(HSO_3^-)$ acts as a nucleophile and attacks the electrophilic carbonyl carbon of the acetaldehyde to form a crystalline bisulphite addition compound.
This reaction is reversible and is commonly used for the purification and separation of aldehydes and ketones.
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4
ChemistryEasyMCQGUJCET · 2011
Which of the following compounds undergoes the Cannizzaro reaction?
A
$CH_2ClCHO$
B
$CCl_3CHO$
C
$CH_3CHO$
D
$CHCl_2CHO$
Solution
(B) The Cannizzaro reaction is given by aldehydes that do not contain an $\alpha$-hydrogen atom.
In $CH_3CHO$,there are three $\alpha$-hydrogens.
In $CH_2ClCHO$,there are two $\alpha$-hydrogens.
In $CHCl_2CHO$,there is one $\alpha$-hydrogen.
In $CCl_3CHO$ (trichloroacetaldehyde),there are no $\alpha$-hydrogen atoms attached to the carbonyl carbon.
Therefore,$CCl_3CHO$ undergoes the Cannizzaro reaction.
In $CH_3CHO$,there are three $\alpha$-hydrogens.
In $CH_2ClCHO$,there are two $\alpha$-hydrogens.
In $CHCl_2CHO$,there is one $\alpha$-hydrogen.
In $CCl_3CHO$ (trichloroacetaldehyde),there are no $\alpha$-hydrogen atoms attached to the carbonyl carbon.
Therefore,$CCl_3CHO$ undergoes the Cannizzaro reaction.
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5
ChemistryEasyMCQGUJCET · 2011
How can you say that glucose is a cyclic compound?
A
Glucose undergoes Tollen's reaction
B
Glucose reacts with phenyl hydrazine
C
Glucose fails to react with sodium hydrogen sulphite
D
Glucose reacts with Nitric acid
Solution
(C) The open-chain structure of glucose contains an aldehyde group $(-CHO)$.
Typically,aldehydes react with sodium hydrogen sulphite $(NaHSO_3)$ to form addition products.
However,glucose does not react with $NaHSO_3$,which indicates that the aldehyde group is not free.
This suggests that the aldehyde group is involved in the formation of a cyclic hemiacetal structure,confirming that glucose exists as a cyclic compound in its stable form.
Typically,aldehydes react with sodium hydrogen sulphite $(NaHSO_3)$ to form addition products.
However,glucose does not react with $NaHSO_3$,which indicates that the aldehyde group is not free.
This suggests that the aldehyde group is involved in the formation of a cyclic hemiacetal structure,confirming that glucose exists as a cyclic compound in its stable form.
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6
ChemistryEasyMCQGUJCET · 2011
How is glucose related to fructose?
A
Functional group isomerism
B
Rotamers
C
Position isomerism
D
Geometrical isomerism
Solution
(A) Glucose $(C_6H_{12}O_6)$ is an aldohexose,meaning it contains an aldehyde group $(-CHO)$.
Fructose $(C_6H_{12}O_6)$ is a ketohexose,meaning it contains a ketone group $(>C=O)$.
Since both have the same molecular formula but differ in the functional group present,they exhibit functional group isomerism.
Therefore,the correct option is $A$.
Fructose $(C_6H_{12}O_6)$ is a ketohexose,meaning it contains a ketone group $(>C=O)$.
Since both have the same molecular formula but differ in the functional group present,they exhibit functional group isomerism.
Therefore,the correct option is $A$.
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7
ChemistryEasyMCQGUJCET · 2011
Give the relation between half-life $(t_{1/2})$ and initial concentration of reactant $([R]_0)$ for an $(n-1)^{th}$ order reaction.
A
$t_{1/2} \propto [R]_0$
B
$t_{1/2} \propto [R]_0^{2-n}$
C
$t_{1/2} \propto [R]_0^{n+1}$
D
$t_{1/2} \propto [R]_0^{n-2}$
Solution
(B) For an $n^{th}$ order reaction,the relationship between half-life $(t_{1/2})$ and initial concentration $([R]_0)$ is given by:
$t_{1/2} \propto [R]_0^{1-n}$
For an $(n-1)^{th}$ order reaction,substitute $n$ with $(n-1)$:
$t_{1/2} \propto [R]_0^{1-(n-1)}$
$t_{1/2} \propto [R]_0^{1-n+1}$
$t_{1/2} \propto [R]_0^{2-n}$
$t_{1/2} \propto [R]_0^{1-n}$
For an $(n-1)^{th}$ order reaction,substitute $n$ with $(n-1)$:
$t_{1/2} \propto [R]_0^{1-(n-1)}$
$t_{1/2} \propto [R]_0^{1-n+1}$
$t_{1/2} \propto [R]_0^{2-n}$
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8
ChemistryEasyMCQGUJCET · 2011
For a first order reaction,the initial concentration of a reactant is $0.05 \ M$. After $45 \ minutes$,it is decreased by $0.015 \ M$. Calculate the half-life period $(t_{1/2})$ (in $min$)?
A
$87.42$
B
$25.90$
C
$78.72$
D
$77.20$
Solution
(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: $[A]_0 = 0.05 \ M$,$[A]_t = 0.05 - 0.015 = 0.035 \ M$,and $t = 45 \ min$.
Substituting the values: $k = \frac{2.303}{45} \log \frac{0.05}{0.035} = \frac{2.303}{45} \log (1.4286) \approx 0.02673 \ min^{-1}$.
The half-life period is: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02673} \approx 25.92 \ min$.
Given: $[A]_0 = 0.05 \ M$,$[A]_t = 0.05 - 0.015 = 0.035 \ M$,and $t = 45 \ min$.
Substituting the values: $k = \frac{2.303}{45} \log \frac{0.05}{0.035} = \frac{2.303}{45} \log (1.4286) \approx 0.02673 \ min^{-1}$.
The half-life period is: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02673} \approx 25.92 \ min$.
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9
ChemistryEasyMCQGUJCET · 2011
Copper exhibits only $+2$ oxidation state in its stable compounds. Why?
A
Copper is a transition metal in the $+2$ state.
B
The high hydration enthalpy of $Cu^{2+}(aq)$ compensates for the second ionization enthalpy of Copper.
C
The electron configuration of Copper in the $+2$ state is $[Ar] 3d^9 4s^0$.
D
Copper forms coloured compounds in the $+2$ state.
Solution
(B) The stability of $Cu^{2+}(aq)$ ions is primarily due to their very high hydration enthalpy,which more than compensates for the energy required to remove the second electron from the $Cu^+(g)$ ion. Therefore,$Cu^{2+}$ is more stable than $Cu^+$ in aqueous solutions. The correct option is $B$.
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10
ChemistryEasyMCQGUJCET · 2011
What metals are in German silver alloy?
A
Zinc,Silver and Copper
B
Nickel,Silver and Copper
C
Germanium,Silver and Copper
D
Zinc,Nickel and Copper
Solution
(D) German silver is an alloy consisting of $Copper$ $(Cu)$,$Zinc$ $(Zn)$,and $Nickel$ $(Ni)$.
Despite its name,it does not contain any $Silver$ $(Ag)$.
Therefore,the correct composition is $Zinc$,$Nickel$,and $Copper$.
Despite its name,it does not contain any $Silver$ $(Ag)$.
Therefore,the correct composition is $Zinc$,$Nickel$,and $Copper$.
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11
ChemistryEasyMCQGUJCET · 2011
When dilute $H_2SO_4$ is added to an aqueous solution of potassium chromate,the yellow color of the solution turns to orange. What does this indicate?
A
Chromate ions are reduced.
B
Chromate ions are oxidized.
C
Monocentric complex is converted into a dicentric complex.
D
Oxygen is removed from chromate ions.
Solution
(C) The reaction between potassium chromate $(K_2CrO_4)$ and dilute $H_2SO_4$ is as follows:
$2CrO_4^{2-} (aq) + 2H^+ (aq) \rightleftharpoons Cr_2O_7^{2-} (aq) + H_2O (l)$
In this reaction,the yellow chromate ion $(CrO_4^{2-})$ is converted into the orange dichromate ion $(Cr_2O_7^{2-})$.
The chromate ion is a monocentric species (containing one $Cr$ atom),while the dichromate ion is a dicentric species (containing two $Cr$ atoms).
Therefore,the color change indicates that the monocentric complex is converted into a dicentric complex.
$2CrO_4^{2-} (aq) + 2H^+ (aq) \rightleftharpoons Cr_2O_7^{2-} (aq) + H_2O (l)$
In this reaction,the yellow chromate ion $(CrO_4^{2-})$ is converted into the orange dichromate ion $(Cr_2O_7^{2-})$.
The chromate ion is a monocentric species (containing one $Cr$ atom),while the dichromate ion is a dicentric species (containing two $Cr$ atoms).
Therefore,the color change indicates that the monocentric complex is converted into a dicentric complex.
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12
ChemistryEasyMCQGUJCET · 2011
What will be the $pH$ of the aqueous solution of the electrolyte in an electrolytic cell during the electrolysis of $CuSO_{4(aq)}$ between graphite electrodes?
A
$pH = 14.0$
B
$pH > 7.0$
C
$pH < 7.0$
D
$pH = 7.0$
Solution
(C) During the electrolysis of $CuSO_{4(aq)}$ using graphite electrodes,the following reactions occur:
At the cathode: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
At the anode: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-$
As the reaction proceeds,$H^+$ ions are produced at the anode,which increases the concentration of $H^+$ ions in the solution.
Since $pH = -\log[H^+]$,an increase in $[H^+]$ leads to a decrease in $pH$.
Therefore,the $pH$ of the solution becomes less than $7.0$.
At the cathode: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
At the anode: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-$
As the reaction proceeds,$H^+$ ions are produced at the anode,which increases the concentration of $H^+$ ions in the solution.
Since $pH = -\log[H^+]$,an increase in $[H^+]$ leads to a decrease in $pH$.
Therefore,the $pH$ of the solution becomes less than $7.0$.
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13
ChemistryEasyMCQGUJCET · 2011
Which of the following reactions is correct for a given electrochemical cell at $25^{\circ} C$ ?
$Pt | Br_{2(g)} | Br_{(aq)}^{-} || Cl_{(aq)}^{-} | Cl_{2(g)} | Pt$
$Pt | Br_{2(g)} | Br_{(aq)}^{-} || Cl_{(aq)}^{-} | Cl_{2(g)} | Pt$
A
$2Br_{(aq)}^{-} + Cl_{2(g)} \rightarrow Br_{2(g)} + 2Cl_{(aq)}^{-}$
B
$Br_{2(g)} + 2Cl_{(aq)}^{-} \rightarrow 2Br_{(aq)}^{-} + Cl_{2(g)}$
C
$Br_{2(g)} + Cl_{2(g)} \rightarrow 2Br_{(aq)}^{-} + 2Cl_{(aq)}^{-}$
D
$2Br_{(aq)}^{-} + 2Cl_{(aq)}^{-} \rightarrow Br_{2(g)} + Cl_{2(g)}$
Solution
(A) In an electrochemical cell representation,the left side represents the anode (oxidation) and the right side represents the cathode (reduction).
Anode reaction (oxidation): $2Br_{(aq)}^{-} \rightarrow Br_{2(g)} + 2e^{-}$
Cathode reaction (reduction): $Cl_{2(g)} + 2e^{-} \rightarrow 2Cl_{(aq)}^{-}$
Adding these two half-reactions gives the overall cell reaction:
$2Br_{(aq)}^{-} + Cl_{2(g)} \rightarrow Br_{2(g)} + 2Cl_{(aq)}^{-}$
Therefore,the correct option is $A$.
Anode reaction (oxidation): $2Br_{(aq)}^{-} \rightarrow Br_{2(g)} + 2e^{-}$
Cathode reaction (reduction): $Cl_{2(g)} + 2e^{-} \rightarrow 2Cl_{(aq)}^{-}$
Adding these two half-reactions gives the overall cell reaction:
$2Br_{(aq)}^{-} + Cl_{2(g)} \rightarrow Br_{2(g)} + 2Cl_{(aq)}^{-}$
Therefore,the correct option is $A$.
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14
ChemistryEasyMCQGUJCET · 2011
Two different electrolytic cells filled with molten $Cu(NO_3)_2$ and molten $Al(NO_3)_3$ respectively are connected in series. When electricity is passed,$2.7 \ g$ of $Al$ is deposited on the electrode. Calculate the weight of $Cu$ deposited on the cathode. [Atomic mass: $Cu = 63.5$,$Al = 27.0 \ g \ mol^{-1}$] (in $g$)
A
$190.5$
B
$9.525$
C
$63.5$
D
$31.75$
Solution
(B) According to Faraday's second law of electrolysis,for cells connected in series,the mass of substances deposited is proportional to their equivalent weights: $\frac{W_{Cu}}{W_{Al}} = \frac{E_{Cu}}{E_{Al}}$.
Equivalent weight of $Cu = \frac{\text{Atomic mass of } Cu}{\text{Valency factor}} = \frac{63.5}{2} = 31.75 \ g \ eq^{-1}$.
Equivalent weight of $Al = \frac{\text{Atomic mass of } Al}{\text{Valency factor}} = \frac{27}{3} = 9 \ g \ eq^{-1}$.
Given $W_{Al} = 2.7 \ g$.
Substituting the values: $\frac{W_{Cu}}{2.7} = \frac{31.75}{9}$.
$W_{Cu} = \frac{31.75 \times 2.7}{9} = 31.75 \times 0.3 = 9.525 \ g$.
Equivalent weight of $Cu = \frac{\text{Atomic mass of } Cu}{\text{Valency factor}} = \frac{63.5}{2} = 31.75 \ g \ eq^{-1}$.
Equivalent weight of $Al = \frac{\text{Atomic mass of } Al}{\text{Valency factor}} = \frac{27}{3} = 9 \ g \ eq^{-1}$.
Given $W_{Al} = 2.7 \ g$.
Substituting the values: $\frac{W_{Cu}}{2.7} = \frac{31.75}{9}$.
$W_{Cu} = \frac{31.75 \times 2.7}{9} = 31.75 \times 0.3 = 9.525 \ g$.
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15
ChemistryEasyMCQGUJCET · 2011
What are the metals and their approximate percentages in German silver?
A
$Cu(52\%), Ni(25\%), Zn(18\%), Fe(5\%)$
B
$Cu(60\%), Ni(40\%)$
C
$Ni(60\%), Fe(25\%), Cr(15\%)$
D
$Cu(50\%), Ni(25\%), Zn(25\%)$
Solution
(D) German silver is an alloy consisting of copper $(Cu)$,nickel $(Ni)$,and zinc $(Zn)$.
Its typical composition is approximately $50\%$ copper,$25\%$ nickel,and $25\%$ zinc.
Therefore,the correct option is $D$.
Its typical composition is approximately $50\%$ copper,$25\%$ nickel,and $25\%$ zinc.
Therefore,the correct option is $D$.
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