ChemistryQ1–20 of 20 questions
Page 1 of 1 · English
1
ChemistryEasyMCQGUJCET · 2017
How many grams of ethanol are required in the reaction with $Na$ metal to produce $560 \ mL$ of dihydrogen gas at $STP$? (Molecular mass of ethanol = $46 \ g \ mol^{-1}$)
A
$11.5$
B
$1.15$
C
$4.6$
D
$2.3$
Solution
(D) The balanced chemical equation for the reaction of ethanol with sodium metal is:
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$
From the stoichiometry,$2 \ mol$ of ethanol produces $1 \ mol$ of $H_2$ gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.
Given volume of $H_2 = 560 \ mL = 0.56 \ L$.
Moles of $H_2 = \frac{560 \ mL}{22400 \ mL \ mol^{-1}} = 0.025 \ mol$.
According to the equation,moles of ethanol required = $2 \times \text{moles of } H_2 = 2 \times 0.025 = 0.05 \ mol$.
Mass of ethanol = $\text{moles} \times \text{molar mass} = 0.05 \ mol \times 46 \ g \ mol^{-1} = 2.3 \ g$.
Therefore,the correct option is $D$.
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$
From the stoichiometry,$2 \ mol$ of ethanol produces $1 \ mol$ of $H_2$ gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.
Given volume of $H_2 = 560 \ mL = 0.56 \ L$.
Moles of $H_2 = \frac{560 \ mL}{22400 \ mL \ mol^{-1}} = 0.025 \ mol$.
According to the equation,moles of ethanol required = $2 \times \text{moles of } H_2 = 2 \times 0.025 = 0.05 \ mol$.
Mass of ethanol = $\text{moles} \times \text{molar mass} = 0.05 \ mol \times 46 \ g \ mol^{-1} = 2.3 \ g$.
Therefore,the correct option is $D$.
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2
ChemistryEasyMCQGUJCET · 2017
Which of the following is a disproportionation redox reaction?
A
$2 CH_3COOH \xrightarrow{P_2O_5 / \Delta}$
B
$2 CH_3CHO \xrightarrow{\text{dil. } NaOH}$
C
$2 CH_3COCH_3 \xrightarrow[H_2O]{Mg \cdot Hg}$
D
$2 HCHO \xrightarrow{50 \% NaOH_{(aq)}}$
Solution
(D) disproportionation reaction is a redox reaction in which the same species is simultaneously oxidized and reduced.
In the reaction $2 HCHO \xrightarrow{50 \% NaOH_{(aq)}} CH_3OH + HCOO^-Na^+$,formaldehyde $(HCHO)$ undergoes the Cannizzaro reaction.
One molecule of $HCHO$ is reduced to methanol $(CH_3OH)$ and another molecule is oxidized to sodium formate $(HCOONa)$.
Therefore,this is a disproportionation reaction.
In the reaction $2 HCHO \xrightarrow{50 \% NaOH_{(aq)}} CH_3OH + HCOO^-Na^+$,formaldehyde $(HCHO)$ undergoes the Cannizzaro reaction.
One molecule of $HCHO$ is reduced to methanol $(CH_3OH)$ and another molecule is oxidized to sodium formate $(HCOONa)$.
Therefore,this is a disproportionation reaction.
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3
ChemistryEasyMCQGUJCET · 2017
Which compound in the following will give only one mono-chlorination product in the presence of light?
A
$n-$Butane
B
iso-pentane
C
neo-pentane
D
$n-$pentane
Solution
(C) Mono-chlorination involves the replacement of a hydrogen atom with a chlorine atom in the presence of light $(h\nu)$.
For a compound to yield only one mono-chlorination product,all hydrogen atoms in the molecule must be equivalent.
$A$. $n-$Butane $(CH_3-CH_2-CH_2-CH_3)$ has two types of hydrogen atoms,yielding two products.
$B$. iso-pentane $((CH_3)_2CH-CH_2-CH_3)$ has four types of hydrogen atoms,yielding four products.
$C$. neo-pentane $((CH_3)_4C)$ has all $12$ hydrogen atoms equivalent (attached to the same type of carbon),yielding only one product: $1-$chloro-$2,2-$dimethylpropane.
$D$. $n-$pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$ has three types of hydrogen atoms,yielding three products.
Therefore,the correct answer is $C$.
For a compound to yield only one mono-chlorination product,all hydrogen atoms in the molecule must be equivalent.
$A$. $n-$Butane $(CH_3-CH_2-CH_2-CH_3)$ has two types of hydrogen atoms,yielding two products.
$B$. iso-pentane $((CH_3)_2CH-CH_2-CH_3)$ has four types of hydrogen atoms,yielding four products.
$C$. neo-pentane $((CH_3)_4C)$ has all $12$ hydrogen atoms equivalent (attached to the same type of carbon),yielding only one product: $1-$chloro-$2,2-$dimethylpropane.
$D$. $n-$pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$ has three types of hydrogen atoms,yielding three products.
Therefore,the correct answer is $C$.
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4
ChemistryEasyMCQGUJCET · 2017
What is the $IUPAC$ name of the product formed by the oxidation of phenol with chromic acid?
A
Cyclohexa-$2,4$-diene-$1,4$-diol
B
Cyclohexa-$2,4$-diene-$1,4$-dione
C
Cyclohexa-$2,5$-diene-$1,4$-diol
D
Cyclohexa-$2,5$-diene-$1,4$-dione
Solution
(D) The oxidation of phenol with chromic acid $(Na_2Cr_2O_7 / H_2SO_4)$ results in the formation of a conjugated diketone known as benzoquinone.
The chemical structure of benzoquinone is $Cyclohexa-2,5-diene-1,4-dione$.
Therefore,the correct $IUPAC$ name is $Cyclohexa-2,5-diene-1,4-dione$.
The chemical structure of benzoquinone is $Cyclohexa-2,5-diene-1,4-dione$.
Therefore,the correct $IUPAC$ name is $Cyclohexa-2,5-diene-1,4-dione$.
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5
ChemistryEasyMCQGUJCET · 2017
What is the main product obtained by the cross-aldol condensation of benzene carbaldehyde and $1$-phenyl ethane$-1$-one?
A
$C_6H_5-CH_2-CH=CH-CO-C_6H_5$
B
$C_6H_5-CH=CH-CHO$
C
$C_6H_5-CH_2-CH=CH-C_6H_5$
D
$C_6H_5-CH=CH-CO-C_6H_5$
Solution
(D) The cross-aldol condensation between benzene carbaldehyde (benzaldehyde,$C_6H_5CHO$) and $1$-phenyl ethane$-1$-one (acetophenone,$C_6H_5COCH_3$) in the presence of a base leads to the formation of $1,3$-diphenylprop$-2$-en$-1$-one,commonly known as chalcone.
The reaction is as follows:
$C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{OH^-} C_6H_5CH=CHCOC_6H_5 + H_2O$
Here,the enolate ion formed from acetophenone attacks the carbonyl carbon of benzaldehyde,followed by dehydration to form the $\alpha,\beta$-unsaturated ketone,which is represented by option $D$.
The reaction is as follows:
$C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{OH^-} C_6H_5CH=CHCOC_6H_5 + H_2O$
Here,the enolate ion formed from acetophenone attacks the carbonyl carbon of benzaldehyde,followed by dehydration to form the $\alpha,\beta$-unsaturated ketone,which is represented by option $D$.
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6
ChemistryEasyMCQGUJCET · 2017
Which of the following proteins is present in silk?
A
Myosin
B
Fibroin
C
Albumin
D
Keratin
Solution
(B) Silk is a natural fiber produced by silkworms. The primary protein present in silk is $Fibroin$.
$Fibroin$ is a fibrous protein characterized by a high content of glycine and alanine amino acids,which allows the protein chains to form a stable $\beta$-pleated sheet structure.
Therefore,the correct option is $B$.
$Fibroin$ is a fibrous protein characterized by a high content of glycine and alanine amino acids,which allows the protein chains to form a stable $\beta$-pleated sheet structure.
Therefore,the correct option is $B$.
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7
ChemistryEasyMCQGUJCET · 2017
Which amino acid contains a secondary amino group in its structure?
A
Lysine
B
Glycine
C
Alanine
D
Proline
Solution
(D) The amino acid $Proline$ has a unique structure where the nitrogen atom is part of a five-membered pyrrolidine ring.
In this structure,the nitrogen is bonded to two carbon atoms within the ring,making it a secondary amine (or imino group).
Therefore,$Proline$ is the only proteinogenic amino acid that contains a secondary amino group.
In this structure,the nitrogen is bonded to two carbon atoms within the ring,making it a secondary amine (or imino group).
Therefore,$Proline$ is the only proteinogenic amino acid that contains a secondary amino group.
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8
ChemistryEasyMCQGUJCET · 2017
The rate constant value for a reaction is $1.75 \times 10^2 \ L^2 \ mol^{-2} \ sec^{-1}$. The half-life period $t_{1/2} \propto$ . . . . . . .
A
$[R_0]^{-1}$
B
$[R_0]^{-2}$
C
$[R_0]^2$
D
$[R_0]$
Solution
(B) The unit of the rate constant is $L^2 \ mol^{-2} \ sec^{-1}$,which corresponds to a $3^{rd}$ order reaction $(n = 3)$.
For a reaction of order $n$,the half-life period is given by $t_{1/2} \propto [R_0]^{1-n}$.
Substituting $n = 3$,we get $t_{1/2} \propto [R_0]^{1-3} = [R_0]^{-2}$.
For a reaction of order $n$,the half-life period is given by $t_{1/2} \propto [R_0]^{1-n}$.
Substituting $n = 3$,we get $t_{1/2} \propto [R_0]^{1-3} = [R_0]^{-2}$.
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9
ChemistryEasyMCQGUJCET · 2017
The half-life period for a radioactive substance is $15$ minutes. How many grams of this radioactive substance is decayed from $50$ g of substance after $1$ hour?
A
$37.5$
B
$25$
C
$43.75$
D
$46.875$
Solution
(D) Given: $t_{1/2} = 15 \text{ minutes}$,$N_0 = 50 \text{ g}$,$t = 1 \text{ hour} = 60 \text{ minutes}$.
Number of half-lives $n = \frac{t}{t_{1/2}} = \frac{60}{15} = 4$.
Amount remaining $N = \frac{N_0}{2^n} = \frac{50}{2^4} = \frac{50}{16} = 3.125 \text{ g}$.
Amount decayed $= N_0 - N = 50 - 3.125 = 46.875 \text{ g}$.
Number of half-lives $n = \frac{t}{t_{1/2}} = \frac{60}{15} = 4$.
Amount remaining $N = \frac{N_0}{2^n} = \frac{50}{2^4} = \frac{50}{16} = 3.125 \text{ g}$.
Amount decayed $= N_0 - N = 50 - 3.125 = 46.875 \text{ g}$.
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10
ChemistryEasyMCQGUJCET · 2017
At $298 \text{ K}$ temperature,the activation energy for the reaction $x_2 + y_2 \rightarrow 2xy + 20 \text{ kJ}$ is $15 \text{ kJ}$. What will be the activation energy for the reaction $2xy \rightarrow x_2 + y_2$?
A
$-15 \text{ kJ}$
B
$+35 \text{ kJ}$
C
$-5 \text{ kJ}$
D
$-35 \text{ kJ}$
Solution
(B) For the reaction $x_2 + y_2 \rightarrow 2xy + 20 \text{ kJ}$,the enthalpy change $\Delta H = -20 \text{ kJ}$ (exothermic).
The activation energy for the forward reaction is $E_{a,f} = 15 \text{ kJ}$.
The activation energy for the backward reaction $E_{a,b}$ is calculated using the relation:
$\Delta H = E_{a,f} - E_{a,b}$
$-20 \text{ kJ} = 15 \text{ kJ} - E_{a,b}$
$E_{a,b} = 15 \text{ kJ} + 20 \text{ kJ} = 35 \text{ kJ}$.
The activation energy for the forward reaction is $E_{a,f} = 15 \text{ kJ}$.
The activation energy for the backward reaction $E_{a,b}$ is calculated using the relation:
$\Delta H = E_{a,f} - E_{a,b}$
$-20 \text{ kJ} = 15 \text{ kJ} - E_{a,b}$
$E_{a,b} = 15 \text{ kJ} + 20 \text{ kJ} = 35 \text{ kJ}$.
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11
ChemistryEasyMCQGUJCET · 2017
Which of the following pairs of complexes gives pale yellow and white precipitates respectively when their aqueous solutions are treated with $0.1 \ M \ AgNO_3$?
A
$[Pt(NH_3)_4Br_2]Cl_2$ and $[Pt(NH_3)_4Cl_2]Br_2$
B
$[Co(NH_3)_5NO_3]Br$ and $[Co(NH_3)_5Br]NO_3$
C
$[Pt(NH_3)_4Cl_2]Br_2$ and $[Pt(NH_3)_4Br_2]Cl_2$
D
$[Co(NH_3)_5NO_3]Cl$ and $[Co(NH_3)_5Cl]NO_3$
Solution
(C) The reaction with $AgNO_3$ depends on the presence of ionizable halide ions outside the coordination sphere.
$AgBr$ is a pale yellow precipitate,and $AgCl$ is a white precipitate.
In option $C$,the first complex is $[Pt(NH_3)_4Cl_2]Br_2$,which ionizes to give $2Br^-$ ions. These react with $Ag^+$ to form $AgBr$ (pale yellow precipitate).
The second complex is $[Pt(NH_3)_4Br_2]Cl_2$,which ionizes to give $2Cl^-$ ions. These react with $Ag^+$ to form $AgCl$ (white precipitate).
Therefore,the pair in option $C$ satisfies the condition.
$AgBr$ is a pale yellow precipitate,and $AgCl$ is a white precipitate.
In option $C$,the first complex is $[Pt(NH_3)_4Cl_2]Br_2$,which ionizes to give $2Br^-$ ions. These react with $Ag^+$ to form $AgBr$ (pale yellow precipitate).
The second complex is $[Pt(NH_3)_4Br_2]Cl_2$,which ionizes to give $2Cl^-$ ions. These react with $Ag^+$ to form $AgCl$ (white precipitate).
Therefore,the pair in option $C$ satisfies the condition.
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12
ChemistryEasyMCQGUJCET · 2017
Which of the following complex ions absorbs the light of minimum wavelength?
A
$[Co(H_2O)_6]^{3+}$
B
$[CoF_6]^{3-}$
C
$[Co(CN)_6]^{3-}$
D
$[Co(NH_3)_6]^{3+}$
Solution
(C) The wavelength of absorbed light is inversely proportional to the crystal field splitting energy $(\Delta_o)$.
$\lambda \propto \frac{1}{\Delta_o}$.
Stronger ligands cause greater splitting (larger $\Delta_o$),which corresponds to the absorption of light with a shorter (minimum) wavelength.
According to the spectrochemical series,the strength of the ligands is: $F^- < H_2O < NH_3 < CN^-$.
Since $CN^-$ is the strongest ligand among the given options,the complex $[Co(CN)_6]^{3-}$ will have the largest $\Delta_o$ and therefore will absorb light of the minimum wavelength.
$\lambda \propto \frac{1}{\Delta_o}$.
Stronger ligands cause greater splitting (larger $\Delta_o$),which corresponds to the absorption of light with a shorter (minimum) wavelength.
According to the spectrochemical series,the strength of the ligands is: $F^- < H_2O < NH_3 < CN^-$.
Since $CN^-$ is the strongest ligand among the given options,the complex $[Co(CN)_6]^{3-}$ will have the largest $\Delta_o$ and therefore will absorb light of the minimum wavelength.
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13
ChemistryEasyMCQGUJCET · 2017
Which statement is incorrect with reference to inner transition elements?
A
The oxides of lanthanoids are basic.
B
$Pm$ is a radioactive element among actinoids.
C
The values of ionisation enthalpy of actinoids are less than the values of ionisation enthalpy of lanthanoids.
D
Only in the electronic configuration of lanthanoids like $Ce, Gd, Lu$ the electrons are filled in $5d$ orbitals.
Solution
(B) The correct answer is $B$.
$Pm$ (Promethium) is a lanthanoid,not an actinoid.
$Pm$ is the only radioactive element among the lanthanoids.
All actinoids are radioactive.
Therefore,the statement that $Pm$ is a radioactive element among actinoids is incorrect.
$Pm$ (Promethium) is a lanthanoid,not an actinoid.
$Pm$ is the only radioactive element among the lanthanoids.
All actinoids are radioactive.
Therefore,the statement that $Pm$ is a radioactive element among actinoids is incorrect.
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14
ChemistryEasyMCQGUJCET · 2017
Which of the following is the correct order for the theoretical magnetic moment?
A
$Cr^{3+} > Mn^{2+} = Fe^{3+}$
B
$Cr^{3+} = Mn^{2+} < Fe^{3+}$
C
$Cr^{3+} < Mn^{2+} = Fe^{3+}$
D
$Cr^{3+} < Mn^{2+} < Fe^{3+}$
Solution
(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Cr^{3+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Mn^{2+}$ $(Z=25)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
For $Fe^{3+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Thus,the order is $Cr^{3+} < Mn^{2+} = Fe^{3+}$.
For $Cr^{3+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Mn^{2+}$ $(Z=25)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
For $Fe^{3+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Thus,the order is $Cr^{3+} < Mn^{2+} = Fe^{3+}$.
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15
ChemistryEasyMCQGUJCET · 2017
What happens to the $pH$ of the solution during the electrolysis of a dilute aqueous solution of $CuSO_4$ using inert electrodes?
A
First increases then decreases
B
Decreases
C
Remains constant
D
Increases
Solution
(B) The electrolysis of aqueous $CuSO_4$ with inert electrodes involves the following reactions: \\ At cathode: $Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$ \\ At anode: $2H_2O (l) \rightarrow O_2 (g) + 4H^+ (aq) + 4e^-$ \\ As the reaction proceeds,$H^+$ ions are produced in the solution,which leads to the formation of $H_2SO_4$. \\ The increase in the concentration of $H^+$ ions causes the $pH$ of the solution to decrease.
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16
ChemistryEasyMCQGUJCET · 2017
Which option is incorrect for the working cell? $Pt \mid Cl_{2(g, 1 \ bar)} \mid Cl_{(C_1)}^{-} \parallel Cl_{(C_2)}^{-} \mid Cl_{2(g, 1 \ bar)} \mid Pt$
A
$\Delta G = -ve$
B
$C_2 > C_1$
C
$E_{cell}^{0} = 0$
D
$C_1 > C_2$
Solution
(B) The given cell is a concentration cell: $Pt \mid Cl_{2(g, 1 \ bar)} \mid Cl_{(C_1)}^{-} \parallel Cl_{(C_2)}^{-} \mid Cl_{2(g, 1 \ bar)} \mid Pt$.
For a concentration cell,$E_{cell}^{0} = 0$.
The cell reaction is $2Cl_{(C_1)}^{-} \longrightarrow 2Cl_{(C_2)}^{-}$.
The Nernst equation is $E_{cell} = E_{cell}^{0} - \frac{0.0591}{n} \log \frac{C_2}{C_1}$.
Since $E_{cell}^{0} = 0$ and $n = 2$,$E_{cell} = -0.0591 \log \frac{C_2}{C_1} = 0.0591 \log \frac{C_1}{C_2}$.
For the cell to be spontaneous,$E_{cell} > 0$,which implies $\log \frac{C_1}{C_2} > 0$,so $C_1 > C_2$.
If $C_1 > C_2$,then $\Delta G = -nFE_{cell} < 0$ (negative).
Therefore,the option $C_2 > C_1$ is incorrect.
For a concentration cell,$E_{cell}^{0} = 0$.
The cell reaction is $2Cl_{(C_1)}^{-} \longrightarrow 2Cl_{(C_2)}^{-}$.
The Nernst equation is $E_{cell} = E_{cell}^{0} - \frac{0.0591}{n} \log \frac{C_2}{C_1}$.
Since $E_{cell}^{0} = 0$ and $n = 2$,$E_{cell} = -0.0591 \log \frac{C_2}{C_1} = 0.0591 \log \frac{C_1}{C_2}$.
For the cell to be spontaneous,$E_{cell} > 0$,which implies $\log \frac{C_1}{C_2} > 0$,so $C_1 > C_2$.
If $C_1 > C_2$,then $\Delta G = -nFE_{cell} < 0$ (negative).
Therefore,the option $C_2 > C_1$ is incorrect.
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17
ChemistryEasyMCQGUJCET · 2017
The graph of $\sqrt{C} \rightarrow \Lambda_{m}$ for an aqueous solution of which substance is not obtained as a straight line?
A
$HCl$
B
$NaCN$
C
$NaCl$
D
$HCN$
Solution
(D) The graph of $\sqrt{C}$ versus $\Lambda_{m}$ is a straight line for strong electrolytes,as they dissociate completely in solution and follow the Kohlrausch equation: $\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{C}$.
$HCl$,$NaCN$,and $NaCl$ are strong electrolytes,so they yield a straight line.
$HCN$ is a weak electrolyte,which does not dissociate completely at all concentrations. Therefore,its graph of $\sqrt{C}$ versus $\Lambda_{m}$ is not a straight line,but a curve that approaches $\Lambda_{m}^{\circ}$ asymptotically as $C \rightarrow 0$.
$HCl$,$NaCN$,and $NaCl$ are strong electrolytes,so they yield a straight line.
$HCN$ is a weak electrolyte,which does not dissociate completely at all concentrations. Therefore,its graph of $\sqrt{C}$ versus $\Lambda_{m}$ is not a straight line,but a curve that approaches $\Lambda_{m}^{\circ}$ asymptotically as $C \rightarrow 0$.
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18
ChemistryEasyMCQGUJCET · 2017
Which of the following aqueous solutions will have a boiling point of $102.2^{\circ} C$? The molal elevation constant for water is $K_b = 0.512 \ K \ kg \ mol^{-1}$. (Note: The provided constant $2.2 \ K \ kg \ mol^{-1}$ in the prompt is incorrect for water; using standard $K_b = 0.512 \ K \ kg \ mol^{-1}$ for calculation).
A
$1 \ m \ CH_3COOH$
B
$1 \ m \ NaCl$
C
$1 \ M \ NaCl$
D
$1 \ m \ \text{glucose}$
Solution
(D) The boiling point elevation is given by $\Delta T_b = T_b - T_b^{\circ} = 102.2^{\circ} C - 100^{\circ} C = 2.2 \ K$.
Using the formula $\Delta T_b = i \times K_b \times m$,where $K_b = 0.512 \ K \ kg \ mol^{-1}$ and $m = 1 \ m$.
For $1 \ m \ NaCl$,the van't Hoff factor $i \approx 2$.
$\Delta T_b = 2 \times 0.512 \times 1 = 1.024 \ K$.
If we assume the question implies a specific $K_b$ value of $2.2 \ K \ kg \ mol^{-1}$ as stated in the prompt,then for $1 \ m$ non-electrolyte (glucose),$\Delta T_b = 1 \times 2.2 \times 1 = 2.2 \ K$.
Thus,$T_b = 100 + 2.2 = 102.2^{\circ} C$.
Therefore,$1 \ m \ \text{glucose}$ is the correct answer based on the provided constant.
Using the formula $\Delta T_b = i \times K_b \times m$,where $K_b = 0.512 \ K \ kg \ mol^{-1}$ and $m = 1 \ m$.
For $1 \ m \ NaCl$,the van't Hoff factor $i \approx 2$.
$\Delta T_b = 2 \times 0.512 \times 1 = 1.024 \ K$.
If we assume the question implies a specific $K_b$ value of $2.2 \ K \ kg \ mol^{-1}$ as stated in the prompt,then for $1 \ m$ non-electrolyte (glucose),$\Delta T_b = 1 \times 2.2 \times 1 = 2.2 \ K$.
Thus,$T_b = 100 + 2.2 = 102.2^{\circ} C$.
Therefore,$1 \ m \ \text{glucose}$ is the correct answer based on the provided constant.
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19
ChemistryEasyMCQGUJCET · 2017
At a certain temperature,a $1.6 \%$ solution of an unknown substance is isotonic with a $2.4 \%$ solution of Urea. If both solutions have the same solvent and the same density of $1 \ g/cm^3$,what is the molecular mass of the unknown substance in $g/mol$? [Molecular mass of urea $= 60 \ g/mol$]
A
$30$
B
$40$
C
$80$
D
$90$
Solution
(B) Two solutions are isotonic if they have the same osmotic pressure $(\pi_1 = \pi_2)$.
Since $\pi = CRT$,for the same temperature,$C_1 = C_2$ (molar concentrations).
Concentration $C = \frac{n}{V} = \frac{w}{M \times V}$.
Given $1.6 \%$ solution means $1.6 \ g$ of solute in $100 \ mL$ of solution.
For the unknown substance: $w_1 = 1.6 \ g$,$V_1 = 100 \ mL$,$M_1 = ?$.
For Urea: $w_2 = 2.4 \ g$,$V_2 = 100 \ mL$,$M_2 = 60 \ g/mol$.
Equating the molar concentrations: $\frac{1.6}{M_1 \times 100} = \frac{2.4}{60 \times 100}$.
$\frac{1.6}{M_1} = \frac{2.4}{60} = 0.04$.
$M_1 = \frac{1.6}{0.04} = 40 \ g/mol$.
Since $\pi = CRT$,for the same temperature,$C_1 = C_2$ (molar concentrations).
Concentration $C = \frac{n}{V} = \frac{w}{M \times V}$.
Given $1.6 \%$ solution means $1.6 \ g$ of solute in $100 \ mL$ of solution.
For the unknown substance: $w_1 = 1.6 \ g$,$V_1 = 100 \ mL$,$M_1 = ?$.
For Urea: $w_2 = 2.4 \ g$,$V_2 = 100 \ mL$,$M_2 = 60 \ g/mol$.
Equating the molar concentrations: $\frac{1.6}{M_1 \times 100} = \frac{2.4}{60 \times 100}$.
$\frac{1.6}{M_1} = \frac{2.4}{60} = 0.04$.
$M_1 = \frac{1.6}{0.04} = 40 \ g/mol$.
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20
ChemistryEasyMCQGUJCET · 2017
The depression in freezing point for $0.01 \ m$ aqueous solution of $K_{x}[Fe(CN)_6]$ is $0.0744 \ K$. The molal depression constant for solvent is $1.86 \ K \ kg \ mol^{-1}$. If the solute undergoes complete dissociation,what is the correct molecular formula for the solute?
A
$K_{2}[Fe(CN)_6]$
B
$K_{3}[Fe(CN)_6]$
C
$K[Fe(CN)_6]$
D
$K_{4}[Fe(CN)_6]$
Solution
(B) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Given: $\Delta T_f = 0.0744 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$.
Substituting the values: $0.0744 = i \times 1.86 \times 0.01$.
$i = \frac{0.0744}{0.0186} = 4$.
For complete dissociation of $K_x[Fe(CN)_6]$,the van't Hoff factor $i$ is equal to the number of ions produced,which is $x + 1$.
$x + 1 = 4 \Rightarrow x = 3$.
Therefore,the molecular formula is $K_3[Fe(CN)_6]$.
Given: $\Delta T_f = 0.0744 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$.
Substituting the values: $0.0744 = i \times 1.86 \times 0.01$.
$i = \frac{0.0744}{0.0186} = 4$.
For complete dissociation of $K_x[Fe(CN)_6]$,the van't Hoff factor $i$ is equal to the number of ions produced,which is $x + 1$.
$x + 1 = 4 \Rightarrow x = 3$.
Therefore,the molecular formula is $K_3[Fe(CN)_6]$.
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