ChemistryQ351–385 of 716 questions
Page 8 of 8 · English
351
ChemistryMediumMCQMHT CET · 2023
If a $0.01 \ m$ aqueous solution of an electrolyte freezes at $-0.056 \ ^{\circ}C$,calculate the van't Hoff factor $(i)$ for the electrolyte. (Cryoscopic constant of water $K_{f} = 1.86 \ K \ kg \ mol^{-1}$)
A
$1.3$
B
$2.33$
C
$3.01$
D
$4.11$
Solution
(C) The formula for depression in freezing point is $\Delta T_{f} = i \times K_{f} \times m$.
Given: $\Delta T_{f} = 0 - (-0.056) = 0.056 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$.
Substituting the values: $0.056 = i \times 1.86 \times 0.01$.
Therefore,$i = \frac{0.056}{1.86 \times 0.01} = \frac{0.056}{0.0186} \approx 3.01$.
Given: $\Delta T_{f} = 0 - (-0.056) = 0.056 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$.
Substituting the values: $0.056 = i \times 1.86 \times 0.01$.
Therefore,$i = \frac{0.056}{1.86 \times 0.01} = \frac{0.056}{0.0186} \approx 3.01$.
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352
ChemistryMediumMCQMHT CET · 2023
What is the molality of a solution of a nonvolatile solute having a boiling point elevation of $7.15 \ K$ and a molal elevation constant of $2.75 \ K \ kg \ mol^{-1}$ (in $m$)?
A
$3.2$
B
$2.0$
C
$2.6$
D
$3.8$
Solution
(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
$\therefore m = \frac{\Delta T_{b}}{K_{b}}$.
Substituting the given values: $m = \frac{7.15 \ K}{2.75 \ K \ kg \ mol^{-1}}$.
$m = 2.6 \ mol \ kg^{-1}$ (or $2.6 \ m$).
$\therefore m = \frac{\Delta T_{b}}{K_{b}}$.
Substituting the given values: $m = \frac{7.15 \ K}{2.75 \ K \ kg \ mol^{-1}}$.
$m = 2.6 \ mol \ kg^{-1}$ (or $2.6 \ m$).
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353
ChemistryEasyMCQMHT CET · 2023
Which among the following is $NOT$ a colligative property?
A
Vapour pressure lowering
B
Boiling point
C
Freezing point depression
D
Osmotic pressure
Solution
(B) Colligative properties are those properties of solutions that depend only on the number of solute particles present in the solution,not on their nature.
These include:
$1$. Relative lowering of vapour pressure
$2$. Elevation of boiling point
$3$. Depression of freezing point
$4$. Osmotic pressure
Among the given options,$A$,$C$,and $D$ are colligative properties.
$B$ (Boiling point) is a physical property of the solvent/solution,whereas 'Elevation of boiling point' is the colligative property.
Therefore,the correct answer is $B$.
These include:
$1$. Relative lowering of vapour pressure
$2$. Elevation of boiling point
$3$. Depression of freezing point
$4$. Osmotic pressure
Among the given options,$A$,$C$,and $D$ are colligative properties.
$B$ (Boiling point) is a physical property of the solvent/solution,whereas 'Elevation of boiling point' is the colligative property.
Therefore,the correct answer is $B$.
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354
ChemistryEasyMCQMHT CET · 2023
Calculate the depression in freezing point of a solution when $4 \,g$ of a nonvolatile solute with a molar mass of $126 \,g \,mol^{-1}$ is dissolved in $80 \,mL$ of water. $[$Cryoscopic constant of water $K_f = 1.86 \,K \,kg \,mol^{-1}]$ (in $\,K$)
A
$0.55$
B
$0.74$
C
$0.86$
D
$0.96$
Solution
(B) Given: Mass of solute $(W_2) = 4 \,g$, Molar mass of solute $(M_2) = 126 \,g \,mol^{-1}$, Volume of water $= 80 \,mL$.
Since the density of water is $1 \,g/mL$, the mass of solvent $(W_1) = 80 \,g$.
The formula for depression in freezing point is $\Delta T_f = \frac{1000 \times K_f \times W_2}{M_2 \times W_1}$.
Substituting the values: $\Delta T_f = \frac{1000 \times 1.86 \times 4}{126 \times 80}$.
$\Delta T_f = \frac{7440}{10080} \approx 0.738 \,K$, which rounds to $0.74 \,K$.
Since the density of water is $1 \,g/mL$, the mass of solvent $(W_1) = 80 \,g$.
The formula for depression in freezing point is $\Delta T_f = \frac{1000 \times K_f \times W_2}{M_2 \times W_1}$.
Substituting the values: $\Delta T_f = \frac{1000 \times 1.86 \times 4}{126 \times 80}$.
$\Delta T_f = \frac{7440}{10080} \approx 0.738 \,K$, which rounds to $0.74 \,K$.
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355
ChemistryMediumMCQMHT CET · 2023
Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \ mmHg$ and $590 \ mmHg$ respectively at the same temperature.
A
$0.078$
B
$0.175$
C
$0.061$
D
$0.092$
Solution
(A) The relative lowering of vapour pressure is defined as the ratio of the decrease in vapour pressure to the vapour pressure of the pure solvent.
The formula is: $\frac{\Delta P}{P_1^0} = \frac{P_1^0 - P_s}{P_1^0}$
Given:
Vapour pressure of pure benzene $(P_1^0)$ = $640 \ mmHg$
Vapour pressure of solution $(P_s)$ = $590 \ mmHg$
Substituting the values:
$\frac{640 - 590}{640} = \frac{50}{640} = 0.078125 \approx 0.078$
Thus,the relative lowering of vapour pressure is $0.078$.
The formula is: $\frac{\Delta P}{P_1^0} = \frac{P_1^0 - P_s}{P_1^0}$
Given:
Vapour pressure of pure benzene $(P_1^0)$ = $640 \ mmHg$
Vapour pressure of solution $(P_s)$ = $590 \ mmHg$
Substituting the values:
$\frac{640 - 590}{640} = \frac{50}{640} = 0.078125 \approx 0.078$
Thus,the relative lowering of vapour pressure is $0.078$.
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356
ChemistryMediumMCQMHT CET · 2023
$A$ solution of a nonvolatile solute is obtained by dissolving $15 \ g$ in $200 \ mL$ of water,which has a depression in freezing point of $0.75 \ K$. Calculate the molar mass of the solute if the cryoscopic constant of water is $1.86 \ K \ kg \ mol^{-1}$.
A
$160 \ g \ mol^{-1}$
B
$172 \ g \ mol^{-1}$
C
$186 \ g \ mol^{-1}$
D
$198 \ g \ mol^{-1}$
Solution
(C) Given: Mass of solute $(W_2)$ = $15 \ g$,Volume of water = $200 \ mL$,so Mass of solvent $(W_1)$ = $200 \ g$ (assuming density = $1 \ g/mL$).
Depression in freezing point $(\Delta T_f)$ = $0.75 \ K$.
Cryoscopic constant $(K_f)$ = $1.86 \ K \ kg \ mol^{-1}$.
The formula for molar mass $(M_2)$ is:
$M_2 = \frac{1000 \times K_f \times W_2}{\Delta T_f \times W_1}$
Substituting the values:
$M_2 = \frac{1000 \times 1.86 \times 15}{0.75 \times 200}$
$M_2 = \frac{27900}{150} = 186 \ g \ mol^{-1}$.
Depression in freezing point $(\Delta T_f)$ = $0.75 \ K$.
Cryoscopic constant $(K_f)$ = $1.86 \ K \ kg \ mol^{-1}$.
The formula for molar mass $(M_2)$ is:
$M_2 = \frac{1000 \times K_f \times W_2}{\Delta T_f \times W_1}$
Substituting the values:
$M_2 = \frac{1000 \times 1.86 \times 15}{0.75 \times 200}$
$M_2 = \frac{27900}{150} = 186 \ g \ mol^{-1}$.
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357
ChemistryMediumMCQMHT CET · 2023
Calculate the number of atoms present in a unit cell for an element having a molar mass of $23 \ g \ mol^{-1}$ and a density of $0.96 \ g \ cm^{-3}$. Given that $a^3 \cdot N_{A} = 48 \ cm^3 \ mol^{-1}$.
A
$1$
B
$2$
C
$4$
D
$6$
Solution
(B) The density formula for a unit cell is given by $\rho = \frac{nM}{a^3 N_{A}}$.
Substituting the given values: $0.96 = \frac{n \times 23}{48}$.
Rearranging for $n$: $n = \frac{0.96 \times 48}{23}$.
$n = \frac{46.08}{23} \approx 2.003$.
Since $n$ represents the number of atoms per unit cell,the value is $2$.
Substituting the given values: $0.96 = \frac{n \times 23}{48}$.
Rearranging for $n$: $n = \frac{0.96 \times 48}{23}$.
$n = \frac{46.08}{23} \approx 2.003$.
Since $n$ represents the number of atoms per unit cell,the value is $2$.
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358
ChemistryMediumMCQMHT CET · 2023
Identify the number of moles of donor atoms in $2n$ moles of $SCN^{-}$.
A
$3n$
B
$6n$
C
$4n$
D
$n$
Solution
(C) The thiocyanate ion,$SCN^{-}$,is an ambidentate ligand that possesses two potential donor atoms: nitrogen $(N)$ and sulfur $(S)$.
In any coordinate bond formation,one of these atoms acts as the donor.
Therefore,$1$ mole of $SCN^{-}$ contains $2$ moles of donor atoms.
For $2n$ moles of $SCN^{-}$,the total number of moles of donor atoms is calculated as: $2n \times 2 = 4n$ moles.
In any coordinate bond formation,one of these atoms acts as the donor.
Therefore,$1$ mole of $SCN^{-}$ contains $2$ moles of donor atoms.
For $2n$ moles of $SCN^{-}$,the total number of moles of donor atoms is calculated as: $2n \times 2 = 4n$ moles.
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359
ChemistryMediumMCQMHT CET · 2023
Calculate the number of atoms present in a unit cell of an element having molar mass $190 \ g \ mol^{-1}$ and density $20 \ g \ cm^{-3}$. Given that $[a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}]$.
A
$1$
B
$2$
C
$6$
D
$4$
Solution
(D) The formula for density $(\rho)$ of a unit cell is given by: $\rho = \frac{M \times n}{a^3 \times N_A}$
Where $M$ is the molar mass,$n$ is the number of atoms per unit cell,$a^3$ is the volume of the unit cell,and $N_A$ is the Avogadro constant.
Given: $\rho = 20 \ g \ cm^{-3}$,$M = 190 \ g \ mol^{-1}$,and $a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}$.
Substituting the values into the formula: $20 = \frac{190 \times n}{38}$
$n = \frac{20 \times 38}{190}$
$n = \frac{760}{190} = 4$
Therefore,the number of atoms present in the unit cell is $4$.
Where $M$ is the molar mass,$n$ is the number of atoms per unit cell,$a^3$ is the volume of the unit cell,and $N_A$ is the Avogadro constant.
Given: $\rho = 20 \ g \ cm^{-3}$,$M = 190 \ g \ mol^{-1}$,and $a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}$.
Substituting the values into the formula: $20 = \frac{190 \times n}{38}$
$n = \frac{20 \times 38}{190}$
$n = \frac{760}{190} = 4$
Therefore,the number of atoms present in the unit cell is $4$.
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360
ChemistryEasyMCQMHT CET · 2023
What is the number of moles of donor atoms in $n$ mole of $NO_2^{-}$?
A
$3 n$
B
$0$
C
$2 n$
D
$n$
Solution
(D) The ligand $NO_2^{-}$ is an ambidentate ligand that can coordinate through either the nitrogen atom or the oxygen atom.
In either case,there is only $1$ donor atom involved in the coordination bond per molecule of $NO_2^{-}$.
Therefore,in $n$ moles of $NO_2^{-}$,the number of moles of donor atoms is $n \times 1 = n$ moles.
In either case,there is only $1$ donor atom involved in the coordination bond per molecule of $NO_2^{-}$.
Therefore,in $n$ moles of $NO_2^{-}$,the number of moles of donor atoms is $n \times 1 = n$ moles.
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361
ChemistryMediumMCQMHT CET · 2023
Identify the product '$B$' in the following reaction.
$Dry \ ice$ $\xrightarrow[Dry \ ether]{CH_3MgBr} A$ $\xrightarrow[dil. \ HCl]{H_2O} B$
$Dry \ ice$ $\xrightarrow[Dry \ ether]{CH_3MgBr} A$ $\xrightarrow[dil. \ HCl]{H_2O} B$
A
Methanoic acid
B
Ethanoic acid
C
Methanol
D
Ethanol
Solution
(B) The reaction of dry ice $(CO_2)$ with a Grignard reagent $(CH_3MgBr)$ in the presence of dry ether forms an intermediate addition product $(A)$,which is $CH_3COOMgBr$.
$CO_2 + CH_3MgBr \xrightarrow{Dry \ ether} CH_3COOMgBr (A)$
This intermediate $(A)$ upon hydrolysis with dilute $HCl$ yields a carboxylic acid $(B)$,which is ethanoic acid $(CH_3COOH)$.
$CH_3COOMgBr + H_2O \xrightarrow{dil. \ HCl} CH_3COOH (B) + Mg(OH)Br$
Therefore,the product $B$ is ethanoic acid.
$CO_2 + CH_3MgBr \xrightarrow{Dry \ ether} CH_3COOMgBr (A)$
This intermediate $(A)$ upon hydrolysis with dilute $HCl$ yields a carboxylic acid $(B)$,which is ethanoic acid $(CH_3COOH)$.
$CH_3COOMgBr + H_2O \xrightarrow{dil. \ HCl} CH_3COOH (B) + Mg(OH)Br$
Therefore,the product $B$ is ethanoic acid.
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362
ChemistryEasyMCQMHT CET · 2023
The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:
A
Dalton's law
B
Avogadro's law
C
Raoult's law
D
Henry's law
Solution
(C) According to Raoult's law,for a solution of volatile liquids,the partial vapour pressure of each component is directly proportional to its mole fraction present in the solution.
Mathematically,$p_i = p_i^0 \times x_i$,where $p_i$ is the partial vapour pressure,$p_i^0$ is the vapour pressure of the pure component,and $x_i$ is the mole fraction.
Mathematically,$p_i = p_i^0 \times x_i$,where $p_i$ is the partial vapour pressure,$p_i^0$ is the vapour pressure of the pure component,and $x_i$ is the mole fraction.
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363
ChemistryEasyMCQMHT CET · 2023
Which of the following pairs of nuclides is an example of isotones?
A
$_{17}^{35}Cl, _{17}^{37}Cl$
B
$_{18}^{40}Ar, _{19}^{40}K$
C
$_{6}^{13}C, _{5}^{11}B$
D
$_{6}^{12}C, _{5}^{11}B$
Solution
(D) Isotones are atoms of different elements that have the same number of neutrons in their nuclei.
To find the number of neutrons,we use the formula: $n = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For $_{6}^{12}C$: $n = 12 - 6 = 6$.
For $_{5}^{11}B$: $n = 11 - 5 = 6$.
Since both have $6$ neutrons,they are isotones.
To find the number of neutrons,we use the formula: $n = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For $_{6}^{12}C$: $n = 12 - 6 = 6$.
For $_{5}^{11}B$: $n = 11 - 5 = 6$.
Since both have $6$ neutrons,they are isotones.
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364
ChemistryMediumMCQMHT CET · 2023
Find the value of spin-only magnetic moment for chromium cation in $+2$ state. (in $BM$)
A
$3.87$
B
$4.90$
C
$2.84$
D
$1.73$
Solution
(B) The electronic configuration of $Cr$ is $[Ar] \ 3d^5 \ 4s^1$.
For $Cr^{2+}$,two electrons are removed,resulting in the configuration $[Ar] \ 3d^4$.
The number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=4$: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
For $Cr^{2+}$,two electrons are removed,resulting in the configuration $[Ar] \ 3d^4$.
The number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=4$: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
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365
ChemistryEasyMCQMHT CET · 2023
What is the calculated value of spin-only magnetic moment in terms of $BM$ if only one unpaired electron is present in a species?
A
$2.76$
B
$2.84$
C
$2.2$
D
$1.73$
Solution
(D) The formula for spin-only magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given that the number of unpaired electrons $n = 1$.
Substituting the value of $n$ into the formula: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
Given that the number of unpaired electrons $n = 1$.
Substituting the value of $n$ into the formula: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
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366
ChemistryEasyMCQMHT CET · 2023
Identify a $CORRECT$ formula for spin only magnetic moment.
A
$\mu = (n^2 + 2) \ BM$
B
$\mu = \sqrt{n^2 + 2} \ BM$
C
$\mu = \sqrt{n(n + 2)} \ BM$
D
$\mu = (n + 2)^2 \ BM$
Solution
(C) The spin-only magnetic moment $(\mu)$ of an atom or ion is calculated using the formula $\mu = \sqrt{n(n + 2)} \ BM$,where $n$ represents the number of unpaired electrons. The unit $BM$ stands for Bohr Magneton.
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367
ChemistryMediumMCQMHT CET · 2023
Which element from the following does $NOT$ exhibit a spin-only magnetic moment in the $+3$ oxidation state?
A
$Cr$
B
$V$
C
$Ti$
D
$Sc$
Solution
(D) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1.$ $Sc$ $(Z=21)$: $[Ar] 3d^1 4s^2$. In the $+3$ state,$Sc^{3+}$ is $[Ar] 3d^0$. Here,$n=0$,so $\mu = 0 \text{ BM}$.
$2.$ $Ti$ $(Z=22)$: $[Ar] 3d^2 4s^2$. In the $+3$ state,$Ti^{3+}$ is $[Ar] 3d^1$. Here,$n=1$,so $\mu = \sqrt{3} \text{ BM}$.
$3.$ $V$ $(Z=23)$: $[Ar] 3d^3 4s^2$. In the $+3$ state,$V^{3+}$ is $[Ar] 3d^2$. Here,$n=2$,so $\mu = \sqrt{8} \text{ BM}$.
$4.$ $Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$. In the $+3$ state,$Cr^{3+}$ is $[Ar] 3d^3$. Here,$n=3$,so $\mu = \sqrt{15} \text{ BM}$.
Therefore,$Sc$ does not exhibit a spin-only magnetic moment in the $+3$ state because it has no unpaired electrons.
$1.$ $Sc$ $(Z=21)$: $[Ar] 3d^1 4s^2$. In the $+3$ state,$Sc^{3+}$ is $[Ar] 3d^0$. Here,$n=0$,so $\mu = 0 \text{ BM}$.
$2.$ $Ti$ $(Z=22)$: $[Ar] 3d^2 4s^2$. In the $+3$ state,$Ti^{3+}$ is $[Ar] 3d^1$. Here,$n=1$,so $\mu = \sqrt{3} \text{ BM}$.
$3.$ $V$ $(Z=23)$: $[Ar] 3d^3 4s^2$. In the $+3$ state,$V^{3+}$ is $[Ar] 3d^2$. Here,$n=2$,so $\mu = \sqrt{8} \text{ BM}$.
$4.$ $Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$. In the $+3$ state,$Cr^{3+}$ is $[Ar] 3d^3$. Here,$n=3$,so $\mu = \sqrt{15} \text{ BM}$.
Therefore,$Sc$ does not exhibit a spin-only magnetic moment in the $+3$ state because it has no unpaired electrons.
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368
ChemistryDifficultMCQMHT CET · 2023
What type of information is collected using scanning electron microscopy $(SEM)$?
A
Structure of material surface
B
Crystal structure
C
Binding nature
D
Particle size
Solution
(A) Scanning electron microscopy $(SEM)$ is a technique used to image the surface of a sample by scanning the surface with a focused beam of electrons.
It provides detailed information about the topography and morphology (structure) of the material surface.
Therefore,the correct option is $A$.
It provides detailed information about the topography and morphology (structure) of the material surface.
Therefore,the correct option is $A$.
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369
ChemistryEasyMCQMHT CET · 2023
Identify the element having general electronic configuration $ns^2 np^4$ from the following.
A
$Se$
B
$Br$
C
$Xe$
D
$Kr$
Solution
(A) The general electronic configuration of group $16$ elements (chalcogens) is $ns^2 np^4$.
Among the given options,$Se$ (Selenium) belongs to group $16$ with the valence shell configuration $4s^2 4p^4$.
$Br$ belongs to group $17$,while $Xe$ and $Kr$ belong to group $18$.
Among the given options,$Se$ (Selenium) belongs to group $16$ with the valence shell configuration $4s^2 4p^4$.
$Br$ belongs to group $17$,while $Xe$ and $Kr$ belong to group $18$.
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370
ChemistryEasyMCQMHT CET · 2023
Identify physisorption from the following.
A
$O_2$ gas deposited on tungsten
B
$H_2$ gas deposited on nickel
C
$N_2$ gas deposited on iron
D
All gases deposited on charcoal
Solution
(D) Physisorption (physical adsorption) is characterized by weak van der Waals forces between the adsorbate and the adsorbent.
It is non-specific in nature.
Chemisorption involves strong chemical bonds and is highly specific.
$O_2$ on tungsten,$H_2$ on nickel,and $N_2$ on iron involve chemical bond formation (chemisorption).
Adsorption of gases on charcoal is a classic example of physisorption due to weak van der Waals forces.
It is non-specific in nature.
Chemisorption involves strong chemical bonds and is highly specific.
$O_2$ on tungsten,$H_2$ on nickel,and $N_2$ on iron involve chemical bond formation (chemisorption).
Adsorption of gases on charcoal is a classic example of physisorption due to weak van der Waals forces.
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371
ChemistryEasyMCQMHT CET · 2023
Which of the following phenomena is inversely proportional to adsorption?
A
Critical temperature of gas
B
Surface area of adsorbent
C
Temperature of process
D
Pressure of gas
Solution
(C) Adsorption is an exothermic process,where $\Delta H < 0$.
According to Le Chatelier's principle,an increase in temperature favors the reverse process,which is desorption.
Therefore,the extent of adsorption is inversely proportional to the temperature of the process.
According to Le Chatelier's principle,an increase in temperature favors the reverse process,which is desorption.
Therefore,the extent of adsorption is inversely proportional to the temperature of the process.
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372
ChemistryEasyMCQMHT CET · 2023
Identify the $FALSE$ statement regarding adsorption from the following.
A
It takes place due to unbalanced forces acting on the surface of solid or liquid.
B
During adsorption,the surface energy of the adsorbent increases.
C
It is caused by van der Waals forces.
D
It is an exothermic process.
Solution
(B) Adsorption is a surface phenomenon that occurs due to unbalanced forces on the surface of a solid or liquid.
During the process of adsorption,the residual forces on the surface decrease,which leads to a decrease in the surface energy of the adsorbent.
Therefore,the statement that surface energy increases is $FALSE$.
Adsorption is generally an exothermic process and can be caused by van der Waals forces (physisorption) or chemical bonds (chemisorption).
During the process of adsorption,the residual forces on the surface decrease,which leads to a decrease in the surface energy of the adsorbent.
Therefore,the statement that surface energy increases is $FALSE$.
Adsorption is generally an exothermic process and can be caused by van der Waals forces (physisorption) or chemical bonds (chemisorption).
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373
ChemistryEasyMCQMHT CET · 2023
Which of the following gases is readily adsorbed by solid adsorbent?
A
$Cl_2$
B
$N_2$
C
$O_2$
D
$H_2$
Solution
(A) The extent of adsorption of a gas on a solid surface depends on the ease of liquefaction of the gas.
Gases with higher critical temperatures are more easily liquefied and thus show greater adsorption.
Among the given options,$Cl_2$ has the highest critical temperature $(417 \ K)$,making it the most readily adsorbed gas compared to $N_2$,$O_2$,and $H_2$.
Gases with higher critical temperatures are more easily liquefied and thus show greater adsorption.
Among the given options,$Cl_2$ has the highest critical temperature $(417 \ K)$,making it the most readily adsorbed gas compared to $N_2$,$O_2$,and $H_2$.
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374
ChemistryEasyMCQMHT CET · 2023
Which of the following is a characteristic of a lyophilic colloid?
A
Particles of the dispersed phase have no affinity for the dispersion medium.
B
Particles are easily detected under an ultramicroscope.
C
Addition of a large amount of electrolyte causes precipitation.
D
These are irreversible colloids.
Solution
(C) Lyophilic colloids are characterized by a strong affinity between the dispersed phase and the dispersion medium.
They are stable and are not easily precipitated by small amounts of electrolytes; a large amount of electrolyte is required to cause coagulation (salting out).
They are reversible in nature,meaning they can be reformed after evaporation of the dispersion medium.
They are not easily visible under an ultramicroscope due to the small size and high solvation of particles.
They are stable and are not easily precipitated by small amounts of electrolytes; a large amount of electrolyte is required to cause coagulation (salting out).
They are reversible in nature,meaning they can be reformed after evaporation of the dispersion medium.
They are not easily visible under an ultramicroscope due to the small size and high solvation of particles.
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375
ChemistryEasyMCQMHT CET · 2023
Which of the following ions has greater coagulating power for a negatively charged sol?
A
$Fe(CN)_6^{4-}$
B
$PO_4^{3-}$
C
$Ba^{2+}$
D
$Al^{3+}$
Solution
(D) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of the charge on the ion of opposite sign to that of the colloidal particle.
For a negatively charged sol,the coagulating power depends on the positive charge of the added cation.
The given cations are $Ba^{2+}$ and $Al^{3+}$.
Since the charge on $Al^{3+}$ $(+3)$ is greater than the charge on $Ba^{2+}$ $(+2)$,$Al^{3+}$ has greater coagulating power.
For a negatively charged sol,the coagulating power depends on the positive charge of the added cation.
The given cations are $Ba^{2+}$ and $Al^{3+}$.
Since the charge on $Al^{3+}$ $(+3)$ is greater than the charge on $Ba^{2+}$ $(+2)$,$Al^{3+}$ has greater coagulating power.
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376
ChemistryEasyMCQMHT CET · 2023
What type of solution is gasoline?
A
Liquid as solute and liquid as solvent
B
Liquid as solute and solid as solvent
C
Solid as solute and liquid as solvent
D
Gas as solute and liquid as solvent
Solution
(A) Gasoline is a mixture of liquid hydrocarbons.
In a liquid-liquid solution,the solute is a liquid and the solvent is also a liquid.
Since gasoline consists of various liquid hydrocarbons mixed together,it is classified as a liquid-in-liquid solution.
Therefore,the correct option is $A$.
In a liquid-liquid solution,the solute is a liquid and the solvent is also a liquid.
Since gasoline consists of various liquid hydrocarbons mixed together,it is classified as a liquid-in-liquid solution.
Therefore,the correct option is $A$.
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377
ChemistryEasyMCQMHT CET · 2023
Which of the following is a positively charged sol?
A
Haemoglobin in blood
B
Sol of starch
C
Gelatin
D
$Ag$ sol
Solution
(A) Haemoglobin in blood is a positively charged sol.
Starch sol,gelatin sol,and silver $(Ag)$ sol are examples of negatively charged sols.
Starch sol,gelatin sol,and silver $(Ag)$ sol are examples of negatively charged sols.
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378
ChemistryEasyMCQMHT CET · 2023
Which of the following colloids is $NOT$ a gel?
A
Cheese
B
Butter
C
Milk
D
Jellies
Solution
(C) gel is a colloidal system in which a liquid is dispersed in a solid.
Cheese,butter,and jellies are examples of gels.
Milk is an emulsion,which is a colloidal system where a liquid is dispersed in another liquid.
Cheese,butter,and jellies are examples of gels.
Milk is an emulsion,which is a colloidal system where a liquid is dispersed in another liquid.
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379
ChemistryEasyMCQMHT CET · 2023
Identify the last step in wet chemical synthesis of nanomaterial.
A
Formation of oxide or alcohol-bridged network
B
Dehydration
C
Aging of the gel
D
Drying of the gel
Solution
(B) The wet chemical synthesis (Sol-Gel process) of nanomaterials typically follows these steps: hydrolysis,condensation,gelation (formation of oxide or alcohol-bridged network),aging,drying,and finally dehydration (or calcination) to remove residual organic groups and water to obtain the final nanomaterial.
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380
ChemistryDifficultMCQMHT CET · 2023
Identify the example of zero-dimensional $(0D)$ nanostructure from the following.
A
Nanotubes
B
Fibres
C
Thin films
D
Quantum dots
Solution
(D) Quantum dots are zero-dimensional $(0D)$ nanostructures because all three dimensions are restricted to the nanoscale $(1-100 \ nm)$.
In contrast,nanotubes and fibres are one-dimensional $(1D)$ nanostructures,and thin films are two-dimensional $(2D)$ nanostructures.
In contrast,nanotubes and fibres are one-dimensional $(1D)$ nanostructures,and thin films are two-dimensional $(2D)$ nanostructures.
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381
ChemistryEasyMCQMHT CET · 2023
Which of the following nanomaterials has one dimension less than $100 \ nm$?
A
Fibres
B
Nanoparticles
C
Thin films
D
Microcapsules
Solution
(C) nanomaterial with one dimension less than $100 \ nm$ is classified as a two-dimensional nanostructure.
Thin films have one dimension in the nanoscale range,while the other two dimensions are much larger.
Therefore,thin films are the correct answer.
Thin films have one dimension in the nanoscale range,while the other two dimensions are much larger.
Therefore,thin films are the correct answer.
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382
ChemistryEasyMCQMHT CET · 2023
Which statement from the following about nanomaterials is $NOT$ correct?
A
As the particle size decreases,the total surface area of particles increases.
B
Nanomaterial-based catalysts exhibit increased catalytic activities.
C
Nanosized $Cu$ and $Pd$ clusters have very less hardness than bulk metal.
D
Carbon nanotubes can act as electrical conductors.
Solution
(C) The statement that nanosized $Cu$ and $Pd$ clusters have less hardness than bulk metal is incorrect.
In reality,nanosized copper and palladium clusters with a diameter in the size range of $5-7 \ nm$ can have hardness up to $500 \%$ greater than the bulk metal due to the Hall-Petch effect.
In reality,nanosized copper and palladium clusters with a diameter in the size range of $5-7 \ nm$ can have hardness up to $500 \%$ greater than the bulk metal due to the Hall-Petch effect.
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383
ChemistryEasyMCQMHT CET · 2023
Which of the following is an example of two-dimensional nanostructures?
A
Nanoparticles
B
Thin Films
C
Quantum dots
D
Nanowires
Solution
(B) Nanostructures are classified based on their dimensions in the nanoscale range $(1-100 \ nm)$:
$1$. Zero-dimensional $(0D)$ nanostructures: All three dimensions are in the nanoscale (e.g.,Quantum dots,Nanoparticles).
$2$. One-dimensional $(1D)$ nanostructures: Two dimensions are in the nanoscale (e.g.,Nanowires,Nanotubes).
$3$. Two-dimensional $(2D)$ nanostructures: One dimension is in the nanoscale (e.g.,Thin films,Layers,Coatings).
Therefore,Thin films are examples of two-dimensional nanostructures.
$1$. Zero-dimensional $(0D)$ nanostructures: All three dimensions are in the nanoscale (e.g.,Quantum dots,Nanoparticles).
$2$. One-dimensional $(1D)$ nanostructures: Two dimensions are in the nanoscale (e.g.,Nanowires,Nanotubes).
$3$. Two-dimensional $(2D)$ nanostructures: One dimension is in the nanoscale (e.g.,Thin films,Layers,Coatings).
Therefore,Thin films are examples of two-dimensional nanostructures.
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384
ChemistryEasyMCQMHT CET · 2023
Which of the following statements is $NOT$ correct?
A
${\Delta}G^{\circ}$ is an extensive property.
B
$E^{\circ}_{cell}$ is an intensive property.
C
Electrical work is equal to $nFE_{cell}$.
D
For a chemical reaction to be spontaneous,$E^{\circ}_{cell}$ must be negative.
Solution
(D) The relationship between Gibbs free energy and cell potential is given by ${\Delta}G = -nFE_{cell}$. For a spontaneous reaction,${\Delta}G$ must be negative,which implies that $E_{cell}$ must be positive. Therefore,the statement that $E^{\circ}_{cell}$ must be negative for a spontaneous reaction is incorrect.
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385
ChemistryMediumMCQMHT CET · 2023
Find the average rate of formation of $O_{2(g)}$ in the following reaction:
$2 NO_{2(g)} \rightarrow 2 NO_{(g)} + O_{2(g)}$
Given that $\left[-\frac{\Delta[NO_2]}{\Delta t}\right] = x \ mol \ dm^{-3} \ s^{-1}$
$2 NO_{2(g)} \rightarrow 2 NO_{(g)} + O_{2(g)}$
Given that $\left[-\frac{\Delta[NO_2]}{\Delta t}\right] = x \ mol \ dm^{-3} \ s^{-1}$
A
$\frac{x}{2} \ mol \ dm^{-3} \ s^{-1}$
B
$x \ mol \ dm^{-3} \ s^{-1}$
C
$2x \ mol \ dm^{-3} \ s^{-1}$
D
$4x \ mol \ dm^{-3} \ s^{-1}$
Solution
(A) The rate of reaction is expressed as:
$Rate = -\frac{1}{2} \frac{\Delta[NO_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[NO]}{\Delta t} = \frac{\Delta[O_2]}{\Delta t}$
Given that the rate of disappearance of $NO_2$ is $-\frac{\Delta[NO_2]}{\Delta t} = x \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression:
$\frac{\Delta[O_2]}{\Delta t} = \frac{1}{2} \left( -\frac{\Delta[NO_2]}{\Delta t} \right)$
$\frac{\Delta[O_2]}{\Delta t} = \frac{1}{2} \times x = \frac{x}{2} \ mol \ dm^{-3} \ s^{-1}$
$Rate = -\frac{1}{2} \frac{\Delta[NO_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[NO]}{\Delta t} = \frac{\Delta[O_2]}{\Delta t}$
Given that the rate of disappearance of $NO_2$ is $-\frac{\Delta[NO_2]}{\Delta t} = x \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression:
$\frac{\Delta[O_2]}{\Delta t} = \frac{1}{2} \left( -\frac{\Delta[NO_2]}{\Delta t} \right)$
$\frac{\Delta[O_2]}{\Delta t} = \frac{1}{2} \times x = \frac{x}{2} \ mol \ dm^{-3} \ s^{-1}$
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