ChemistryQ301–323 of 563 questions
Page 7 of 7 · English
301
ChemistryMediumMCQMHT CET · 2021
Which of the following formulae is used to calculate the depression in freezing point?
A
$ \Delta T_{f} = \frac{T_{f}^{\circ}}{T_{f}} $
B
$ \Delta T_{f} = T_{f}^{\circ} - T_{f} $
C
$ \Delta T_{f} = T_{f} - T_{f}^{\circ} $
D
$ \Delta T_{f} = \frac{T_{f}}{T_{f}^{\circ}} $
Solution
(B) The depression in freezing point $( \Delta T_{f} )$ is defined as the difference between the freezing point of the pure solvent $( T_{f}^{\circ} )$ and the freezing point of the solution $( T_{f} )$.
Therefore,the correct formula is $ \Delta T_{f} = T_{f}^{\circ} - T_{f} $.
Therefore,the correct formula is $ \Delta T_{f} = T_{f}^{\circ} - T_{f} $.
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302
ChemistryMediumMCQMHT CET · 2021
$5 \ g$ sucrose (molar mass $= 342 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of solvent,decreasing the freezing point by $2.15 \ K$. What is the cryoscopic constant $(K_{f})$ of the solvent?
A
$14.7 \ K \ kg \ mol^{-1}$
B
$2.15 \ K \ kg \ mol^{-1}$
C
$4.30 \ K \ kg \ mol^{-1}$
D
$7.35 \ K \ kg \ mol^{-1}$
Solution
(A) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \cdot m$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{5 / 342}{0.1 \ kg} = \frac{50}{342} \ mol \ kg^{-1}$.
Now,substitute the values into the depression in freezing point equation:
$2.15 = K_{f} \cdot \frac{50}{342}$.
Solving for $K_{f}$:
$K_{f} = \frac{2.15 \times 342}{50} = 14.7 \ K \ kg \ mol^{-1}$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{5 / 342}{0.1 \ kg} = \frac{50}{342} \ mol \ kg^{-1}$.
Now,substitute the values into the depression in freezing point equation:
$2.15 = K_{f} \cdot \frac{50}{342}$.
Solving for $K_{f}$:
$K_{f} = \frac{2.15 \times 342}{50} = 14.7 \ K \ kg \ mol^{-1}$.
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303
ChemistryMediumMCQMHT CET · 2021
Calculate the osmotic pressure exerted by a solution containing $0.822 \ g$ of solute in $300 \ mL$ of water at $300 \ K$. (Molar mass of solute $= 340 \ g \ mol^{-1}, R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$) (in $atm$)
A
$0.5$
B
$0.2$
C
$0.1$
D
$0.4$
Solution
(B) The formula for osmotic pressure is $\pi = CRT$.
First,calculate the number of moles of solute: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.822 \ g}{340 \ g \ mol^{-1}} \approx 0.0024176 \ mol$.
Next,calculate the molar concentration $C$ in $mol \ L^{-1}$: $C = \frac{n}{V(L)} = \frac{0.0024176 \ mol}{0.300 \ L} \approx 0.0080588 \ M$.
Now,substitute the values into the osmotic pressure equation: $\pi = 0.0080588 \ mol \ L^{-1} \times 0.0821 \ L \ atm \ mol^{-1} \ K^{-1} \times 300 \ K$.
$\pi \approx 0.1985 \ atm \approx 0.2 \ atm$.
First,calculate the number of moles of solute: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.822 \ g}{340 \ g \ mol^{-1}} \approx 0.0024176 \ mol$.
Next,calculate the molar concentration $C$ in $mol \ L^{-1}$: $C = \frac{n}{V(L)} = \frac{0.0024176 \ mol}{0.300 \ L} \approx 0.0080588 \ M$.
Now,substitute the values into the osmotic pressure equation: $\pi = 0.0080588 \ mol \ L^{-1} \times 0.0821 \ L \ atm \ mol^{-1} \ K^{-1} \times 300 \ K$.
$\pi \approx 0.1985 \ atm \approx 0.2 \ atm$.
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304
ChemistryEasyMCQMHT CET · 2021
What is the freezing point of a $1 \ molal$ aqueous solution of a non-volatile solute (in $^{\circ} C$)? $(K_{f} = 1.86 \ K \ kg \ mol^{-1}, T_{f}^{\circ} \text{ for water } = 0^{\circ} C)$
A
$-0.93$
B
$-2.43$
C
$-3.72$
D
$-1.86$
Solution
(D) The depression in freezing point is given by the formula: $\Delta T_{f} = K_{f} \cdot m$
Substituting the given values: $\Delta T_{f} = 1.86 \ K \ kg \ mol^{-1} \times 1 \ mol \ kg^{-1} = 1.86 \ K$
Since the change in temperature in Celsius is the same as in Kelvin,$\Delta T_{f} = 1.86^{\circ} C$
The freezing point of the solution is calculated as: $T_{f} = T_{f}^{\circ} - \Delta T_{f}$
$T_{f} = 0^{\circ} C - 1.86^{\circ} C = -1.86^{\circ} C$
Substituting the given values: $\Delta T_{f} = 1.86 \ K \ kg \ mol^{-1} \times 1 \ mol \ kg^{-1} = 1.86 \ K$
Since the change in temperature in Celsius is the same as in Kelvin,$\Delta T_{f} = 1.86^{\circ} C$
The freezing point of the solution is calculated as: $T_{f} = T_{f}^{\circ} - \Delta T_{f}$
$T_{f} = 0^{\circ} C - 1.86^{\circ} C = -1.86^{\circ} C$
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305
ChemistryMediumMCQMHT CET · 2021
What is the vapour pressure of a solution containing $1.8 \ g$ of glucose in $16.2 \ g$ of water (in $mm \ Hg$)? ($P_1^0 = 24 \ mm \ Hg$ and molar mass of glucose $= 180 \ g \ mol^{-1}$)
A
$18.1$
B
$15.7$
C
$12.4$
D
$23.8$
Solution
(D) According to Raoult's law for non-volatile solutes: $\frac{P_1^0 - P_s}{P_1^0} = X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$
$n_{\text{glucose}} = \frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$
$n_{\text{water}} = \frac{16.2 \ g}{18 \ g \ mol^{-1}} = 0.9 \ mol$
Substituting the values: $\frac{24 - P_s}{24} = \frac{0.01}{0.01 + 0.9} = \frac{0.01}{0.91} = \frac{1}{91}$
$24 - P_s = \frac{24}{91} \approx 0.2637$
$P_s = 24 - 0.2637 = 23.736 \ mm \ Hg \approx 23.8 \ mm \ Hg$
$n_{\text{glucose}} = \frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$
$n_{\text{water}} = \frac{16.2 \ g}{18 \ g \ mol^{-1}} = 0.9 \ mol$
Substituting the values: $\frac{24 - P_s}{24} = \frac{0.01}{0.01 + 0.9} = \frac{0.01}{0.91} = \frac{1}{91}$
$24 - P_s = \frac{24}{91} \approx 0.2637$
$P_s = 24 - 0.2637 = 23.736 \ mm \ Hg \approx 23.8 \ mm \ Hg$
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306
ChemistryMediumMCQMHT CET · 2021
The solution containing $6 \ g$ urea (molar mass $60$) per $dm^3$ of water and another solution containing $9 \ g$ of solute $A$ per $dm^3$ water freeze at the same temperature. What is the molar mass of $A$?
A
$90$
B
$180$
C
$54$
D
$120$
Solution
(A) Since both solutions freeze at the same temperature,their freezing point depression $(\Delta T_f)$ is equal.
For dilute aqueous solutions,$\Delta T_f = K_f \times m$,where $m$ is the molality.
Since $K_f$ is the same for both,$m_1 = m_2$.
Given $1 \ dm^3$ of water $\approx 1 \ kg$ of water.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
For urea: $m_1 = \frac{6 \ g / 60 \ g \cdot mol^{-1}}{1 \ kg} = 0.1 \ mol \cdot kg^{-1}$.
For solute $A$: $m_2 = \frac{9 \ g / M_A}{1 \ kg} = \frac{9}{M_A} \ mol \cdot kg^{-1}$.
Equating $m_1 = m_2$: $0.1 = \frac{9}{M_A}$.
Therefore,$M_A = \frac{9}{0.1} = 90 \ g \cdot mol^{-1}$.
For dilute aqueous solutions,$\Delta T_f = K_f \times m$,where $m$ is the molality.
Since $K_f$ is the same for both,$m_1 = m_2$.
Given $1 \ dm^3$ of water $\approx 1 \ kg$ of water.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
For urea: $m_1 = \frac{6 \ g / 60 \ g \cdot mol^{-1}}{1 \ kg} = 0.1 \ mol \cdot kg^{-1}$.
For solute $A$: $m_2 = \frac{9 \ g / M_A}{1 \ kg} = \frac{9}{M_A} \ mol \cdot kg^{-1}$.
Equating $m_1 = m_2$: $0.1 = \frac{9}{M_A}$.
Therefore,$M_A = \frac{9}{0.1} = 90 \ g \cdot mol^{-1}$.
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307
ChemistryEasyMCQMHT CET · 2021
The solution containing $3 \ g$ urea (molar mass $60 \ g \ mol^{-1}$) per $dm^3$ of water and another solution containing $4.5 \ g$ of solute $A$ per $dm^3$ of water boil at the same temperature. What is the molar mass of $A$?
A
$54 \ g \ mol^{-1}$
B
$180 \ g \ mol^{-1}$
C
$120 \ g \ mol^{-1}$
D
$90 \ g \ mol^{-1}$
Solution
(D) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$.
Since both solutions boil at the same temperature,their boiling point elevation $\Delta T_{b}$ must be equal.
Assuming $K_{b}$ is the same for both dilute aqueous solutions,the molality $m$ must be equal.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.
For $1 \ dm^3$ of water,the mass is $1 \ kg$.
For urea: $\text{moles} = \frac{3 \ g}{60 \ g \ mol^{-1}} = 0.05 \ mol$.
For solute $A$: $\text{moles} = \frac{4.5 \ g}{M_A}$.
Equating the molalities: $0.05 = \frac{4.5}{M_A}$.
$M_A = \frac{4.5}{0.05} = 90 \ g \ mol^{-1}$.
Since both solutions boil at the same temperature,their boiling point elevation $\Delta T_{b}$ must be equal.
Assuming $K_{b}$ is the same for both dilute aqueous solutions,the molality $m$ must be equal.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.
For $1 \ dm^3$ of water,the mass is $1 \ kg$.
For urea: $\text{moles} = \frac{3 \ g}{60 \ g \ mol^{-1}} = 0.05 \ mol$.
For solute $A$: $\text{moles} = \frac{4.5 \ g}{M_A}$.
Equating the molalities: $0.05 = \frac{4.5}{M_A}$.
$M_A = \frac{4.5}{0.05} = 90 \ g \ mol^{-1}$.
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308
ChemistryEasyMCQMHT CET · 2021
According to Raoult's law,the mole fraction of solute in a solution is given by which formula?
A
$\frac{\Delta P}{P_1^0}$
B
$\frac{P_1^0}{P_1}$
C
$\frac{P_1^0}{\Delta P}$
D
$\frac{P_1}{P_1^0}$
Solution
(A) According to Raoult's law for solutions containing non-volatile solutes,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
$P_1 =$ Vapour pressure of solution
$P_1^0 =$ Vapour pressure of pure solvent
$\Delta P = P_1^0 - P_1$ (Lowering of vapour pressure)
Relative lowering of vapour pressure $= \frac{P_1^0 - P_1}{P_1^0} = \frac{\Delta P}{P_1^0}$
Thus,the mole fraction of solute $x_2 = \frac{\Delta P}{P_1^0}$.
$P_1 =$ Vapour pressure of solution
$P_1^0 =$ Vapour pressure of pure solvent
$\Delta P = P_1^0 - P_1$ (Lowering of vapour pressure)
Relative lowering of vapour pressure $= \frac{P_1^0 - P_1}{P_1^0} = \frac{\Delta P}{P_1^0}$
Thus,the mole fraction of solute $x_2 = \frac{\Delta P}{P_1^0}$.
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309
ChemistryEasyMCQMHT CET · 2021
Vapour pressure of a solution and of a pure solvent are $P_1$ and $P_1^0$ respectively. If $\frac{P_1}{P_1^0}$ is $0.15$,find the mole fraction of the solute.
A
$0.66$
B
$0.85$
C
$0.15$
D
$0.33$
Solution
(B) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute $(x_{\text{solute}})$.
The formula is: $\frac{P_1^0 - P_1}{P_1^0} = x_{\text{solute}}$.
This can be rewritten as: $1 - \frac{P_1}{P_1^0} = x_{\text{solute}}$.
Given that $\frac{P_1}{P_1^0} = 0.15$,we substitute this value into the equation:
$x_{\text{solute}} = 1 - 0.15 = 0.85$.
The formula is: $\frac{P_1^0 - P_1}{P_1^0} = x_{\text{solute}}$.
This can be rewritten as: $1 - \frac{P_1}{P_1^0} = x_{\text{solute}}$.
Given that $\frac{P_1}{P_1^0} = 0.15$,we substitute this value into the equation:
$x_{\text{solute}} = 1 - 0.15 = 0.85$.
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310
ChemistryEasyMCQMHT CET · 2021
Which among the following gases is adsorbed to a greater extent at similar conditions of temperature and pressure if the adsorbent remains same?
A
$N_2$
B
$Cl_2$
C
$H_2$
D
$O_2$
Solution
(B) The extent of adsorption of a gas on a solid depends on the ease of liquefaction of the gas,which is directly related to its critical temperature $(T_c)$.
Easily liquefiable gases have higher critical temperatures and are adsorbed to a greater extent.
The critical temperatures $(T_c)$ of the given gases are:
$Cl_2$: $417 \ K$
$O_2$: $154 \ K$
$N_2$: $126 \ K$
$H_2$: $33 \ K$
Since $Cl_2$ has the highest critical temperature,it is adsorbed to the greatest extent.
Easily liquefiable gases have higher critical temperatures and are adsorbed to a greater extent.
The critical temperatures $(T_c)$ of the given gases are:
$Cl_2$: $417 \ K$
$O_2$: $154 \ K$
$N_2$: $126 \ K$
$H_2$: $33 \ K$
Since $Cl_2$ has the highest critical temperature,it is adsorbed to the greatest extent.
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311
ChemistryDifficultMCQMHT CET · 2021
What is the spin-only magnetic moment of an element having one unpaired electron (in $BM$)?
A
$0.34$
B
$1.0$
C
$1.73$
D
$3.1$
Solution
(C) The spin-only magnetic moment is calculated using the formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For an element with $n = 1$ unpaired electron:
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
For an element with $n = 1$ unpaired electron:
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
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312
ChemistryMediumMCQMHT CET · 2021
What is the value of the spin-only magnetic moment of $Ni$ $(Z=28)$ in the $+2$ oxidation state (in $BM$)?
A
$3.1$
B
$0.0$
C
$2.8$
D
$1.7$
Solution
(C) The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.
In the $3d$ subshell,there are $8$ electrons,which results in $2$ unpaired electrons $(n=2)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83 \ BM$.
Thus,the value is approximately $2.8 \ BM$.
For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.
In the $3d$ subshell,there are $8$ electrons,which results in $2$ unpaired electrons $(n=2)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83 \ BM$.
Thus,the value is approximately $2.8 \ BM$.
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313
ChemistryMediumMCQMHT CET · 2021
Which of the following elements in their respective oxidation states does not develop spin only magnetic moment? $[Ti (Z=22), Zn(Z=30), V(Z=23), Cu(Z=29)]$
A
$Cu^{2+}$
B
$Zn^{2+}$
C
$Ti^{3+}$
D
$V^{3+}$
Solution
(B) The spin only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
For a species to have no spin only magnetic moment,it must have $n = 0$ (i.e.,all electrons are paired).
Let us analyze the electronic configurations:
$1. Cu^{2+} (Z=29): [Ar] 3d^9$. It has $1$ unpaired electron.
$2. Zn^{2+} (Z=30): [Ar] 3d^{10}$. It has $0$ unpaired electrons.
$3. Ti^{3+} (Z=22): [Ar] 3d^1$. It has $1$ unpaired electron.
$4. V^{3+} (Z=23): [Ar] 3d^2$. It has $2$ unpaired electrons.
Since $Zn^{2+}$ has $0$ unpaired electrons,its spin only magnetic moment is $0$.
For a species to have no spin only magnetic moment,it must have $n = 0$ (i.e.,all electrons are paired).
Let us analyze the electronic configurations:
$1. Cu^{2+} (Z=29): [Ar] 3d^9$. It has $1$ unpaired electron.
$2. Zn^{2+} (Z=30): [Ar] 3d^{10}$. It has $0$ unpaired electrons.
$3. Ti^{3+} (Z=22): [Ar] 3d^1$. It has $1$ unpaired electron.
$4. V^{3+} (Z=23): [Ar] 3d^2$. It has $2$ unpaired electrons.
Since $Zn^{2+}$ has $0$ unpaired electrons,its spin only magnetic moment is $0$.
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314
ChemistryEasyMCQMHT CET · 2021
Which of the following statements is correct for physisorption?
A
It involves the formation of covalent or ionic bonds.
B
It is favoured at high temperature.
C
It is reversible.
D
It is very specific.
Solution
(C) Physisorption:
$-$ It arises because of weak van der Waals' forces.
$-$ It is not specific in nature.
$-$ It is reversible in nature.
$-$ Low temperature is favourable for adsorption. It decreases with an increase in temperature.
$-$ It arises because of weak van der Waals' forces.
$-$ It is not specific in nature.
$-$ It is reversible in nature.
$-$ Low temperature is favourable for adsorption. It decreases with an increase in temperature.
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315
ChemistryEasyMCQMHT CET · 2021
Which among the following is true for chemisorption?
A
Heat of adsorption is in the range of $20-40 \ kJ \ mol^{-1}$
B
It is multimolecular layered
C
Van der Waals forces are involved
D
It is favored at high temperature up to a certain limit
Solution
(D) Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent.
- The heat of adsorption is high,typically in the range of $80-240 \ kJ \ mol^{-1}$.
- It is unimolecular in nature,as chemical bonds are specific.
- It involves chemical forces,not $Van \ der \ Waals$ forces.
- It is favored at high temperatures because it requires activation energy to form chemical bonds.
- The heat of adsorption is high,typically in the range of $80-240 \ kJ \ mol^{-1}$.
- It is unimolecular in nature,as chemical bonds are specific.
- It involves chemical forces,not $Van \ der \ Waals$ forces.
- It is favored at high temperatures because it requires activation energy to form chemical bonds.
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316
ChemistryEasyMCQMHT CET · 2021
Which of the following statements is true for adsorption?
A
It is accompanied by evolution of heat.
B
It is a bulk phenomenon.
C
It is independent of surface area.
D
It is independent of temperature.
Solution
(A) Adsorption is an exothermic process,meaning it is accompanied by the evolution of heat.
Adsorption is a surface phenomenon,not a bulk phenomenon.
Adsorption depends on temperature and the surface area of the adsorbent.
It increases with an increase in the surface area of the adsorbent.
Adsorption is a surface phenomenon,not a bulk phenomenon.
Adsorption depends on temperature and the surface area of the adsorbent.
It increases with an increase in the surface area of the adsorbent.
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317
ChemistryDifficultMCQMHT CET · 2021
What is the value of the intercept on the $y$-axis when $\log \frac{x}{m}$ is plotted against $\log P$ in the Freundlich isotherm?
A
$\frac{1}{n}$
B
$\log k$
C
$n$
D
$k$
Solution
(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k P^{\frac{1}{n}}$.
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log k$.
Therefore,the intercept on the $y$-axis is $\log k$.
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log k$.
Therefore,the intercept on the $y$-axis is $\log k$.
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318
ChemistryDifficultMCQMHT CET · 2021
Which of the following is the first step in the mechanism of heterogeneous catalysis?
A
Desorption of reaction product from the catalyst surface
B
Adsorption of reactant molecules on the catalyst surface to form an intermediate
C
Occurrence of chemical reaction on the catalyst surface to form an intermediate
D
Diffusion of reactants towards the catalyst surface
Solution
(D) The catalytic activity is localised on the surface of the catalyst. The mechanism involves five steps:
$(i)$ Diffusion of reactants to the surface of the catalyst.
$(ii)$ Adsorption of reactant molecules on the surface of the catalyst.
$(iii)$ Occurrence of chemical reaction on the catalyst surface,through the formation of an intermediate.
$(iv)$ Desorption of reaction products away from the catalyst surface.
$(v)$ Diffusion of reaction products away from the catalyst surface.
Therefore,the first step is the diffusion of reactants towards the catalyst surface.
$(i)$ Diffusion of reactants to the surface of the catalyst.
$(ii)$ Adsorption of reactant molecules on the surface of the catalyst.
$(iii)$ Occurrence of chemical reaction on the catalyst surface,through the formation of an intermediate.
$(iv)$ Desorption of reaction products away from the catalyst surface.
$(v)$ Diffusion of reaction products away from the catalyst surface.
Therefore,the first step is the diffusion of reactants towards the catalyst surface.
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319
ChemistryEasyMCQMHT CET · 2021
Which among the following is a multi-molecular colloid?
A
Nylon
B
Gold sol
C
Cellulose
D
Soap
Solution
(B) multi-molecular colloid is formed by the aggregation of a large number of atoms or smaller molecules with diameters less than $1 \ nm$. $Gold \ sol$ is a classic example of a multi-molecular colloid.
$Nylon$ and $Cellulose$ are examples of macromolecular colloids.
$Soap$ is an example of an associated colloid (micelle).
$Nylon$ and $Cellulose$ are examples of macromolecular colloids.
$Soap$ is an example of an associated colloid (micelle).
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320
ChemistryMediumMCQMHT CET · 2021
Air is an example of a solution of
A
gas in solid
B
liquid in gas
C
gas in liquid
D
gas in gas
Solution
(D) Air is a homogeneous mixture of various gases like $N_2$,$O_2$,$Ar$,$CO_2$,etc.
Since both the solute and the solvent are in the gaseous state,it is classified as a solution of $gas$ in $gas$.
Since both the solute and the solvent are in the gaseous state,it is classified as a solution of $gas$ in $gas$.
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321
ChemistryEasyMCQMHT CET · 2021
Identify the negatively charged sol from the following.
A
Sol of clay
B
$Fe_2O_3 \cdot xH_2O$
C
$TiO_2$ sol
D
Haemoglobin
Solution
(A) The charge on colloidal particles depends on the nature of the dispersed phase and the dispersion medium.
$1$. Sol of clay is a negatively charged sol.
$2$. $Fe_2O_3 \cdot xH_2O$ is a positively charged sol.
$3$. $TiO_2$ sol is a positively charged sol.
$4$. Haemoglobin is a positively charged sol.
Therefore,the correct option is $A$.
$1$. Sol of clay is a negatively charged sol.
$2$. $Fe_2O_3 \cdot xH_2O$ is a positively charged sol.
$3$. $TiO_2$ sol is a positively charged sol.
$4$. Haemoglobin is a positively charged sol.
Therefore,the correct option is $A$.
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322
ChemistryDifficultMCQMHT CET · 2021
Identify the correct decreasing order of precipitation power of the flocculating ion added from the following.
A
$Al^{3+} > Ba^{2+} > Na^{+}$
B
$Ba^{2+} > Al^{3+} > Na^{+}$
C
$Al^{3+} > Ba^{2+} > Na^{+}$
D
$Na^{+} > Ba^{2+} > Al^{3+}$
Solution
(A) According to the Hardy-Schulze rule,the precipitating power of an ion is directly proportional to its valency (charge).
Greater the valency of the flocculating ion,the greater will be its power to cause coagulation.
Comparing the charges: $Al^{3+}$ $(+3)$ > $Ba^{2+}$ $(+2)$ > $Na^{+}$ $(+1)$.
Therefore,the correct decreasing order is $Al^{3+} > Ba^{2+} > Na^{+}$.
Greater the valency of the flocculating ion,the greater will be its power to cause coagulation.
Comparing the charges: $Al^{3+}$ $(+3)$ > $Ba^{2+}$ $(+2)$ > $Na^{+}$ $(+1)$.
Therefore,the correct decreasing order is $Al^{3+} > Ba^{2+} > Na^{+}$.
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323
ChemistryMediumMCQMHT CET · 2021
Which among the following is $NOT$ an example of a one-dimensional nanostructure?
A
Nano shells
B
Nanowires
C
Nanotubes
D
Fibres
Solution
(A) one-dimensional nanostructure is defined as a structure where two dimensions are in the nanometer range and one dimension is larger.
Nanowires,nanotubes,and fibres are examples of one-dimensional nanostructures.
Nano shells are considered zero-dimensional nanostructures because all three dimensions are in the nanometer range.
Therefore,nano shells are not an example of a one-dimensional nanostructure.
Nanowires,nanotubes,and fibres are examples of one-dimensional nanostructures.
Nano shells are considered zero-dimensional nanostructures because all three dimensions are in the nanometer range.
Therefore,nano shells are not an example of a one-dimensional nanostructure.
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