ChemistryQ451–498 of 772 questions
Page 10 of 10 · English
451
ChemistryEasyMCQMHT CET · 2020
Which among the following is true for the value of Henry's law constant $K_H$?
A
It is greater for gases with higher solubilities.
B
It first increases and then decreases with an increase in temperature.
C
It increases with an increase in temperature.
D
It is the same for all gases.
Solution
(C) Henry's law is given by the relation $P = K_H \cdot x$,where $P$ is the partial pressure of the gas,$K_H$ is Henry's law constant,and $x$ is the mole fraction of the gas in the solution.
For a given partial pressure,the solubility of a gas $(x)$ is inversely proportional to $K_H$ $(x = P / K_H)$.
As the temperature increases,the kinetic energy of gas molecules increases,making it harder for them to remain dissolved in the solvent,thus decreasing solubility.
Since solubility decreases as temperature increases,the value of $K_H$ must increase with an increase in temperature for a given pressure.
For a given partial pressure,the solubility of a gas $(x)$ is inversely proportional to $K_H$ $(x = P / K_H)$.
As the temperature increases,the kinetic energy of gas molecules increases,making it harder for them to remain dissolved in the solvent,thus decreasing solubility.
Since solubility decreases as temperature increases,the value of $K_H$ must increase with an increase in temperature for a given pressure.
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452
ChemistryEasyMCQMHT CET · 2020
Henry's law is a relation between
A
pressure and solubility
B
temperature and pressure
C
volume and solubility
D
pressure and volume
Solution
(A) According to Henry's Law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution.
Mathematically,this is expressed as $S = K_{H} \times P$.
Here,$S$ represents the solubility of the gas,$P$ represents the partial pressure of the gas,and $K_{H}$ is the Henry's law constant.
Mathematically,this is expressed as $S = K_{H} \times P$.
Here,$S$ represents the solubility of the gas,$P$ represents the partial pressure of the gas,and $K_{H}$ is the Henry's law constant.
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453
ChemistryEasyMCQMHT CET · 2020
Which of the following pairs of solutions is isotonic? (Molar mass: urea $= 60 \ g \ mol^{-1}$,sucrose $= 342 \ g \ mol^{-1}$)
A
$3.0 \ g \ L^{-1}$ urea and $17.19 \ g \ L^{-1}$ sucrose
B
$0.3 \ g \ L^{-1}$ urea and $1.719 \ g \ L^{-1}$ sucrose
C
$3.0 \ g \ L^{-1}$ urea and $1.719 \ g \ L^{-1}$ sucrose
D
$0.3 \ g \ L^{-1}$ urea and $17.19 \ g \ L^{-1}$ sucrose
Solution
(A) For isotonic solutions,the molar concentrations must be equal: $C_{\text{urea}} = C_{\text{sucrose}}$.
First,calculate the molarity for each option.
For option $A$: Molarity of urea $= \frac{3.0 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = 0.05 \ mol \ L^{-1}$.
Molarity of sucrose $= \frac{17.19 \ g \ L^{-1}}{342 \ g \ mol^{-1}} = 0.05026 \ mol \ L^{-1} \approx 0.05 \ mol \ L^{-1}$.
Since the molar concentrations are equal,the solutions are isotonic.
First,calculate the molarity for each option.
For option $A$: Molarity of urea $= \frac{3.0 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = 0.05 \ mol \ L^{-1}$.
Molarity of sucrose $= \frac{17.19 \ g \ L^{-1}}{342 \ g \ mol^{-1}} = 0.05026 \ mol \ L^{-1} \approx 0.05 \ mol \ L^{-1}$.
Since the molar concentrations are equal,the solutions are isotonic.
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454
ChemistryEasyMCQMHT CET · 2020
According to Raoult's law,the relative lowering of vapour pressure for a solution containing a non-volatile solute is equal to:
A
mole fraction of solvent
B
moles of solute
C
moles of solvent
D
mole fraction of solute
Solution
(D) According to Raoult's law,for a solution containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P_1^o - P_1}{P_1^o} = x_2$.
Here,$\frac{P_1^o - P_1}{P_1^o}$ represents the relative lowering of vapour pressure,and $x_2$ is the mole fraction of the solute.
Therefore,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
Here,$\frac{P_1^o - P_1}{P_1^o}$ represents the relative lowering of vapour pressure,and $x_2$ is the mole fraction of the solute.
Therefore,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
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455
ChemistryMediumMCQMHT CET · 2020
$0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20 \%$ ionized. If $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$,the lowering in freezing point of the solution is: (in $K$)
A
$0.56$
B
$1.12$
C
$0.28$
D
$0.84$
Solution
(B) The dissociation of the weak acid is represented as: $HX \rightleftharpoons H^{+} + X^{-}$.
For the dissociation of an electrolyte,the van't Hoff factor $i$ is given by $i = 1 + \alpha(n-1)$,where $n$ is the number of ions produced per molecule.
Here,$n = 2$ and $\alpha = 0.20$ $(20 \%)$.
So,$i = 1 + 0.20(2-1) = 1.2$.
The lowering in freezing point is given by $\Delta T_{f} = i \times K_{f} \times m$.
Substituting the values: $\Delta T_{f} = 1.2 \times 1.86 \times 0.5 = 1.116 \ K \approx 1.12 \ K$.
For the dissociation of an electrolyte,the van't Hoff factor $i$ is given by $i = 1 + \alpha(n-1)$,where $n$ is the number of ions produced per molecule.
Here,$n = 2$ and $\alpha = 0.20$ $(20 \%)$.
So,$i = 1 + 0.20(2-1) = 1.2$.
The lowering in freezing point is given by $\Delta T_{f} = i \times K_{f} \times m$.
Substituting the values: $\Delta T_{f} = 1.2 \times 1.86 \times 0.5 = 1.116 \ K \approx 1.12 \ K$.
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456
ChemistryEasyMCQMHT CET · 2020
The osmotic pressure of a $1 \ M$ solution at $27^{\circ}C$ is $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$: (in $atm$)
A
$2.46$
B
$12.1$
C
$24.6$
D
$1.21$
Solution
(C) The formula for osmotic pressure is $\pi = MRT$,where $M$ is the molarity,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
Given: $M = 1 \ M$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 27 + 273 = 300 \ K$.
Substituting the values: $\pi = 1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 24.6 \ atm$.
Given: $M = 1 \ M$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 27 + 273 = 300 \ K$.
Substituting the values: $\pi = 1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 24.6 \ atm$.
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457
ChemistryEasyMCQMHT CET · 2020
What is the value of $K_{f}$ if $30 \ g$ urea (molar mass $60$) dissolved in $0.5 \ dm^{3}$ of water decreases the freezing point by $0.15^{\circ}C$?
A
$0.15 \ K \ kg \ mol^{-1}$
B
$0.030 \ K \ kg \ mol^{-1}$
C
$0.30 \ K \ kg \ mol^{-1}$
D
$0.015 \ K \ kg \ mol^{-1}$
Solution
(A) Mass of water $= 0.5 \ dm^{3} = 0.5 \ kg$
Moles of urea $= \frac{30 \ g}{60 \ g \ mol^{-1}} = 0.5 \ mol$
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.5 \ mol}{0.5 \ kg} = 1 \ mol \ kg^{-1}$
The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$
Given $\Delta T_{f} = 0.15 \ K$
$K_{f} = \frac{\Delta T_{f}}{m} = \frac{0.15 \ K}{1 \ mol \ kg^{-1}} = 0.15 \ K \ kg \ mol^{-1}$
Moles of urea $= \frac{30 \ g}{60 \ g \ mol^{-1}} = 0.5 \ mol$
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.5 \ mol}{0.5 \ kg} = 1 \ mol \ kg^{-1}$
The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$
Given $\Delta T_{f} = 0.15 \ K$
$K_{f} = \frac{\Delta T_{f}}{m} = \frac{0.15 \ K}{1 \ mol \ kg^{-1}} = 0.15 \ K \ kg \ mol^{-1}$
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458
ChemistryDifficultMCQMHT CET · 2020
Vapour pressure of solvent $A$ is $0.90 \text{ atm}$. When a non-volatile solute is added,the vapour pressure drops to $0.60 \text{ atm}$. What is the mole fraction of $A$ in the solution?
A
$0.3$
B
$0.333$
C
$0.5$
D
$0.667$
Solution
(D) According to Raoult's law,$P_s = P^o_A \cdot X_A$,where $P_s$ is the vapour pressure of the solution,$P^o_A$ is the vapour pressure of the pure solvent,and $X_A$ is the mole fraction of the solvent.
Given: $P^o_A = 0.90 \text{ atm}$ and $P_s = 0.60 \text{ atm}$.
$X_A = \frac{P_s}{P^o_A} = \frac{0.60}{0.90} = \frac{2}{3} \approx 0.667$.
Given: $P^o_A = 0.90 \text{ atm}$ and $P_s = 0.60 \text{ atm}$.
$X_A = \frac{P_s}{P^o_A} = \frac{0.60}{0.90} = \frac{2}{3} \approx 0.667$.
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459
ChemistryDifficultMCQMHT CET · 2020
Which of the following $0.1 \ m$ aqueous solutions exhibits the highest osmotic pressure at $25^{\circ}C$?
A
Urea
B
$CoCl_{2}$
C
$KCl$
D
Glucose
Solution
(B) The osmotic pressure $(\pi)$ is a colligative property given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For a given concentration and temperature,$\pi$ is directly proportional to the van't Hoff factor $(i)$.
$1$. Urea: $i = 1$ (non-electrolyte)
$2$. $CoCl_{2}$: $i = 3$ $(Co^{2+} + 2Cl^{-})$
$3$. $KCl$: $i = 2$ $(K^{+} + Cl^{-})$
$4$. Glucose: $i = 1$ (non-electrolyte)
Since $CoCl_{2}$ has the highest van't Hoff factor $(i = 3)$,it will exhibit the highest osmotic pressure.
For a given concentration and temperature,$\pi$ is directly proportional to the van't Hoff factor $(i)$.
$1$. Urea: $i = 1$ (non-electrolyte)
$2$. $CoCl_{2}$: $i = 3$ $(Co^{2+} + 2Cl^{-})$
$3$. $KCl$: $i = 2$ $(K^{+} + Cl^{-})$
$4$. Glucose: $i = 1$ (non-electrolyte)
Since $CoCl_{2}$ has the highest van't Hoff factor $(i = 3)$,it will exhibit the highest osmotic pressure.
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460
ChemistryEasyMCQMHT CET · 2020
Identify the correct relation between depression in freezing point and the freezing point of a pure solvent.
A
$T^{\circ} = T \times \Delta T_{f}$
B
$T^{\circ} = \Delta T_{f} - T$
C
$T^{\circ} = T - \Delta T_{f}$
D
$T^{\circ} = \Delta T_{f} + T$
Solution
(D) The depression in freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.
$\Delta T_{f} = T^{\circ} - T$
Rearranging the equation to solve for the freezing point of the pure solvent $(T^{\circ})$:
$T^{\circ} = \Delta T_{f} + T$
Where:
$T^{\circ} = \text{freezing point of pure solvent}$
$T = \text{freezing point of solution}$
$\Delta T_{f} = \text{depression in freezing point}$
$\Delta T_{f} = T^{\circ} - T$
Rearranging the equation to solve for the freezing point of the pure solvent $(T^{\circ})$:
$T^{\circ} = \Delta T_{f} + T$
Where:
$T^{\circ} = \text{freezing point of pure solvent}$
$T = \text{freezing point of solution}$
$\Delta T_{f} = \text{depression in freezing point}$
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461
ChemistryEasyMCQMHT CET · 2020
Which of the following is $NOT$ used as a semipermeable membrane?
A
Cellulose nitrate
B
Copper ferrocyanide
C
Ammonium chloride
D
Cellulose
Solution
(C) semipermeable membrane allows the passage of solvent molecules but blocks solute particles.
$Cellulose \ nitrate$,$Copper \ ferrocyanide$,and $Cellulose$ are commonly used as semipermeable membranes.
$Ammonium \ chloride$ $(NH_4Cl)$ is a salt and does not possess the structural properties required to act as a semipermeable membrane.
$Cellulose \ nitrate$,$Copper \ ferrocyanide$,and $Cellulose$ are commonly used as semipermeable membranes.
$Ammonium \ chloride$ $(NH_4Cl)$ is a salt and does not possess the structural properties required to act as a semipermeable membrane.
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462
ChemistryEasyMCQMHT CET · 2020
$A$ solution has an osmotic pressure of $x \ kPa$ at $300 \ K$ having $1 \ mole$ of solute in $10.5 \ m^{3}$ of solution. If its osmotic pressure is reduced to $(\frac{1}{10})$th of its initial value,what is the new volume of solution (in $m^{3}$)?
A
$30$
B
$105$
C
$110$
D
$11.0$
Solution
(B) The osmotic pressure formula is given by $\pi = \frac{n}{V} RT$.
For the initial state: $\pi = x = \frac{1}{10.5} RT$.
For the final state: $\pi' = \frac{x}{10} = \frac{1}{V'} RT$.
Dividing the two equations: $\frac{x}{x/10} = \frac{\frac{1}{10.5} RT}{\frac{1}{V'} RT}$.
This simplifies to $10 = \frac{V'}{10.5}$.
Therefore,$V' = 10 \times 10.5 = 105 \ m^{3}$.
For the initial state: $\pi = x = \frac{1}{10.5} RT$.
For the final state: $\pi' = \frac{x}{10} = \frac{1}{V'} RT$.
Dividing the two equations: $\frac{x}{x/10} = \frac{\frac{1}{10.5} RT}{\frac{1}{V'} RT}$.
This simplifies to $10 = \frac{V'}{10.5}$.
Therefore,$V' = 10 \times 10.5 = 105 \ m^{3}$.
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463
ChemistryEasyMCQMHT CET · 2020
If a centimolal aqueous solution of $K_{3}[Fe(CN)_{6}]$ has a degree of dissociation of $0.78$,what is the value of the van't Hoff factor?
A
$3.34$
B
$1.2$
C
$2.5$
D
$4$
Solution
(A) The dissociation of $K_{3}[Fe(CN)_{6}]$ is given by:
$K_{3}[Fe(CN)_{6}] \longrightarrow 3 K^{+} + [Fe(CN)_{6}]^{3-}$
Here,the number of ions produced per formula unit is $n = 3 + 1 = 4$.
The van't Hoff factor $(i)$ is calculated using the formula:
$i = 1 + \alpha(n - 1)$
Given the degree of dissociation $\alpha = 0.78$ and $n = 4$:
$i = 1 + 0.78(4 - 1) = 1 + 0.78(3) = 1 + 2.34 = 3.34$
$K_{3}[Fe(CN)_{6}] \longrightarrow 3 K^{+} + [Fe(CN)_{6}]^{3-}$
Here,the number of ions produced per formula unit is $n = 3 + 1 = 4$.
The van't Hoff factor $(i)$ is calculated using the formula:
$i = 1 + \alpha(n - 1)$
Given the degree of dissociation $\alpha = 0.78$ and $n = 4$:
$i = 1 + 0.78(4 - 1) = 1 + 0.78(3) = 1 + 2.34 = 3.34$
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464
ChemistryMediumMCQMHT CET · 2020
Solutions $A, B, C$ and $D$ are respectively $0.2 \ M$ urea,$0.10 \ M$ $NaCl$,$0.05 \ M$ $BaCl_{2}$ and $0.05 \ M$ $AlCl_{3}$. All solutions are isotonic with each other except:
A
$B$
B
$A$
C
$D$
D
$C$
Solution
(D) The osmotic pressure is given by the formula $\pi = iMRT$,where $i$ is the van't Hoff factor,$M$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.
For isotonic solutions,the osmotic pressure $\pi$ must be equal.
$(A)$ For urea: $i = 1$,$\pi = 1 \times 0.2 \ RT = 0.2 \ RT$
$(B)$ For $NaCl$: $i = 2$,$\pi = 2 \times 0.1 \ RT = 0.2 \ RT$
$(C)$ For $BaCl_{2}$: $i = 3$,$\pi = 3 \times 0.05 \ RT = 0.15 \ RT$
$(D)$ For $AlCl_{3}$: $i = 4$,$\pi = 4 \times 0.05 \ RT = 0.2 \ RT$
Since the osmotic pressure of solution $C$ is $0.15 \ RT$ while others are $0.2 \ RT$,solution $C$ is not isotonic with the others.
For isotonic solutions,the osmotic pressure $\pi$ must be equal.
$(A)$ For urea: $i = 1$,$\pi = 1 \times 0.2 \ RT = 0.2 \ RT$
$(B)$ For $NaCl$: $i = 2$,$\pi = 2 \times 0.1 \ RT = 0.2 \ RT$
$(C)$ For $BaCl_{2}$: $i = 3$,$\pi = 3 \times 0.05 \ RT = 0.15 \ RT$
$(D)$ For $AlCl_{3}$: $i = 4$,$\pi = 4 \times 0.05 \ RT = 0.2 \ RT$
Since the osmotic pressure of solution $C$ is $0.15 \ RT$ while others are $0.2 \ RT$,solution $C$ is not isotonic with the others.
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465
ChemistryEasyMCQMHT CET · 2020
What is the osmotic pressure of a semi-molar solution at $27^{\circ} C$ (in $atm$)?
$(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
$(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
A
$4.96$
B
$2.46$
C
$20.5$
D
$12.3$
Solution
(D) For a semi-molar solution,the molarity $(M)$ is $0.5 \ M$.
Temperature $(T) = 27^{\circ} C = 27 + 273 = 300 \ K$.
The formula for osmotic pressure $(\pi)$ is $\pi = MRT$.
Substituting the values: $\pi = 0.5 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 12.3 \ atm$.
Temperature $(T) = 27^{\circ} C = 27 + 273 = 300 \ K$.
The formula for osmotic pressure $(\pi)$ is $\pi = MRT$.
Substituting the values: $\pi = 0.5 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 12.3 \ atm$.
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466
ChemistryEasyMCQMHT CET · 2020
Relative lowering in vapour pressure of a solution containing a non-volatile solute is the ratio of:
A
Number of moles of solute to number of moles of solvent.
B
Number of moles of solvent to total number of moles of solution.
C
Number of moles of solvent to number of moles of solute.
D
Number of moles of solute to total number of moles of solution.
Solution
(D) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P_0 - P}{P_0} = x_2$.
Here,$x_2$ is the mole fraction of the solute.
The mole fraction of the solute is defined as the ratio of the number of moles of solute $(n_2)$ to the total number of moles of the solution $(n_1 + n_2)$.
Therefore,$\frac{P_0 - P}{P_0} = \frac{n_2}{n_1 + n_2}$.
Here,$x_2$ is the mole fraction of the solute.
The mole fraction of the solute is defined as the ratio of the number of moles of solute $(n_2)$ to the total number of moles of the solution $(n_1 + n_2)$.
Therefore,$\frac{P_0 - P}{P_0} = \frac{n_2}{n_1 + n_2}$.
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467
ChemistryEasyMCQMHT CET · 2020
The vapour pressure of a solvent decreases by $10 \ mm \ Hg$ if the mole fraction of a non-volatile solute is $0.2$. Calculate the vapour pressure of the pure solvent. (in $mm \ Hg$)
A
$50$
B
$70$
C
$40$
D
$60$
Solution
(A) Given: The decrease in vapour pressure $\Delta P = P_{0} - P = 10 \ mm \ Hg$.
The mole fraction of the solute $x_{2} = 0.2$.
According to Raoult's Law for non-volatile solutes:
$\frac{P_{0} - P}{P_{0}} = x_{2}$
Substituting the values:
$\frac{10 \ mm \ Hg}{P_{0}} = 0.2$
$P_{0} = \frac{10}{0.2} \ mm \ Hg = 50 \ mm \ Hg$.
Therefore,the vapour pressure of the pure solvent is $50 \ mm \ Hg$.
The mole fraction of the solute $x_{2} = 0.2$.
According to Raoult's Law for non-volatile solutes:
$\frac{P_{0} - P}{P_{0}} = x_{2}$
Substituting the values:
$\frac{10 \ mm \ Hg}{P_{0}} = 0.2$
$P_{0} = \frac{10}{0.2} \ mm \ Hg = 50 \ mm \ Hg$.
Therefore,the vapour pressure of the pure solvent is $50 \ mm \ Hg$.
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468
ChemistryDifficultMCQMHT CET · 2020
Which of the following $0.10 \ m$ aqueous solutions will have the maximum $\Delta T_{f}$ value?
A
$Al_{2}(SO_{4})_{3}$
B
$KI$
C
$C_{12}H_{22}O_{11}$
D
$NH_{2}CONH_{2}$
Solution
(A) The depression in freezing point is given by the formula $\Delta T_{f} = i \times K_{f} \times m$.
Since $K_{f}$ and $m$ $(0.10 \ m)$ are constant for all solutions,$\Delta T_{f}$ depends directly on the van't Hoff factor $(i)$.
$i$ represents the number of particles produced upon dissociation.
For $Al_{2}(SO_{4})_{3}$,$i = 5$ $(2Al^{3+} + 3SO_{4}^{2-})$.
For $KI$,$i = 2$ $(K^{+} + I^{-})$.
For $C_{12}H_{22}O_{11}$ (sucrose) and $NH_{2}CONH_{2}$ (urea),$i = 1$ as they are non-electrolytes.
Since $Al_{2}(SO_{4})_{3}$ has the highest $i$ value,it will have the maximum $\Delta T_{f}$ value.
Since $K_{f}$ and $m$ $(0.10 \ m)$ are constant for all solutions,$\Delta T_{f}$ depends directly on the van't Hoff factor $(i)$.
$i$ represents the number of particles produced upon dissociation.
For $Al_{2}(SO_{4})_{3}$,$i = 5$ $(2Al^{3+} + 3SO_{4}^{2-})$.
For $KI$,$i = 2$ $(K^{+} + I^{-})$.
For $C_{12}H_{22}O_{11}$ (sucrose) and $NH_{2}CONH_{2}$ (urea),$i = 1$ as they are non-electrolytes.
Since $Al_{2}(SO_{4})_{3}$ has the highest $i$ value,it will have the maximum $\Delta T_{f}$ value.
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469
ChemistryMediumMCQMHT CET · 2020
If the boiling point of a urea solution is $100.18^{\circ}C$ and $K_b$ for water is $0.512 \ K \ kg \ mol^{-1}$,what is the molality of the solution? (Boiling point of water $= 100^{\circ}C$)
A
$0.25 \ mol \ kg^{-1}$
B
$0.6 \ mol \ kg^{-1}$
C
$0.45 \ mol \ kg^{-1}$
D
$0.35 \ mol \ kg^{-1}$
Solution
(D) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^{\circ} = 100.18^{\circ}C - 100^{\circ}C = 0.18 \ K$.
Given $K_b = 0.512 \ K \ kg \ mol^{-1}$.
Using the formula $\Delta T_b = K_b \times m$,where $m$ is the molality.
$m = \frac{\Delta T_b}{K_b} = \frac{0.18}{0.512} \approx 0.35 \ mol \ kg^{-1}$.
Given $K_b = 0.512 \ K \ kg \ mol^{-1}$.
Using the formula $\Delta T_b = K_b \times m$,where $m$ is the molality.
$m = \frac{\Delta T_b}{K_b} = \frac{0.18}{0.512} \approx 0.35 \ mol \ kg^{-1}$.
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470
ChemistryEasyMCQMHT CET · 2020
Which of the following formulas correctly gives the value of the ebullioscopic constant $(K_{b})$?
A
$\frac{W_{1} \times 1000}{\Delta T_{b} \times W_{2} \times M_{2}}$
B
$\frac{W_{2} \times 1000}{\Delta T_{b} \times W_{1} \times M_{2}}$
C
$\frac{M_{2} \times \Delta T_{b} \times W_{2}}{W_{1}}$
D
$\frac{\Delta T_{b} \times M_{2} \times W_{1}}{W_{2}}$
Solution
(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = K_{b} \times m$.
Here,$m$ is the molality,which is defined as $m = \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$.
Substituting this into the equation: $\Delta T_{b} = K_{b} \times \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$.
Rearranging for $K_{b}$: $K_{b} = \frac{\Delta T_{b} \times M_{2} \times W_{1}}{W_{2} \times 1000}$.
Note: If the molality is defined as moles per kg of solvent,the $1000$ factor is omitted. Given the options provided,option $D$ represents the relationship where $W_{1}$ is in kg.
Here,$m$ is the molality,which is defined as $m = \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$.
Substituting this into the equation: $\Delta T_{b} = K_{b} \times \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$.
Rearranging for $K_{b}$: $K_{b} = \frac{\Delta T_{b} \times M_{2} \times W_{1}}{W_{2} \times 1000}$.
Note: If the molality is defined as moles per kg of solvent,the $1000$ factor is omitted. Given the options provided,option $D$ represents the relationship where $W_{1}$ is in kg.
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471
ChemistryDifficultMCQMHT CET · 2020
What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)
A
$135.1 \ g \ mol^{-1}$
B
$226.1 \ g \ mol^{-1}$
C
$328.4 \ g \ mol^{-1}$
D
$117.0 \ g \ mol^{-1}$
Solution
(D) According to Raoult's law for non-volatile solutes: $\frac{P^{0} - P}{P^{0}} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}} = \frac{W_{2} \times M_{1}}{M_{2} \times W_{1}}$
Given: $P^{0} = 450 \ mm \ Hg$,$P = 400 \ mm \ Hg$,$W_{2} = 1.5 \ g$,$W_{1} = 30 \ g$,$M_{1} (C_{6}H_{6}) = (6 \times 12) + (6 \times 1) = 78 \ g \ mol^{-1}$.
Rearranging the formula for $M_{2}$:
$M_{2} = \frac{W_{2} \times M_{1} \times P^{0}}{W_{1} \times (P^{0} - P)}$
$M_{2} = \frac{1.5 \times 78 \times 450}{30 \times (450 - 400)}$
$M_{2} = \frac{1.5 \times 78 \times 450}{30 \times 50} = \frac{1.5 \times 78 \times 9}{30} = \frac{1053}{9} = 117.0 \ g \ mol^{-1}$.
Given: $P^{0} = 450 \ mm \ Hg$,$P = 400 \ mm \ Hg$,$W_{2} = 1.5 \ g$,$W_{1} = 30 \ g$,$M_{1} (C_{6}H_{6}) = (6 \times 12) + (6 \times 1) = 78 \ g \ mol^{-1}$.
Rearranging the formula for $M_{2}$:
$M_{2} = \frac{W_{2} \times M_{1} \times P^{0}}{W_{1} \times (P^{0} - P)}$
$M_{2} = \frac{1.5 \times 78 \times 450}{30 \times (450 - 400)}$
$M_{2} = \frac{1.5 \times 78 \times 450}{30 \times 50} = \frac{1.5 \times 78 \times 9}{30} = \frac{1053}{9} = 117.0 \ g \ mol^{-1}$.
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472
ChemistryDifficultMCQMHT CET · 2020
If the van't Hoff factor of mono-$1$-fluoroacetic acid in water is $1.076$,what is its degree of dissociation?
A
$0.76$
B
$0.076$
C
$0.924$
D
$0.538$
Solution
(B) Given: van't Hoff factor $i = 1.076$.
For the dissociation of mono-$1$-fluoroacetic acid: $CH_{2}FCOOH \rightleftharpoons CH_{2}FCOO^{-} + H^{+}$.
The number of ions produced per molecule is $n = 2$.
The formula for the degree of dissociation $\alpha$ is $\alpha = \frac{i-1}{n-1}$.
Substituting the values: $\alpha = \frac{1.076-1}{2-1} = \frac{0.076}{1} = 0.076$.
For the dissociation of mono-$1$-fluoroacetic acid: $CH_{2}FCOOH \rightleftharpoons CH_{2}FCOO^{-} + H^{+}$.
The number of ions produced per molecule is $n = 2$.
The formula for the degree of dissociation $\alpha$ is $\alpha = \frac{i-1}{n-1}$.
Substituting the values: $\alpha = \frac{1.076-1}{2-1} = \frac{0.076}{1} = 0.076$.
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473
ChemistryEasyMCQMHT CET · 2020
Molal elevation constant is the elevation in boiling point produced by
A
$1$ mole of solute in one litre of solvent
B
$1 \ g$ of solute in $100 \ g$ of solvent
C
$100 \ g$ of solute in $1000 \ g$ of solvent
D
$1$ mole of solute in one $kg$ of solvent
Solution
(D) The molal elevation constant $(K_b)$ is defined as the elevation in boiling point when $1 \ mole$ of a non-volatile solute is dissolved in $1 \ kg$ $(1000 \ g)$ of solvent.
This is based on the formula $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
When $m = 1 \ mol/kg$,then $\Delta T_b = K_b$.
This is based on the formula $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
When $m = 1 \ mol/kg$,then $\Delta T_b = K_b$.
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474
ChemistryEasyMCQMHT CET · 2020
Which of the following factors affect the molarity of a solution?
A
Nature of solute dissolved
B
Temperature
C
Mass of solvent
D
Molar mass of solvent
Solution
(B) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
$M = \frac{n_{solute}}{V_{solution(L)}}$
Since the volume of a solution $(V)$ is dependent on temperature (due to thermal expansion or contraction),a change in temperature will change the volume of the solution.
Therefore,molarity is affected by a change in temperature.
$M = \frac{n_{solute}}{V_{solution(L)}}$
Since the volume of a solution $(V)$ is dependent on temperature (due to thermal expansion or contraction),a change in temperature will change the volume of the solution.
Therefore,molarity is affected by a change in temperature.
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475
ChemistryMediumMCQMHT CET · 2020
$3.42 \times 10^{-2} \ kg$ of sugar (molar mass $342 \ g \ mol^{-1}$) is dissolved in water to produce $234.2 \ g$ of sugar syrup. What is the molality of the sugar syrup?
A
$0.50 \ mol \ kg^{-1}$
B
$0.76 \ mol \ kg^{-1}$
C
$0.67 \ mol \ kg^{-1}$
D
$0.85 \ mol \ kg^{-1}$
Solution
(A) Mass of sugar $= 3.42 \times 10^{-2} \ kg = 34.2 \ g$.
Mass of sugar syrup $= 234.2 \ g$.
Mass of solvent (water) $= 234.2 \ g - 34.2 \ g = 200 \ g = 0.2 \ kg$.
Moles of sugar $= \frac{\text{Mass}}{\text{Molar mass}} = \frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$.
Mass of sugar syrup $= 234.2 \ g$.
Mass of solvent (water) $= 234.2 \ g - 34.2 \ g = 200 \ g = 0.2 \ kg$.
Moles of sugar $= \frac{\text{Mass}}{\text{Molar mass}} = \frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$.
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476
ChemistryEasyMCQMHT CET · 2020
Mole fraction of solute in its $2$ molal aqueous solution is
A
$0.028775$
B
$0.0347$
C
$0.054$
D
$0.018$
Solution
(B) $2$ molal aqueous solution means $2$ moles of solute are present in $1 \ kg$ $(1000 \ g)$ of water.
Moles of solute $(n_{solute})$ $= 2 \ mol$.
Moles of solvent (water,$n_{solvent}$) $= \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Total moles $= n_{solute} + n_{solvent} = 2 + 55.56 = 57.56 \ mol$.
Mole fraction of solute $(x_{solute})$ $= \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{2}{57.56} \approx 0.0347$.
Moles of solute $(n_{solute})$ $= 2 \ mol$.
Moles of solvent (water,$n_{solvent}$) $= \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Total moles $= n_{solute} + n_{solvent} = 2 + 55.56 = 57.56 \ mol$.
Mole fraction of solute $(x_{solute})$ $= \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{2}{57.56} \approx 0.0347$.
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477
ChemistryEasyMCQMHT CET · 2020
$A$ solution is $0.25 \%$ by mass. What is the weight of solvent containing $1.25 \ g$ solute (in $g$)?
A
$498.75$
B
$300$
C
$600$
D
$200$
Solution
(A) Mass percentage of solute $= 0.25 \%$. This means $0.25 \ g$ of solute is present in $100 \ g$ of solution.
Mass of solvent $= \text{Mass of solution} - \text{Mass of solute} = 100 \ g - 0.25 \ g = 99.75 \ g$.
Thus,$0.25 \ g$ of solute is present in $99.75 \ g$ of solvent.
For $1.25 \ g$ of solute,the mass of solvent required is $\frac{99.75 \ g \text{ solvent}}{0.25 \ g \text{ solute}} \times 1.25 \ g \text{ solute} = 498.75 \ g$.
Mass of solvent $= \text{Mass of solution} - \text{Mass of solute} = 100 \ g - 0.25 \ g = 99.75 \ g$.
Thus,$0.25 \ g$ of solute is present in $99.75 \ g$ of solvent.
For $1.25 \ g$ of solute,the mass of solvent required is $\frac{99.75 \ g \text{ solvent}}{0.25 \ g \text{ solute}} \times 1.25 \ g \text{ solute} = 498.75 \ g$.
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478
ChemistryMediumMCQMHT CET · 2020
What is the molality of a solution containing $300 \ mg$ of urea (molar mass $60 \ g/mol$) dissolved in $30 \ g$ of water (in $m$)?
A
$0.133$
B
$0.825$
C
$0.498$
D
$0.166$
Solution
(D) Step $1$: Calculate the number of moles of urea. Given mass $= 300 \ mg = 0.3 \ g$. Molar mass $= 60 \ g/mol$. Moles $(n) = \frac{0.3 \ g}{60 \ g/mol} = 0.005 \ mol$.
Step $2$: Calculate the mass of the solvent in $kg$. Mass of water $= 30 \ g = 0.03 \ kg$.
Step $3$: Calculate molality $(m)$. Molality $= \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{0.005 \ mol}{0.03 \ kg} = 0.166 \ m$.
Step $2$: Calculate the mass of the solvent in $kg$. Mass of water $= 30 \ g = 0.03 \ kg$.
Step $3$: Calculate molality $(m)$. Molality $= \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{0.005 \ mol}{0.03 \ kg} = 0.166 \ m$.
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479
ChemistryEasyMCQMHT CET · 2020
What is the molarity of a solution containing $3.2 \ g$ of $NaOH$ (molar mass $40 \ g \ mol^{-1}$) in $250 \ cm^{3}$ of water?
A
$0.512 \ mol \ dm^{-3}$
B
$0.32 \ mol \ dm^{-3}$
C
$0.032 \ mol \ dm^{-3}$
D
$0.02 \ mol \ dm^{-3}$
Solution
(B) Mass of $NaOH$ $(m) = 3.2 \ g$
Volume of solution $(V) = 250 \ cm^{3} = 0.25 \ L$
Molar mass of $NaOH$ $(M) = 40 \ g \ mol^{-1}$
Number of moles $(n) = \frac{m}{M} = \frac{3.2 \ g}{40 \ g \ mol^{-1}} = 0.08 \ mol$
Molarity $(M) = \frac{n}{V(L)} = \frac{0.08 \ mol}{0.25 \ L} = 0.32 \ mol \ dm^{-3}$
Volume of solution $(V) = 250 \ cm^{3} = 0.25 \ L$
Molar mass of $NaOH$ $(M) = 40 \ g \ mol^{-1}$
Number of moles $(n) = \frac{m}{M} = \frac{3.2 \ g}{40 \ g \ mol^{-1}} = 0.08 \ mol$
Molarity $(M) = \frac{n}{V(L)} = \frac{0.08 \ mol}{0.25 \ L} = 0.32 \ mol \ dm^{-3}$
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480
ChemistryMediumMCQMHT CET · 2020
What is the molarity of a solution containing $0.8 \ g$ of $NaOH$ (molar mass $40 \ g \ mol^{-1}$) in $150 \ cm^{3}$ of water?
A
$0.02 \ mol \ dm^{-3}$
B
$0.12 \ mol \ dm^{-3}$
C
$5.33 \ mol \ dm^{-3}$
D
$0.1333 \ mol \ dm^{-3}$
Solution
(D) Step $1$: Calculate the number of moles of $NaOH$ $(n)$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.8 \ g}{40 \ g \ mol^{-1}} = 0.02 \ mol$
Step $2$: Convert the volume of the solution from $cm^{3}$ to $L$:
$V = 150 \ cm^{3} = 0.15 \ L$
Step $3$: Calculate the molarity $(M)$:
$M = \frac{n}{V(L)} = \frac{0.02 \ mol}{0.15 \ L} = 0.1333 \ mol \ dm^{-3}$
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.8 \ g}{40 \ g \ mol^{-1}} = 0.02 \ mol$
Step $2$: Convert the volume of the solution from $cm^{3}$ to $L$:
$V = 150 \ cm^{3} = 0.15 \ L$
Step $3$: Calculate the molarity $(M)$:
$M = \frac{n}{V(L)} = \frac{0.02 \ mol}{0.15 \ L} = 0.1333 \ mol \ dm^{-3}$
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481
ChemistryMediumMCQMHT CET · 2020
$34.2 \ g$ of sugar is dissolved in $1.8 \times 10^{2} \ g$ of water to form sugar syrup. Calculate the mole fraction of sugar. (Molar mass of sugar $= 342 \ g/mol$,water $= 18 \ g/mol$)
A
$0.0099$
B
$0.001$
C
$0.1$
D
$0.9$
Solution
(A) Given: Mass of sugar $= 34.2 \ g$,Molar mass of sugar $= 342 \ g/mol$.
Number of moles of sugar $(n_{sugar}) = \frac{34.2}{342} = 0.1 \ mol$.
Mass of water $= 1.8 \times 10^{2} \ g = 180 \ g$,Molar mass of water $= 18 \ g/mol$.
Number of moles of water $(n_{water}) = \frac{180}{18} = 10 \ mol$.
Mole fraction of sugar $(x_{sugar}) = \frac{n_{sugar}}{n_{sugar} + n_{water}} = \frac{0.1}{0.1 + 10} = \frac{0.1}{10.1} \approx 0.0099$.
Number of moles of sugar $(n_{sugar}) = \frac{34.2}{342} = 0.1 \ mol$.
Mass of water $= 1.8 \times 10^{2} \ g = 180 \ g$,Molar mass of water $= 18 \ g/mol$.
Number of moles of water $(n_{water}) = \frac{180}{18} = 10 \ mol$.
Mole fraction of sugar $(x_{sugar}) = \frac{n_{sugar}}{n_{sugar} + n_{water}} = \frac{0.1}{0.1 + 10} = \frac{0.1}{10.1} \approx 0.0099$.
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482
ChemistryEasyMCQMHT CET · 2020
Freundlich's equation for adsorption of gas on solid is represented as
A
$m/x = k P^n$
B
$x/m = k P^n$
C
$m/x = k P^{1/n}$
D
$x/m = k P^{1/n}$
Solution
(D) The Freundlich adsorption isotherm is an empirical relationship between the amount of gas adsorbed per unit mass of solid adsorbent $(x/m)$ and the equilibrium pressure $(P)$ of the gas.
The mathematical expression is given by: $x/m = k P^{1/n}$,where $k$ and $n$ are constants that depend on the nature of the adsorbent and the gas at a particular temperature $(n > 1)$.
The mathematical expression is given by: $x/m = k P^{1/n}$,where $k$ and $n$ are constants that depend on the nature of the adsorbent and the gas at a particular temperature $(n > 1)$.
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483
ChemistryMediumMCQMHT CET · 2020
Which of the following is a characteristic of a catalyst?
A
It changes the position of equilibrium.
B
It increases the rates of both forward and backward reactions equally in a reversible reaction.
C
It affects the energies of reactants and products of the reaction.
D
It increases the activation energy of reactants.
Solution
(B) catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
In a reversible reaction,a catalyst increases the rates of both the forward and backward reactions to the same extent.
This allows the system to reach the state of equilibrium more quickly.
However,a catalyst does not alter the position of the equilibrium,nor does it change the thermodynamic properties like the energies of reactants and products.
It works by providing an alternative reaction pathway with a lower activation energy.
In a reversible reaction,a catalyst increases the rates of both the forward and backward reactions to the same extent.
This allows the system to reach the state of equilibrium more quickly.
However,a catalyst does not alter the position of the equilibrium,nor does it change the thermodynamic properties like the energies of reactants and products.
It works by providing an alternative reaction pathway with a lower activation energy.
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484
ChemistryEasyMCQMHT CET · 2020
Identify the enzyme that catalyses the reaction of $CO_{2}$ with water in the human body.
A
Ferroxidase
B
Catalase
C
Nitrogenase
D
Carbonic anhydrase
Solution
(D) The correct answer is Carbonic anhydrase.
In the human body,the enzyme carbonic anhydrase catalyses the hydration of $CO_{2}$.
It facilitates the reaction that converts $CO_{2}$ and water into carbonic acid $(H_{2}CO_{3})$,which subsequently dissociates into protons $(H^{+})$ and bicarbonate ions $(HCO_{3}^{-})$.
This enzyme is primarily found in red blood cells and is essential for the transport of $CO_{2}$ from tissues to the lungs.
The reaction is: $CO_{2} + H_{2}O \xrightarrow{\text{Carbonic anhydrase}} H_{2}CO_{3}$
In the human body,the enzyme carbonic anhydrase catalyses the hydration of $CO_{2}$.
It facilitates the reaction that converts $CO_{2}$ and water into carbonic acid $(H_{2}CO_{3})$,which subsequently dissociates into protons $(H^{+})$ and bicarbonate ions $(HCO_{3}^{-})$.
This enzyme is primarily found in red blood cells and is essential for the transport of $CO_{2}$ from tissues to the lungs.
The reaction is: $CO_{2} + H_{2}O \xrightarrow{\text{Carbonic anhydrase}} H_{2}CO_{3}$
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485
ChemistryEasyMCQMHT CET · 2020
Identify the enzyme '$X$' in the following reaction.
$H_2O_{2(aq)} \xrightarrow{X} H_2O_{(l)} + \frac{1}{2} O_{2(g)}$
$H_2O_{2(aq)} \xrightarrow{X} H_2O_{(l)} + \frac{1}{2} O_{2(g)}$
A
amylase
B
ferroxidase
C
Catalase
D
carbonic anhydrase
Solution
(C) The given reaction is the decomposition of hydrogen peroxide $(H_2O_2)$ into water and oxygen.
This reaction is biologically catalyzed by the enzyme $Catalase$ in living organisms to prevent the accumulation of toxic $H_2O_2$ in cells.
Therefore,the enzyme '$X$' is $Catalase$.
This reaction is biologically catalyzed by the enzyme $Catalase$ in living organisms to prevent the accumulation of toxic $H_2O_2$ in cells.
Therefore,the enzyme '$X$' is $Catalase$.
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486
ChemistryEasyMCQMHT CET · 2020
Which of the following can form a colloidal sol with water?
A
Common salt
B
Starch
C
Glucose
D
Ammonium sulphate
Solution
(B) colloidal sol is formed by substances that are macromolecules or aggregates of molecules with sizes in the range of $1 \ nm$ to $1000 \ nm$.
$1$. Common salt $(NaCl)$,glucose,and ammonium sulphate are electrolytes or small molecules that form true solutions in water.
$2$. Starch is a macromolecule that,when dispersed in water,forms a lyophilic colloidal sol.
Therefore,the correct option is $B$.
$1$. Common salt $(NaCl)$,glucose,and ammonium sulphate are electrolytes or small molecules that form true solutions in water.
$2$. Starch is a macromolecule that,when dispersed in water,forms a lyophilic colloidal sol.
Therefore,the correct option is $B$.
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487
ChemistryEasyMCQMHT CET · 2020
Which of the following is an example of a hydrophobic sol?
A
Rubber in benzene
B
Cellulose acetate in acetone
C
Metal sulphide
D
Starch in water
Solution
(C) Hydrophobic sols are those in which the dispersed phase has little or no affinity for the dispersion medium. These are also known as lyophobic sols.
Metal sulphides (e.g.,$As_2S_3$) are classic examples of hydrophobic (lyophobic) sols.
Rubber in benzene,cellulose acetate in acetone,and starch in water are examples of hydrophilic (lyophilic) sols,as they have a strong affinity for the dispersion medium.
Metal sulphides (e.g.,$As_2S_3$) are classic examples of hydrophobic (lyophobic) sols.
Rubber in benzene,cellulose acetate in acetone,and starch in water are examples of hydrophilic (lyophilic) sols,as they have a strong affinity for the dispersion medium.
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488
ChemistryEasyMCQMHT CET · 2020
An amalgam of mercury with sodium is an example of:
A
solid in liquid solution
B
solid in solid solution
C
liquid in liquid solution
D
liquid in solid solution
Solution
(D) An amalgam is a mixture of mercury $(Hg)$ with another metal. In the case of sodium amalgam $(Na-Hg)$,sodium $(Na)$ is the solid solvent and mercury $(Hg)$ is the liquid solute. Therefore,it is an example of a liquid in solid solution.
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489
ChemistryEasyMCQMHT CET · 2020
$A$ solution of chloroform in nitrogen is an example of:
A
liquid in gas
B
liquid in solid
C
liquid in liquid
D
gas in liquid
Solution
(A) In a solution of chloroform in nitrogen,chloroform $(CHCl_3)$ is the solute (liquid state) and nitrogen $(N_2)$ is the solvent (gaseous state).
Therefore,it is an example of a liquid in gas type of solution.
Therefore,it is an example of a liquid in gas type of solution.
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490
ChemistryEasyMCQMHT CET · 2020
Which of the following ionic species has the highest precipitating power?
A
$Cu^{2+}$
B
$SO_4^{2-}$
C
$Na^{+}$
D
$Al^{3+}$
Solution
(D) According to the Hardy-Schulze rule,the precipitating power of an ion is directly proportional to its valency (charge).
Greater the magnitude of the charge on the ion,the higher is its precipitating power.
Comparing the charges: $Na^{+}$ $(+1)$,$Cu^{2+}$ $(+2)$,$Al^{3+}$ $(+3)$,$SO_4^{2-}$ $(-2)$.
Since $Al^{3+}$ has the highest magnitude of charge $(+3)$,it has the highest precipitating power.
Greater the magnitude of the charge on the ion,the higher is its precipitating power.
Comparing the charges: $Na^{+}$ $(+1)$,$Cu^{2+}$ $(+2)$,$Al^{3+}$ $(+3)$,$SO_4^{2-}$ $(-2)$.
Since $Al^{3+}$ has the highest magnitude of charge $(+3)$,it has the highest precipitating power.
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491
ChemistryEasyMCQMHT CET · 2020
Electrophoresis is used
A
to count number of particles in colloidal dispersions
B
for stability of colloids
C
to determine charge on colloidal particles
D
to determine size of colloidal particles
Solution
(C) Electrophoresis is a phenomenon in which colloidal particles move towards oppositely charged electrodes under the influence of an electric field.
Since the direction of movement of the particles depends on the nature of the charge they carry,this technique is used to determine the charge on colloidal particles.
Since the direction of movement of the particles depends on the nature of the charge they carry,this technique is used to determine the charge on colloidal particles.
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492
ChemistryEasyMCQMHT CET · 2020
Tyndall effect is observed due to
A
neutralization of charge on colloidal particles
B
Zig-zag motion of colloidal particles
C
precipitation of colloidal particles
D
Scattering of light by colloidal particles
Solution
(D) The Tyndall effect is the phenomenon in which the particles in a colloid scatter the beams of light that are directed at them. This occurs because the size of colloidal particles is comparable to the wavelength of light,allowing them to scatter the light in all directions,making the path of the light beam visible.
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493
ChemistryMediumMCQMHT CET · 2020
Which among the following ionic species has the least precipitating power?
A
$Cl^{-}$
B
$SO_{4}^{2-}$
C
$Mg^{2+}$
D
$Al^{3+}$
Solution
(A) According to the Hardy-Schulze rule,the precipitating power of an ion is directly proportional to its valency.
Since the precipitating power increases with the increase in the magnitude of the charge on the ion,the ion with the lowest valency will have the least precipitating power.
Comparing the valencies: $Cl^{-}$ $(1)$,$SO_{4}^{2-}$ $(2)$,$Mg^{2+}$ $(2)$,$Al^{3+}$ $(3)$.
Therefore,$Cl^{-}$ has the least precipitating power.
Since the precipitating power increases with the increase in the magnitude of the charge on the ion,the ion with the lowest valency will have the least precipitating power.
Comparing the valencies: $Cl^{-}$ $(1)$,$SO_{4}^{2-}$ $(2)$,$Mg^{2+}$ $(2)$,$Al^{3+}$ $(3)$.
Therefore,$Cl^{-}$ has the least precipitating power.
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494
ChemistryEasyMCQMHT CET · 2020
Which of the following is a multimolecular colloid?
A
Silver sol
B
Solution of rubber in organic solvent
C
Aqueous solution of protein
D
Aqueous polyvinyl alcohol
Solution
(A) Multimolecular colloids are formed by the aggregation of a large number of atoms or smaller molecules (diameter $< 1 \ nm$).
Silver sol is a classic example of a multimolecular colloid,where many silver atoms aggregate to form particles of colloidal size.
Rubber,proteins,and polyvinyl alcohol are examples of macromolecular colloids,where the molecules themselves are of colloidal dimensions.
Silver sol is a classic example of a multimolecular colloid,where many silver atoms aggregate to form particles of colloidal size.
Rubber,proteins,and polyvinyl alcohol are examples of macromolecular colloids,where the molecules themselves are of colloidal dimensions.
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495
ChemistryMediumMCQMHT CET · 2020
If a dilute solution of $AgNO_3$ is added to a dilute solution of excess $NaI$,then the species adsorbed on $AgI$ colloidal particles is:
A
$NO_3^-$
B
$Na^+$
C
$I^-$
D
$Ag^+$
Solution
(C) When $AgNO_3$ is added to an excess of $NaI$,the $AgI$ precipitate is formed.
According to the preferential adsorption theory,the common ion present in excess in the dispersion medium is adsorbed on the surface of the colloidal particles.
Here,$I^-$ ions are in excess,so they get adsorbed on the $AgI$ surface,forming a negatively charged sol: $[AgI]I^- / Na^+$.
According to the preferential adsorption theory,the common ion present in excess in the dispersion medium is adsorbed on the surface of the colloidal particles.
Here,$I^-$ ions are in excess,so they get adsorbed on the $AgI$ surface,forming a negatively charged sol: $[AgI]I^- / Na^+$.
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496
ChemistryEasyMCQMHT CET · 2020
When soap is added to an oily part of cloth,the hydrocarbon part of soap dissolves in
A
Water,keeping the tail away from the oil
B
Oil,keeping the head away from the oil
C
Water,keeping the head away from the oil
D
Oil,keeping the tail away from the oil
Solution
(B) Soap molecules have a hydrophobic hydrocarbon tail and a hydrophilic ionic head.
When soap is added to an oily or greasy part of cloth,the hydrophobic hydrocarbon tail dissolves in the oil,while the hydrophilic head remains in the water.
This arrangement forms a micelle,where the oil droplet is surrounded by soap molecules.
Consequently,the oil is emulsified and washed away by the stream of water.
When soap is added to an oily or greasy part of cloth,the hydrophobic hydrocarbon tail dissolves in the oil,while the hydrophilic head remains in the water.
This arrangement forms a micelle,where the oil droplet is surrounded by soap molecules.
Consequently,the oil is emulsified and washed away by the stream of water.
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497
ChemistryEasyMCQMHT CET · 2020
Which of the following is an emulsion?
A
Butter
B
Jellies
C
Milk
D
Mist
Solution
(C) An emulsion is a colloidal system in which both the dispersed phase and the dispersion medium are liquids.
Milk is a classic example of an emulsion where liquid fat is dispersed in water.
Butter is a gel (solid-in-liquid),jellies are gels (liquid-in-solid),and mist is an aerosol (liquid-in-gas).
Therefore,the correct answer is $C$.
Milk is a classic example of an emulsion where liquid fat is dispersed in water.
Butter is a gel (solid-in-liquid),jellies are gels (liquid-in-solid),and mist is an aerosol (liquid-in-gas).
Therefore,the correct answer is $C$.
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498
ChemistryMediumMCQMHT CET · 2020
Identify the element having the highest enthalpy of atomization from the following.
A
$Cu$ $(Z=29)$
B
$Fe$ $(Z=26)$
C
$Zn$ $(Z=30)$
D
$Sc$ $(Z=21)$
Solution
(B) The enthalpy of atomization of transition elements depends on the strength of metallic bonding,which is directly related to the number of unpaired electrons in the $d$-orbital.
More unpaired electrons lead to stronger metallic bonding and higher enthalpy of atomization.
Let us analyze the number of unpaired electrons in the given elements:
$Sc$ $(Z=21)$: $[Ar] 3d^1 4s^2$ ($1$ unpaired electron)
$Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$ ($4$ unpaired electrons)
$Cu$ $(Z=29)$: $[Ar] 3d^{10} 4s^1$ ($1$ unpaired electron)
$Zn$ $(Z=30)$: $[Ar] 3d^{10} 4s^2$ ($0$ unpaired electrons)
Among the given options,$Fe$ has the maximum number of unpaired electrons $(4)$,resulting in the strongest metallic bonding and the highest enthalpy of atomization.
More unpaired electrons lead to stronger metallic bonding and higher enthalpy of atomization.
Let us analyze the number of unpaired electrons in the given elements:
$Sc$ $(Z=21)$: $[Ar] 3d^1 4s^2$ ($1$ unpaired electron)
$Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$ ($4$ unpaired electrons)
$Cu$ $(Z=29)$: $[Ar] 3d^{10} 4s^1$ ($1$ unpaired electron)
$Zn$ $(Z=30)$: $[Ar] 3d^{10} 4s^2$ ($0$ unpaired electrons)
Among the given options,$Fe$ has the maximum number of unpaired electrons $(4)$,resulting in the strongest metallic bonding and the highest enthalpy of atomization.
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