ChemistryQ301–350 of 627 questions
Page 7 of 8 · English
301
ChemistryMediumMCQMHT CET · 2022
What is the void volume in the crystal lattice formed by a $BCC$ unit cell (in $\%$)?
A
$32$
B
$30$
C
$47.64$
D
$26$
Solution
(A) The packing efficiency of a $BCC$ (Body-Centered Cubic) unit cell is $68 \%$.
The void volume is calculated as $100 \% - \text{Packing Efficiency} = 100 \% - 68 \% = 32 \%$.
The void volume is calculated as $100 \% - \text{Packing Efficiency} = 100 \% - 68 \% = 32 \%$.
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302
ChemistryEasyMCQMHT CET · 2022
Calculate the volume of a unit cell having four particles in it with a density of $19.0 \ g \ cm^{-3}$ [molar mass of element $= 190 \ g \ mol^{-1}$].
A
$3.32 \times 10^{-23} \ cm^3$
B
$5.0 \times 10^{-23} \ cm^3$
C
$6.64 \times 10^{-23} \ cm^3$
D
$2.4 \times 10^{-23} \ cm^3$
Solution
(C) The density formula for a unit cell is given by $d = \frac{Z \times M}{V \times N_A}$.
Given: $Z = 4$,$d = 19.0 \ g \ cm^{-3}$,$M = 190 \ g \ mol^{-1}$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $19.0 = \frac{4 \times 190}{V \times 6.022 \times 10^{23}}$.
Rearranging for $V$: $V = \frac{4 \times 190}{19.0 \times 6.022 \times 10^{23}}$.
$V = \frac{40}{6.022} \times 10^{-23} \ cm^3$.
$V \approx 6.64 \times 10^{-23} \ cm^3$.
Given: $Z = 4$,$d = 19.0 \ g \ cm^{-3}$,$M = 190 \ g \ mol^{-1}$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $19.0 = \frac{4 \times 190}{V \times 6.022 \times 10^{23}}$.
Rearranging for $V$: $V = \frac{4 \times 190}{19.0 \times 6.022 \times 10^{23}}$.
$V = \frac{40}{6.022} \times 10^{-23} \ cm^3$.
$V \approx 6.64 \times 10^{-23} \ cm^3$.
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303
ChemistryEasyMCQMHT CET · 2022
Calculate the molar mass of a metal with density $1 \ g \ cm^{-3}$ forming a $bcc$ structure with an edge length of $420 \ pm$.
A
$32.2 \ g \ mol^{-1}$
B
$22.3 \ g \ mol^{-1}$
C
$25.5 \ g \ mol^{-1}$
D
$43.3 \ g \ mol^{-1}$
Solution
(B) The formula for density is $d = \frac{Z \times M}{a^3 \times N_A}$.
For a $bcc$ structure, the number of atoms per unit cell $Z = 2$.
The edge length $a = 420 \ pm = 420 \times 10^{-10} \ cm = 4.20 \times 10^{-8} \ cm$.
The Avogadro number $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $1 = \frac{2 \times M}{(4.20 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = \frac{1 \times (74.088 \times 10^{-24}) \times 6.022 \times 10^{23}}{2}$.
$M = \frac{44.61}{2} \approx 22.3 \ g \ mol^{-1}$.
For a $bcc$ structure, the number of atoms per unit cell $Z = 2$.
The edge length $a = 420 \ pm = 420 \times 10^{-10} \ cm = 4.20 \times 10^{-8} \ cm$.
The Avogadro number $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $1 = \frac{2 \times M}{(4.20 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = \frac{1 \times (74.088 \times 10^{-24}) \times 6.022 \times 10^{23}}{2}$.
$M = \frac{44.61}{2} \approx 22.3 \ g \ mol^{-1}$.
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304
ChemistryEasyMCQMHT CET · 2022
Calculate the density of a metal having a unit cell volume of $64 \times 10^{-24} \ cm^3$ and a molar mass of $192 \ g \ mol^{-1}$,containing $4$ particles per unit cell. (in $g \ cm^{-3}$)
A
$16.00$
B
$19.93$
C
$14.92$
D
$18.00$
Solution
(B) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{a^3 \times N_A}$
Where:
$Z = 4$ (number of particles per unit cell)
$M = 192 \ g \ mol^{-1}$ (molar mass)
$a^3 = 64 \times 10^{-24} \ cm^3$ (volume of unit cell)
$N_A = 6.022 \times 10^{23} \ mol^{-1}$ (Avogadro constant)
Substituting the values:
$d = \frac{4 \times 192}{64 \times 10^{-24} \times 6.022 \times 10^{23}}$
$d = \frac{768}{38.5408} \approx 19.93 \ g \ cm^{-3}$
Where:
$Z = 4$ (number of particles per unit cell)
$M = 192 \ g \ mol^{-1}$ (molar mass)
$a^3 = 64 \times 10^{-24} \ cm^3$ (volume of unit cell)
$N_A = 6.022 \times 10^{23} \ mol^{-1}$ (Avogadro constant)
Substituting the values:
$d = \frac{4 \times 192}{64 \times 10^{-24} \times 6.022 \times 10^{23}}$
$d = \frac{768}{38.5408} \approx 19.93 \ g \ cm^{-3}$
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305
ChemistryEasyMCQMHT CET · 2022
Calculate the number of atoms in $5 \ g$ of a metal that crystallises in a simple cubic unit cell structure with an edge length of $336 \ pm$. (Density of the metal $= 9.4 \ g \ cm^{-3}$)
A
$1.4 \times 10^{22}$
B
$1.8 \times 10^{22}$
C
$1.0 \times 10^{22}$
D
$2.1 \times 10^{22}$
Solution
(A) For a simple cubic unit cell, the number of atoms per unit cell $(z)$ $= 1$.
Edge length $(a)$ $= 336 \ pm = 3.36 \times 10^{-8} \ cm$.
Density $(d)$ $= \frac{z \times M}{a^3 \times N_A}$, where $M$ is the molar mass and $N_A$ is Avogadro's number $(6.022 \times 10^{23} \ mol^{-1})$.
$9.4 = \frac{1 \times M}{(3.36 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = 9.4 \times (3.793 \times 10^{-23}) \times 6.022 \times 10^{23} \approx 214.65 \ g \ mol^{-1}$.
Number of atoms in $5 \ g = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{5}{214.65} \times 6.022 \times 10^{23} \approx 1.4 \times 10^{22}$ atoms.
Edge length $(a)$ $= 336 \ pm = 3.36 \times 10^{-8} \ cm$.
Density $(d)$ $= \frac{z \times M}{a^3 \times N_A}$, where $M$ is the molar mass and $N_A$ is Avogadro's number $(6.022 \times 10^{23} \ mol^{-1})$.
$9.4 = \frac{1 \times M}{(3.36 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = 9.4 \times (3.793 \times 10^{-23}) \times 6.022 \times 10^{23} \approx 214.65 \ g \ mol^{-1}$.
Number of atoms in $5 \ g = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{5}{214.65} \times 6.022 \times 10^{23} \approx 1.4 \times 10^{22}$ atoms.
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306
ChemistryEasyMCQMHT CET · 2022
What is the total volume occupied by atoms in a $bcc$ unit cell (in $\%$)?
A
$68$
B
$80$
C
$74$
D
$52.36$
Solution
(A) In a $bcc$ (body-centered cubic) unit cell,the number of atoms per unit cell is $Z = 2$.
The relationship between the edge length $a$ and the atomic radius $r$ is $4r = \sqrt{3}a$,or $a = \frac{4r}{\sqrt{3}}$.
The volume of the unit cell is $V_{cell} = a^3 = (\frac{4r}{\sqrt{3}})^3 = \frac{64r^3}{3\sqrt{3}}$.
The volume occupied by $2$ atoms is $V_{atoms} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3$.
The packing efficiency is $\frac{V_{atoms}}{V_{cell}} \times 100 = \frac{(8/3) \pi r^3}{(64/3\sqrt{3}) r^3} \times 100 = \frac{\sqrt{3} \pi}{8} \times 100 \approx 68 \%$.
The relationship between the edge length $a$ and the atomic radius $r$ is $4r = \sqrt{3}a$,or $a = \frac{4r}{\sqrt{3}}$.
The volume of the unit cell is $V_{cell} = a^3 = (\frac{4r}{\sqrt{3}})^3 = \frac{64r^3}{3\sqrt{3}}$.
The volume occupied by $2$ atoms is $V_{atoms} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3$.
The packing efficiency is $\frac{V_{atoms}}{V_{cell}} \times 100 = \frac{(8/3) \pi r^3}{(64/3\sqrt{3}) r^3} \times 100 = \frac{\sqrt{3} \pi}{8} \times 100 \approx 68 \%$.
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307
ChemistryEasyMCQMHT CET · 2022
What is the coordination number of a particle in an $fcc$ crystal lattice?
A
$4$
B
$8$
C
$12$
D
$6$
Solution
(C) In a face-centered cubic $(fcc)$ crystal lattice,each atom is in contact with $12$ nearest neighbors.
Therefore,the coordination number of a particle in an $fcc$ lattice is $12$.
Therefore,the coordination number of a particle in an $fcc$ lattice is $12$.
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308
ChemistryEasyMCQMHT CET · 2022
Calculate the mass of a $bcc$ unit cell if the metal has a molar mass of $56 \ g \ mol^{-1}$.
A
$1.86 \times 10^{-22} \ g$
B
$9.3 \times 10^{-24} \ g$
C
$2.79 \times 10^{-24} \ g$
D
$3.72 \times 10^{-22} \ g$
Solution
(A) For a $bcc$ unit cell,the number of atoms per unit cell $(Z)$ is $2$.
The mass of one atom is given by $\frac{\text{Molar mass}}{N_A} = \frac{56}{6.022 \times 10^{23}} \ g$.
The mass of the unit cell is $Z \times \text{mass of one atom} = 2 \times \frac{56}{6.022 \times 10^{23}} \ g$.
$= \frac{112}{6.022 \times 10^{23}} \ g \approx 1.86 \times 10^{-22} \ g$.
The mass of one atom is given by $\frac{\text{Molar mass}}{N_A} = \frac{56}{6.022 \times 10^{23}} \ g$.
The mass of the unit cell is $Z \times \text{mass of one atom} = 2 \times \frac{56}{6.022 \times 10^{23}} \ g$.
$= \frac{112}{6.022 \times 10^{23}} \ g \approx 1.86 \times 10^{-22} \ g$.
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309
ChemistryDifficultMCQMHT CET · 2022
Which of the following is an example of an ionic solid?
A
Calcium fluoride
B
Ice
C
Silica
D
Sodium metal
Solution
(A) $CaF_2$ is an ionic solid.
$Ice$ is a molecular solid.
$SiO_2$ is a covalent network (atomic) solid.
$Na$ is a metallic solid.
$Ice$ is a molecular solid.
$SiO_2$ is a covalent network (atomic) solid.
$Na$ is a metallic solid.
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310
ChemistryMediumMCQMHT CET · 2022
Which of the following statements is not true about ionic solids?
A
The constituent particles of ionic solids are cations and anions.
B
Ionic solids are hard and brittle.
C
In these,constituent particles are held together by electrostatic forces of attraction.
D
Pure ionic solids are good conductors of electricity in the solid state.
Solution
(D) Ionic solids consist of ions held together by strong electrostatic forces of attraction,making them hard and brittle. However,in the solid state,the ions are fixed in their lattice positions and are not free to move. Therefore,they are poor conductors of electricity in the solid state. They only conduct electricity when molten or dissolved in water,as the ions become free to move. Thus,statement $D$ is incorrect.
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311
ChemistryEasyMCQMHT CET · 2022
Which among the following statements is $NOT$ true about Frenkel defect?
A
It occurs in ionic compounds with sizes of cations and anions almost equal.
B
The density of solid crystal and its chemical properties remain unchanged as no ions are missing from the crystal lattice as a whole.
C
The crystal as whole remains electrically neutral because equal number of cations and anions are present.
D
The ions of ionic compound must have low co-ordination number.
Solution
(A) Frenkel defect occurs in ionic compounds where there is a large difference in the size of ions,specifically where the cation is much smaller than the anion. Option $A$ is incorrect because Frenkel defect is characteristic of compounds with a large difference in ionic sizes,whereas Schottky defect occurs in compounds where cation and anion sizes are almost equal.
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312
ChemistryEasyMCQMHT CET · 2022
Which of the following statements is $NOT$ true about Schottky defect?
A
The electrical neutrality of the compound is not preserved.
B
It occurs in ionic compounds with sizes of cation and anion almost equal.
C
The ions of ionic compound must have high co-ordination number.
D
The density of the substance decreases.
Solution
(A) Schottky defect is a type of point defect in ionic crystals where an equal number of cations and anions are missing from their lattice sites.
$1$. Electrical neutrality is maintained because the number of missing cations equals the number of missing anions.
$2$. It typically occurs in ionic compounds where the cation and anion sizes are similar.
$3$. It is favored in compounds with high coordination numbers.
$4$. Due to the loss of ions from the lattice,the density of the crystal decreases.
Therefore,the statement that electrical neutrality is not preserved is incorrect.
$1$. Electrical neutrality is maintained because the number of missing cations equals the number of missing anions.
$2$. It typically occurs in ionic compounds where the cation and anion sizes are similar.
$3$. It is favored in compounds with high coordination numbers.
$4$. Due to the loss of ions from the lattice,the density of the crystal decreases.
Therefore,the statement that electrical neutrality is not preserved is incorrect.
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313
ChemistryEasyMCQMHT CET · 2022
Which among the following is $NOT$ ferromagnetic in nature?
A
$Fe$
B
$Zn$
C
$Ni$
D
$Co$
Solution
(B) $Fe$,$Co$,and $Ni$ are ferromagnetic substances.
$Zn$ is diamagnetic in nature.
$Zn$ is diamagnetic in nature.
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314
ChemistryEasyMCQMHT CET · 2022
Calculate the pressure of the gas if the solubility of the gas in water at $25^{\circ} C$ is $6.85 \times 10^{-4} \ mol \ dm^{-3}$. (Henry's law constant $K_H$ is $6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$) (in $bar$)
A
$0.853$
B
$1.5$
C
$0.5$
D
$1$
Solution
(D) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to the partial pressure $(P)$ of the gas above the solution:
$S = K_H \times P$
Given:
Solubility $(S)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3}$
Henry's law constant $(K_H)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$
Rearranging the formula to solve for pressure $(P)$:
$P = S / K_H$
$P = (6.85 \times 10^{-4} \ mol \ dm^{-3}) / (6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1})$
$P = 1.0 \ bar$
$S = K_H \times P$
Given:
Solubility $(S)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3}$
Henry's law constant $(K_H)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$
Rearranging the formula to solve for pressure $(P)$:
$P = S / K_H$
$P = (6.85 \times 10^{-4} \ mol \ dm^{-3}) / (6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1})$
$P = 1.0 \ bar$
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315
ChemistryEasyMCQMHT CET · 2022
Calculate the solubility of gas in water at $1.2 \text{ atm}$ and $25^{\circ} C$ if Henry's law constant is $0.45 \text{ mol dm}^{-3} \text{ atm}^{-1}$ at $25^{\circ} C$.
A
$0.45 \text{ mol dm}^{-3}$
B
$0.54 \text{ mol dm}^{-3}$
C
$0.25 \text{ mol dm}^{-3}$
D
$0.31 \text{ mol dm}^{-3}$
Solution
(B) According to Henry's law,the solubility of a gas is given by the formula: $S = k \times p$
Here,$k = 0.45 \text{ mol dm}^{-3} \text{ atm}^{-1}$ and $p = 1.2 \text{ atm}$.
Substituting the values: $S = 0.45 \text{ mol dm}^{-3} \text{ atm}^{-1} \times 1.2 \text{ atm} = 0.54 \text{ mol dm}^{-3}$.
Thus,the solubility of the gas is $0.54 \text{ mol dm}^{-3}$.
Here,$k = 0.45 \text{ mol dm}^{-3} \text{ atm}^{-1}$ and $p = 1.2 \text{ atm}$.
Substituting the values: $S = 0.45 \text{ mol dm}^{-3} \text{ atm}^{-1} \times 1.2 \text{ atm} = 0.54 \text{ mol dm}^{-3}$.
Thus,the solubility of the gas is $0.54 \text{ mol dm}^{-3}$.
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316
ChemistryEasyMCQMHT CET · 2022
Which of the following laws represents the quantitative relationship between the solubility of gas in liquid and its pressure?
A
Charles's law
B
Avogadro's law
C
Raoult's law
D
Henry's law
Solution
(D) Henry's law states that the partial pressure of the gas in the vapor phase $(p)$ is proportional to the mole fraction of the gas $(x)$ in the solution.
Mathematically,it is expressed as $p = K_H \cdot x$,where $K_H$ is Henry's law constant.
This law directly describes the quantitative relationship between the solubility of a gas in a liquid and its pressure.
Mathematically,it is expressed as $p = K_H \cdot x$,where $K_H$ is Henry's law constant.
This law directly describes the quantitative relationship between the solubility of a gas in a liquid and its pressure.
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317
ChemistryEasyMCQMHT CET · 2022
What is the unit of Henry's law constant $(K_H)$?
A
$mol \ dm^{-3}$
B
$mol \ dm^3 \ bar^{-1}$
C
$mol \ dm^{-3} \ bar^{-1}$
D
$mol \ dm^{-3} \ bar$
Solution
(C) According to Henry's law,the solubility $(S)$ of a gas in a liquid is directly proportional to the partial pressure $(P)$ of the gas above the liquid: $S = K_H \times P$.
Here,$S$ is expressed in $mol \ dm^{-3}$ and $P$ is expressed in $bar$.
Therefore,the unit of Henry's law constant $(K_H)$ is given by: $K_H = \frac{S}{P} = \frac{mol \ dm^{-3}}{bar} = mol \ dm^{-3} \ bar^{-1}$.
Here,$S$ is expressed in $mol \ dm^{-3}$ and $P$ is expressed in $bar$.
Therefore,the unit of Henry's law constant $(K_H)$ is given by: $K_H = \frac{S}{P} = \frac{mol \ dm^{-3}}{bar} = mol \ dm^{-3} \ bar^{-1}$.
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318
ChemistryEasyMCQMHT CET · 2022
Calculate the solubility of a gas in water at $0.8 \ atm$ and $25^{\circ} C$. [Henry's law constant is $6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$]
A
$5.48 \times 10^{-4} \ mol \ dm^{-3}$
B
$3.94 \times 10^{-4} \ mol \ dm^{-3}$
C
$6.858 \times 10^{-4} \ mol \ dm^{-3}$
D
$2.74 \times 10^{-4} \ mol \ dm^{-3}$
Solution
(A) According to Henry's law,the solubility $(S)$ of a gas is given by the formula: $S = K_H \times P$
Given:
Pressure $(P)$ = $0.8 \ atm$
Henry's law constant $(K_H)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$
Substituting the values:
$S = 6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1} \times 0.8 \ atm$
$S = 5.48 \times 10^{-4} \ mol \ dm^{-3}$
Given:
Pressure $(P)$ = $0.8 \ atm$
Henry's law constant $(K_H)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$
Substituting the values:
$S = 6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1} \times 0.8 \ atm$
$S = 5.48 \times 10^{-4} \ mol \ dm^{-3}$
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319
ChemistryEasyMCQMHT CET · 2022
Calculate the solubility of a gas in water at $260 \ mm \ Hg$ and $25^{\circ} C$ if Henry's law constant of the gas is $0.159 \ mol \ dm^{-3} \ atm^{-1}$ at $25^{\circ} C$.
A
$3.8 \times 10^{-2} \ mol \ dm^{-3}$
B
$5.4 \times 10^{-2} \ mol \ dm^{-3}$
C
$2.7 \times 10^{-2} \ mol \ dm^{-3}$
D
$1.2 \times 10^{-2} \ mol \ dm^{-3}$
Solution
(B) According to Henry's law,the solubility $(S)$ of a gas is given by $S = K_H \times p$.
First,convert the pressure from $mm \ Hg$ to $atm$:
$p = \frac{260 \ mm \ Hg}{760 \ mm \ Hg \ atm^{-1}} \approx 0.3421 \ atm$.
Now,calculate the solubility:
$S = 0.159 \ mol \ dm^{-3} \ atm^{-1} \times 0.3421 \ atm \approx 0.05439 \ mol \ dm^{-3}$.
Rounding to two significant figures,we get $S = 5.4 \times 10^{-2} \ mol \ dm^{-3}$.
First,convert the pressure from $mm \ Hg$ to $atm$:
$p = \frac{260 \ mm \ Hg}{760 \ mm \ Hg \ atm^{-1}} \approx 0.3421 \ atm$.
Now,calculate the solubility:
$S = 0.159 \ mol \ dm^{-3} \ atm^{-1} \times 0.3421 \ atm \approx 0.05439 \ mol \ dm^{-3}$.
Rounding to two significant figures,we get $S = 5.4 \times 10^{-2} \ mol \ dm^{-3}$.
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320
ChemistryEasyMCQMHT CET · 2022
What is the relation between the molar mass of a solute and the boiling point elevation?
A
$M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}$
B
$M_2 = \frac{1000 \times \Delta T_b \times W_2}{K_b \times W_1}$
C
$M_2 = \frac{\Delta T_b \times W_1}{1000 \times K_b \times W_2}$
D
$M_2 = \frac{1000 \times K_b \times W_1}{\Delta T_b \times W_2}$
Solution
(A) The boiling point elevation is given by the formula: $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent ($W_1$ in grams): $m = \frac{n_2 \times 1000}{W_1}$.
Since $n_2 = \frac{W_2}{M_2}$,where $W_2$ is the mass of solute and $M_2$ is the molar mass of solute,we substitute this into the molality equation:
$m = \frac{W_2 \times 1000}{M_2 \times W_1}$.
Substituting this into the boiling point elevation formula:
$\Delta T_b = K_b \times \frac{W_2 \times 1000}{M_2 \times W_1}$.
Rearranging for $M_2$,we get: $M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}$.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent ($W_1$ in grams): $m = \frac{n_2 \times 1000}{W_1}$.
Since $n_2 = \frac{W_2}{M_2}$,where $W_2$ is the mass of solute and $M_2$ is the molar mass of solute,we substitute this into the molality equation:
$m = \frac{W_2 \times 1000}{M_2 \times W_1}$.
Substituting this into the boiling point elevation formula:
$\Delta T_b = K_b \times \frac{W_2 \times 1000}{M_2 \times W_1}$.
Rearranging for $M_2$,we get: $M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}$.
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321
ChemistryEasyMCQMHT CET · 2022
What is the molar mass of a solute when $2.3 \ g$ of a non-volatile solute is dissolved in $46 \ g$ of benzene at $30^{\circ} C$? (Relative lowering of vapour pressure is $0.06$ and the molar mass of benzene is $78 \ g \ mol^{-1}$)
A
$65 \ g \ mol^{-1}$
B
$80 \ g \ mol^{-1}$
C
$72 \ g \ mol^{-1}$
D
$48 \ g \ mol^{-1}$
Solution
(A) The formula for relative lowering of vapour pressure is: $\frac{\Delta P}{P_{A}^0} = \frac{n_{B}}{n_{A}} = \frac{W_{B}}{M_{B}} \times \frac{M_{A}}{W_{A}}$
Given: $\frac{\Delta P}{P_{A}^0} = 0.06$,$W_{B} = 2.3 \ g$,$W_{A} = 46 \ g$,$M_{A} = 78 \ g \ mol^{-1}$.
Substituting the values: $0.06 = \frac{2.3}{M_{B}} \times \frac{78}{46}$
$0.06 = \frac{2.3}{M_{B}} \times 1.6956$
$M_{B} = \frac{2.3 \times 1.6956}{0.06} = \frac{3.9}{0.06} = 65 \ g \ mol^{-1}$.
Given: $\frac{\Delta P}{P_{A}^0} = 0.06$,$W_{B} = 2.3 \ g$,$W_{A} = 46 \ g$,$M_{A} = 78 \ g \ mol^{-1}$.
Substituting the values: $0.06 = \frac{2.3}{M_{B}} \times \frac{78}{46}$
$0.06 = \frac{2.3}{M_{B}} \times 1.6956$
$M_{B} = \frac{2.3 \times 1.6956}{0.06} = \frac{3.9}{0.06} = 65 \ g \ mol^{-1}$.
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322
ChemistryMediumMCQMHT CET · 2022
What is the unit of cryoscopic constant?
A
$K \ kg \ mol^{-1}$
B
$K \ kg \ mol^{3}$
C
$K \ kg \ mol$
D
$K \ kg \ dm^{-3}$
Solution
(A) The cryoscopic constant $(K_f)$,also known as the molal freezing point depression constant,is defined by the equation: $\Delta T_f = K_f \times m$,where $\Delta T_f$ is the depression in freezing point $(K)$ and $m$ is the molality $(mol \ kg^{-1})$.
Rearranging for $K_f$,we get: $K_f = \frac{\Delta T_f}{m} = \frac{K}{mol \ kg^{-1}} = K \ kg \ mol^{-1}$.
Rearranging for $K_f$,we get: $K_f = \frac{\Delta T_f}{m} = \frac{K}{mol \ kg^{-1}} = K \ kg \ mol^{-1}$.
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323
ChemistryMediumMCQMHT CET · 2022
What is the mass of solute having molar mass $60 \ g \ mol^{-1}$ when dissolved in $98 \ g$ of solvent decreases its freezing point by $0.2 \ K$ (in $g$)? (The numerical value of cryoscopic constant of solvent is $1.71 \ K \ kg \ mol^{-1}$)
A
$0.5$
B
$1.5$
C
$0.687$
D
$2.0$
Solution
(C) The formula for depression in freezing point is $\Delta T_f = K_f \times m$.
Here,$\Delta T_f = 0.2 \ K$,$K_f = 1.71 \ K \ kg \ mol^{-1}$,$M_B = 60 \ g \ mol^{-1}$,and $W_A = 98 \ g$.
The molality $m$ is given by $\frac{W_B \times 1000}{M_B \times W_A}$.
Substituting the values: $0.2 = 1.71 \times \frac{W_B \times 1000}{60 \times 98}$.
Solving for $W_B$: $W_B = \frac{0.2 \times 60 \times 98}{1.71 \times 1000} = \frac{1176}{1710} \approx 0.687 \ g$.
Here,$\Delta T_f = 0.2 \ K$,$K_f = 1.71 \ K \ kg \ mol^{-1}$,$M_B = 60 \ g \ mol^{-1}$,and $W_A = 98 \ g$.
The molality $m$ is given by $\frac{W_B \times 1000}{M_B \times W_A}$.
Substituting the values: $0.2 = 1.71 \times \frac{W_B \times 1000}{60 \times 98}$.
Solving for $W_B$: $W_B = \frac{0.2 \times 60 \times 98}{1.71 \times 1000} = \frac{1176}{1710} \approx 0.687 \ g$.
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324
ChemistryMediumMCQMHT CET · 2022
What is the molar mass of a solute when $5 \ g$ of solute dissolved in $70 \ g$ of solvent lowers its freezing point by $2.5 \ K$? Given $K_f = 3.5 \ K \ kg \ mol^{-1}$.
A
$100 \ g \ mol^{-1}$
B
$120 \ g \ mol^{-1}$
C
$160 \ g \ mol^{-1}$
D
$140 \ g \ mol^{-1}$
Solution
(A) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2 = 5 \ g$ (solute mass),$w_1 = 70 \ g$ (solvent mass),and $M_2$ is the molar mass of the solute.
Substituting the values: $2.5 = 3.5 \times \frac{5 \times 1000}{M_2 \times 70}$.
$M_2 = \frac{3.5 \times 5 \times 1000}{2.5 \times 70} = \frac{17500}{175} = 100 \ g \ mol^{-1}$.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2 = 5 \ g$ (solute mass),$w_1 = 70 \ g$ (solvent mass),and $M_2$ is the molar mass of the solute.
Substituting the values: $2.5 = 3.5 \times \frac{5 \times 1000}{M_2 \times 70}$.
$M_2 = \frac{3.5 \times 5 \times 1000}{2.5 \times 70} = \frac{17500}{175} = 100 \ g \ mol^{-1}$.
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325
ChemistryMediumMCQMHT CET · 2022
Calculate the molality of a solution having freezing point depression $3.6 \ K$ and freezing point depression constant $4.8 \ K \ kg \ mol^{-1}$.
A
$0.3 \ mol \ kg^{-1}$
B
$0.9 \ mol \ kg^{-1}$
C
$0.75 \ mol \ kg^{-1}$
D
$0.5 \ mol \ kg^{-1}$
Solution
(C) The formula for freezing point depression is given by: $\Delta T_{f} = K_{f} \times m$
Where $\Delta T_{f}$ is the freezing point depression,$K_{f}$ is the freezing point depression constant,and $m$ is the molality.
Given: $\Delta T_{f} = 3.6 \ K$ and $K_{f} = 4.8 \ K \ kg \ mol^{-1}$.
Substituting the values in the formula: $3.6 = 4.8 \times m$
$m = \frac{3.6}{4.8} = 0.75 \ mol \ kg^{-1}$.
Where $\Delta T_{f}$ is the freezing point depression,$K_{f}$ is the freezing point depression constant,and $m$ is the molality.
Given: $\Delta T_{f} = 3.6 \ K$ and $K_{f} = 4.8 \ K \ kg \ mol^{-1}$.
Substituting the values in the formula: $3.6 = 4.8 \times m$
$m = \frac{3.6}{4.8} = 0.75 \ mol \ kg^{-1}$.
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326
ChemistryEasyMCQMHT CET · 2022
Which among the following colligative properties is useful to determine molar masses of very expensive solutes?
A
Vapour pressure lowering
B
Osmotic pressure
C
Freezing point depression
D
Boiling point elevation
Solution
(B) Osmotic pressure is the most suitable colligative property for determining the molar masses of expensive solutes.
This is because it can be measured at room temperature,and it requires only a very small amount of the solute to produce a measurable change in pressure,making it ideal for costly or biologically sensitive substances.
This is because it can be measured at room temperature,and it requires only a very small amount of the solute to produce a measurable change in pressure,making it ideal for costly or biologically sensitive substances.
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327
ChemistryMediumMCQMHT CET · 2022
Calculate the molar mass of the solute when $1.5 \ g$ of a non-volatile solute is dissolved in $100 \ mL$ of a solvent having a density of $0.8 \ g \ mL^{-1}$,which lowers its freezing point by $0.75 \ K$. (Freezing point depression constant for the solvent is $5 \ K \ kg \ mol^{-1}$).
A
$125 \ g \ mol^{-1}$
B
$110 \ g \ mol^{-1}$
C
$100 \ g \ mol^{-1}$
D
$75 \ g \ mol^{-1}$
Solution
(A) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
The mass of the solvent $(W_A)$ is calculated as: $W_A = \text{density} \times \text{volume} = 0.8 \ g \ mL^{-1} \times 100 \ mL = 80 \ g$.
Substituting the given values into the formula: $\Delta T_f = K_f \times \frac{W_B}{M_B} \times \frac{1000}{W_A(g)}$.
$0.75 = 5 \times \frac{1.5}{M_B} \times \frac{1000}{80}$.
$0.75 = 5 \times \frac{1.5}{M_B} \times 12.5$.
$M_B = \frac{5 \times 1.5 \times 12.5}{0.75} = \frac{93.75}{0.75} = 125 \ g \ mol^{-1}$.
The mass of the solvent $(W_A)$ is calculated as: $W_A = \text{density} \times \text{volume} = 0.8 \ g \ mL^{-1} \times 100 \ mL = 80 \ g$.
Substituting the given values into the formula: $\Delta T_f = K_f \times \frac{W_B}{M_B} \times \frac{1000}{W_A(g)}$.
$0.75 = 5 \times \frac{1.5}{M_B} \times \frac{1000}{80}$.
$0.75 = 5 \times \frac{1.5}{M_B} \times 12.5$.
$M_B = \frac{5 \times 1.5 \times 12.5}{0.75} = \frac{93.75}{0.75} = 125 \ g \ mol^{-1}$.
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328
ChemistryMediumMCQMHT CET · 2022
Calculate the amount of solute dissolved in $3 \ dm^3$ water having osmotic pressure $0.3 \ atm$ at $300 \ K$. (Molar mass of solute $= 108 \ g \ mol^{-1}, R = 0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$) (in $g$)
A
$4.51$
B
$3.95$
C
$3.45$
D
$5.26$
Solution
(B) The formula for osmotic pressure is $\pi = CRT$,where $C = \frac{n}{V} = \frac{W}{M \times V}$.
Given: $\pi = 0.3 \ atm$,$V = 3 \ dm^3$,$T = 300 \ K$,$M = 108 \ g \ mol^{-1}$,$R = 0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.3 = \frac{W}{108 \times 3} \times 0.0821 \times 300$.
$0.3 = \frac{W \times 0.0821 \times 300}{324}$.
$W = \frac{0.3 \times 324}{0.0821 \times 300} = \frac{97.2}{24.63} \approx 3.946 \ g \approx 3.95 \ g$.
Given: $\pi = 0.3 \ atm$,$V = 3 \ dm^3$,$T = 300 \ K$,$M = 108 \ g \ mol^{-1}$,$R = 0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.3 = \frac{W}{108 \times 3} \times 0.0821 \times 300$.
$0.3 = \frac{W \times 0.0821 \times 300}{324}$.
$W = \frac{0.3 \times 324}{0.0821 \times 300} = \frac{97.2}{24.63} \approx 3.946 \ g \approx 3.95 \ g$.
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329
ChemistryEasyMCQMHT CET · 2022
Which of the following solutes dissolved in water,having the same concentration,exhibits the highest value of colligative property?
A
Urea
B
Glucose
C
Sucrose
D
Sodium chloride
Solution
(D) Colligative properties depend on the number of particles in the solution. The relationship is given by the van't Hoff factor $(i)$.
For non-electrolytes like urea,glucose,and sucrose,the value of $i$ is $1$.
For sodium chloride $(NaCl)$,which is a strong electrolyte,it dissociates as $NaCl \rightarrow Na^+ + Cl^-$,resulting in $i = 2$.
Since the colligative property is directly proportional to the van't Hoff factor $(i)$,the solute with the highest $i$ value will exhibit the highest colligative property.
For non-electrolytes like urea,glucose,and sucrose,the value of $i$ is $1$.
For sodium chloride $(NaCl)$,which is a strong electrolyte,it dissociates as $NaCl \rightarrow Na^+ + Cl^-$,resulting in $i = 2$.
Since the colligative property is directly proportional to the van't Hoff factor $(i)$,the solute with the highest $i$ value will exhibit the highest colligative property.
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330
ChemistryEasyMCQMHT CET · 2022
What is the relation between the depression in freezing point and the molar mass of a non-volatile solute?
A
$M_2 = \frac{1000 \cdot K_f \cdot W_1}{\Delta T_f \cdot W_2}$
B
$M_2 = \frac{\Delta T_f \cdot W_1}{1000 \cdot K_f \cdot W_2}$
C
$M_2 = \frac{1000 \cdot \Delta T_f \cdot W_2}{K_f \cdot W_1}$
D
$M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$
Solution
(D) The depression in freezing point $(\Delta T_f)$ is given by the formula: $\Delta T_f = K_f \cdot m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent $(W_1 \text{ in grams})$: $m = \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Substituting this into the depression in freezing point equation: $\Delta T_f = K_f \cdot \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Rearranging the formula to solve for the molar mass of the solute $(M_2)$: $M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent $(W_1 \text{ in grams})$: $m = \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Substituting this into the depression in freezing point equation: $\Delta T_f = K_f \cdot \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Rearranging the formula to solve for the molar mass of the solute $(M_2)$: $M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$.
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331
ChemistryEasyMCQMHT CET · 2022
Find the freezing point depression of a solution having molality $0.25 \ mol \ kg^{-1}$. $(K_{f} = 4.0 \ K \ kg \ mol^{-1})$ (in $K$)
A
$2.5$
B
$3.0$
C
$2.0$
D
$1.0$
Solution
(D) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$.
Given:
Molality $(m) = 0.25 \ mol \ kg^{-1}$
Cryoscopic constant $(K_{f}) = 4.0 \ K \ kg \ mol^{-1}$
Calculation:
$\Delta T_{f} = 4.0 \times 0.25 = 1.0 \ K$
Thus,the freezing point depression is $1.0 \ K$.
Given:
Molality $(m) = 0.25 \ mol \ kg^{-1}$
Cryoscopic constant $(K_{f}) = 4.0 \ K \ kg \ mol^{-1}$
Calculation:
$\Delta T_{f} = 4.0 \times 0.25 = 1.0 \ K$
Thus,the freezing point depression is $1.0 \ K$.
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332
ChemistryEasyMCQMHT CET · 2022
Find the molar mass of solute when $2 \ g$ is dissolved in $60 \ g$ of benzene,and the relative lowering of vapour pressure is $0.06$. (Molar mass of benzene is $78 \ g \ mol^{-1}$)
A
$17.4 \ g \ mol^{-1}$
B
$35.2 \ g \ mol^{-1}$
C
$43.3 \ g \ mol^{-1}$
D
$24.2 \ g \ mol^{-1}$
Solution
(C) The formula for relative lowering of vapour pressure is given by: $\frac{\Delta P}{P_{A}^{\circ}} = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A} = \frac{W_B}{M_B} \times \frac{M_A}{W_A}$
Given: $\frac{\Delta P}{P_{A}^{\circ}} = 0.06$,$W_B = 2 \ g$,$W_A = 60 \ g$,$M_A = 78 \ g \ mol^{-1}$.
Substituting the values: $0.06 = \frac{2}{M_B} \times \frac{78}{60}$
$0.06 = \frac{156}{60 \times M_B}$
$M_B = \frac{156}{60 \times 0.06} = \frac{156}{3.6} = 43.33 \ g \ mol^{-1}$
Thus,the molar mass of the solute is $43.3 \ g \ mol^{-1}$.
Given: $\frac{\Delta P}{P_{A}^{\circ}} = 0.06$,$W_B = 2 \ g$,$W_A = 60 \ g$,$M_A = 78 \ g \ mol^{-1}$.
Substituting the values: $0.06 = \frac{2}{M_B} \times \frac{78}{60}$
$0.06 = \frac{156}{60 \times M_B}$
$M_B = \frac{156}{60 \times 0.06} = \frac{156}{3.6} = 43.33 \ g \ mol^{-1}$
Thus,the molar mass of the solute is $43.3 \ g \ mol^{-1}$.
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333
ChemistryEasyMCQMHT CET · 2022
Which of the following is $NOT$ a colligative property?
A
Freezing point depression.
B
Osmotic pressure.
C
Vapour pressure of pure benzene.
D
Boiling point elevation.
Solution
(C) Colligative properties are properties of solutions that depend on the number of solute particles in a given amount of solvent and not on the nature of the solute particles. The four main colligative properties are:
$1$. Relative lowering of vapour pressure
$2$. Elevation of boiling point
$3$. Depression of freezing point
$4$. Osmotic pressure
Since 'vapour pressure of pure benzene' is a property of the pure solvent and not a property dependent on the number of solute particles,it is $NOT$ a colligative property.
$1$. Relative lowering of vapour pressure
$2$. Elevation of boiling point
$3$. Depression of freezing point
$4$. Osmotic pressure
Since 'vapour pressure of pure benzene' is a property of the pure solvent and not a property dependent on the number of solute particles,it is $NOT$ a colligative property.
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334
ChemistryEasyMCQMHT CET · 2022
Calculate the freezing point of a $0.05 \ m$ aqueous solution of a non-electrolyte. (in $K$)
A
$186$
B
$272.9$
C
$93$
D
$278$
Solution
(B) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$.
Given $K_f = 1.86 \ K \ kg \ mol^{-1}$ for water and molality $m = 0.05 \ m$.
$\Delta T_f = 1.86 \times 0.05 = 0.093 \ K$.
The freezing point of the solution $T_f$ is calculated as: $T_f = T_f^{\circ} - \Delta T_f$.
Since the freezing point of pure water $T_f^{\circ} = 273 \ K$,we have: $T_f = 273 - 0.093 = 272.907 \ K \approx 272.9 \ K$.
Given $K_f = 1.86 \ K \ kg \ mol^{-1}$ for water and molality $m = 0.05 \ m$.
$\Delta T_f = 1.86 \times 0.05 = 0.093 \ K$.
The freezing point of the solution $T_f$ is calculated as: $T_f = T_f^{\circ} - \Delta T_f$.
Since the freezing point of pure water $T_f^{\circ} = 273 \ K$,we have: $T_f = 273 - 0.093 = 272.907 \ K \approx 272.9 \ K$.
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335
ChemistryEasyMCQMHT CET · 2022
What is the vapour pressure of a solution containing $1.8 \ g$ of glucose in $16.2 \ g$ of water,if the vapour pressure of pure water is $32 \ mm \ Hg$ (in $mm \ Hg$)?
A
$22.2$
B
$26.6$
C
$24.6$
D
$31.7$
Solution
(D) The relative lowering of vapour pressure is given by the formula: $\frac{P_A^{\circ} - P_A}{P_A^{\circ}} = x_B = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A}$.
Given: $W_B$ (glucose) $= 1.8 \ g$,$M_B$ (glucose) $= 180 \ g/mol$,$W_A$ (water) $= 16.2 \ g$,$M_A$ (water) $= 18 \ g/mol$,$P_A^{\circ} = 32 \ mm \ Hg$.
Moles of glucose $(n_B)$ $= \frac{1.8}{180} = 0.01 \ mol$.
Moles of water $(n_A)$ $= \frac{16.2}{18} = 0.9 \ mol$.
Substituting values: $\frac{32 - P_A}{32} = \frac{0.01}{0.9 + 0.01} = \frac{0.01}{0.91} \approx 0.010989$.
$32 - P_A = 32 \times 0.010989 = 0.3516$.
$P_A = 32 - 0.3516 = 31.648 \ mm \ Hg \approx 31.7 \ mm \ Hg$.
Given: $W_B$ (glucose) $= 1.8 \ g$,$M_B$ (glucose) $= 180 \ g/mol$,$W_A$ (water) $= 16.2 \ g$,$M_A$ (water) $= 18 \ g/mol$,$P_A^{\circ} = 32 \ mm \ Hg$.
Moles of glucose $(n_B)$ $= \frac{1.8}{180} = 0.01 \ mol$.
Moles of water $(n_A)$ $= \frac{16.2}{18} = 0.9 \ mol$.
Substituting values: $\frac{32 - P_A}{32} = \frac{0.01}{0.9 + 0.01} = \frac{0.01}{0.91} \approx 0.010989$.
$32 - P_A = 32 \times 0.010989 = 0.3516$.
$P_A = 32 - 0.3516 = 31.648 \ mm \ Hg \approx 31.7 \ mm \ Hg$.
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336
ChemistryMediumMCQMHT CET · 2022
Which among the following equations represents the relation between the cryoscopic constant,depression in freezing point,and molality?
A
$K_f = \frac{m}{\Delta T_f}$
B
$K_f = \Delta T_f \times m$
C
$K_f = \frac{\Delta T_f}{m}$
D
$K_f = \frac{1}{\Delta T_f \times m}$
Solution
(C) The depression in freezing point $(\Delta T_f)$ is directly proportional to the molality $(m)$ of the solution.
$\Delta T_f = K_f \times m$
Where $K_f$ is the cryoscopic constant (molal depression constant).
Rearranging the equation to solve for $K_f$ gives:
$K_f = \frac{\Delta T_f}{m}$
Therefore,the correct option is $C$.
$\Delta T_f = K_f \times m$
Where $K_f$ is the cryoscopic constant (molal depression constant).
Rearranging the equation to solve for $K_f$ gives:
$K_f = \frac{\Delta T_f}{m}$
Therefore,the correct option is $C$.
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337
ChemistryMediumMCQMHT CET · 2022
Calculate the amount of solute dissolved in $612 \ g$ of water at $30^{\circ} C$ if molar mass of solute is $342 \ g \ mol^{-1}$ (Relative vapour pressure lowering is $0.025$ and molar mass of water $18 \ g \ mol^{-1}$ ). (in $g$)
A
$142.5$
B
$270.6$
C
$240.2$
D
$290.7$
Solution
(D) The relative lowering of vapour pressure is given by the formula: $\frac{\Delta P}{P_A^{\circ}} = x_B = \frac{n_B}{n_A + n_B}$.
For dilute solutions,$n_B \ll n_A$,so $\frac{\Delta P}{P_A^{\circ}} \approx \frac{n_B}{n_A} = \frac{W_B \times M_A}{M_B \times W_A}$.
Given: $\frac{\Delta P}{P_A^{\circ}} = 0.025$,$W_A = 612 \ g$,$M_A = 18 \ g \ mol^{-1}$,$M_B = 342 \ g \ mol^{-1}$.
Substituting the values: $0.025 = \frac{W_B \times 18}{342 \times 612}$.
$W_B = \frac{0.025 \times 342 \times 612}{18}$.
$W_B = 0.025 \times 342 \times 34 = 290.7 \ g$.
For dilute solutions,$n_B \ll n_A$,so $\frac{\Delta P}{P_A^{\circ}} \approx \frac{n_B}{n_A} = \frac{W_B \times M_A}{M_B \times W_A}$.
Given: $\frac{\Delta P}{P_A^{\circ}} = 0.025$,$W_A = 612 \ g$,$M_A = 18 \ g \ mol^{-1}$,$M_B = 342 \ g \ mol^{-1}$.
Substituting the values: $0.025 = \frac{W_B \times 18}{342 \times 612}$.
$W_B = \frac{0.025 \times 342 \times 612}{18}$.
$W_B = 0.025 \times 342 \times 34 = 290.7 \ g$.
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338
ChemistryEasyMCQMHT CET · 2022
What is the molar mass of a solute if a solution is prepared by dissolving $4.5 \ g$ of solute in $3 \ dm^3$ of water having an osmotic pressure of $0.25 \ atm$ at $300 \ K$? $(R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1})$
A
$148 \ g \ mol^{-1}$
B
$160 \ g \ mol^{-1}$
C
$172 \ g \ mol^{-1}$
D
$136 \ g \ mol^{-1}$
Solution
(A) The formula for osmotic pressure is $\pi = CRT = \frac{n}{V} RT = \frac{w}{M_B \times V} RT$.
Given: $\pi = 0.25 \ atm$,$w = 4.5 \ g$,$V = 3 \ dm^3$,$T = 300 \ K$,and $R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.25 = \frac{4.5}{M_B \times 3} \times 0.0821 \times 300$.
$0.25 = \frac{4.5 \times 0.0821 \times 100}{M_B}$.
$M_B = \frac{4.5 \times 8.21}{0.25} = 18 \times 8.21 = 147.78 \ g \ mol^{-1}$.
Rounding to the nearest integer,$M_B \approx 148 \ g \ mol^{-1}$.
Given: $\pi = 0.25 \ atm$,$w = 4.5 \ g$,$V = 3 \ dm^3$,$T = 300 \ K$,and $R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.25 = \frac{4.5}{M_B \times 3} \times 0.0821 \times 300$.
$0.25 = \frac{4.5 \times 0.0821 \times 100}{M_B}$.
$M_B = \frac{4.5 \times 8.21}{0.25} = 18 \times 8.21 = 147.78 \ g \ mol^{-1}$.
Rounding to the nearest integer,$M_B \approx 148 \ g \ mol^{-1}$.
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339
ChemistryEasyMCQMHT CET · 2022
Find the amount of solute in a solution obtained by dissolving it in $2.5 \ dm^3$ of water that generates an osmotic pressure of $0.245 \ atm$ at $300 \ K$. (Molar mass of solute $= 58 \ g \ mol^{-1}$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$) (in $g$)
A
$1.75$
B
$1.0$
C
$0.72$
D
$1.44$
Solution
(D) The osmotic pressure formula is $\pi = CRT$,where $C = \frac{n}{V} = \frac{W}{M \times V}$.
Given: $\pi = 0.245 \ atm$,$V = 2.5 \ dm^3$,$T = 300 \ K$,$M = 58 \ g \ mol^{-1}$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.245 = \frac{W}{58 \times 2.5} \times 0.0821 \times 300$.
$0.245 = \frac{W \times 24.63}{145}$.
$W = \frac{0.245 \times 145}{24.63} \approx 1.44 \ g$.
Given: $\pi = 0.245 \ atm$,$V = 2.5 \ dm^3$,$T = 300 \ K$,$M = 58 \ g \ mol^{-1}$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.245 = \frac{W}{58 \times 2.5} \times 0.0821 \times 300$.
$0.245 = \frac{W \times 24.63}{145}$.
$W = \frac{0.245 \times 145}{24.63} \approx 1.44 \ g$.
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340
ChemistryEasyMCQMHT CET · 2022
Calculate the cryoscopic constant $(K_f)$ when $0.8 \ g$ of a non-volatile solute with a molar mass of $64 \ g \ mol^{-1}$ is dissolved in $43 \ g$ of a solvent,lowering the freezing point by $0.34 \ K$.
A
$2.5 \ K \ kg \ mol^{-1}$
B
$2.0 \ K \ kg \ mol^{-1}$
C
$0.85 \ K \ kg \ mol^{-1}$
D
$1.17 \ K \ kg \ mol^{-1}$
Solution
(D) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.8 \ g / 64 \ g \ mol^{-1}}{43 \ g / 1000} = \frac{0.0125 \ mol}{0.043 \ kg} \approx 0.2907 \ mol \ kg^{-1}$.
Given $\Delta T_f = 0.34 \ K$,we have $0.34 = K_f \times 0.2907$.
$K_f = \frac{0.34}{0.2907} \approx 1.169 \ K \ kg \ mol^{-1}$.
Rounding to two decimal places,$K_f \approx 1.17 \ K \ kg \ mol^{-1}$.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.8 \ g / 64 \ g \ mol^{-1}}{43 \ g / 1000} = \frac{0.0125 \ mol}{0.043 \ kg} \approx 0.2907 \ mol \ kg^{-1}$.
Given $\Delta T_f = 0.34 \ K$,we have $0.34 = K_f \times 0.2907$.
$K_f = \frac{0.34}{0.2907} \approx 1.169 \ K \ kg \ mol^{-1}$.
Rounding to two decimal places,$K_f \approx 1.17 \ K \ kg \ mol^{-1}$.
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341
ChemistryEasyMCQMHT CET · 2022
Identify the concentration of the solution from the following such that the values of $\Delta T_{f}$ and $K_{f}$ are the same.
A
$1 \ M$
B
$1 \ m$
C
$\frac{N}{10}$
D
$1 \ N$
Solution
(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Given that $\Delta T_{f} = K_{f}$,we can substitute this into the equation: $K_{f} = K_{f} \times m$.
Dividing both sides by $K_{f}$,we get $m = 1$.
Therefore,the concentration of the solution must be $1 \ m$ (molal).
Given that $\Delta T_{f} = K_{f}$,we can substitute this into the equation: $K_{f} = K_{f} \times m$.
Dividing both sides by $K_{f}$,we get $m = 1$.
Therefore,the concentration of the solution must be $1 \ m$ (molal).
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342
ChemistryEasyMCQMHT CET · 2022
Calculate the amount of solute dissolved in $160 \ g$ of solvent that boils at $85^{\circ}C$. The molar mass of the solute is $120 \ g \ mol^{-1}$. $(K_{b}$ for solvent $= 2.7^{\circ}C \ kg \ mol^{-1}$ and boiling point of pure solvent $= 76^{\circ}C)$ (in $g$)
A
$64$
B
$42$
C
$50$
D
$60$
Solution
(A) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 85^{\circ}C - 76^{\circ}C = 9^{\circ}C$.
Using the formula $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality:
$9 = 2.7 \times \frac{W_{B}}{120} \times \frac{1000}{160}$.
Solving for $W_{B}$:
$W_{B} = \frac{9 \times 120 \times 160}{2.7 \times 1000} = \frac{172800}{2700} = 64 \ g$.
Using the formula $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality:
$9 = 2.7 \times \frac{W_{B}}{120} \times \frac{1000}{160}$.
Solving for $W_{B}$:
$W_{B} = \frac{9 \times 120 \times 160}{2.7 \times 1000} = \frac{172800}{2700} = 64 \ g$.
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343
ChemistryEasyMCQMHT CET · 2022
What is the molality of a solution that has a freezing point depression of $3 \ K$ and a freezing point depression constant of $5 \ K \ kg \ mol^{-1}$ (in $m$)?
A
$0.6$
B
$0.85$
C
$0.7$
D
$0.5$
Solution
(A) The formula for freezing point depression is $\Delta T_f = K_f \cdot m$.
Given: $\Delta T_f = 3 \ K$ and $K_f = 5 \ K \ kg \ mol^{-1}$.
Substituting the values into the equation: $3 = 5 \cdot m$.
Solving for molality $(m)$: $m = \frac{3}{5} = 0.6 \ m$.
Given: $\Delta T_f = 3 \ K$ and $K_f = 5 \ K \ kg \ mol^{-1}$.
Substituting the values into the equation: $3 = 5 \cdot m$.
Solving for molality $(m)$: $m = \frac{3}{5} = 0.6 \ m$.
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344
ChemistryEasyMCQMHT CET · 2022
What is the relation between molality $(m)$ of the solution and molar mass $(M_2)$ of the solute,where $W_2$ is the mass of solute in grams and $W_1$ is the mass of solvent in grams?
A
$m = \frac{M_2 \times W_2}{1000 \times W_1}$
B
$m = \frac{1000 \times W_2}{M_2 \times W_1}$
C
$m = \frac{1000 \times W_1}{M_2 \times W_2}$
D
$m = \frac{M_2 \times W_1}{1000 \times W_2}$
Solution
(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Number of moles of solute $(n_2)$ = $\frac{W_2}{M_2}$.
Mass of solvent in kg = $\frac{W_1}{1000}$.
Therefore,$m = \frac{n_2}{\text{mass of solvent in kg}} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 \times W_2}{M_2 \times W_1}$.
Number of moles of solute $(n_2)$ = $\frac{W_2}{M_2}$.
Mass of solvent in kg = $\frac{W_1}{1000}$.
Therefore,$m = \frac{n_2}{\text{mass of solvent in kg}} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 \times W_2}{M_2 \times W_1}$.
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345
ChemistryEasyMCQMHT CET · 2022
Which among the following statements is $TRUE$ about isotones?
A
These occupy same positions in the modern periodic table.
B
These have different number of neutrons and same number of protons.
C
These have same chemical properties.
D
They are nuclides having the same number of neutrons but different number of protons and hence different mass numbers.
Solution
(D) Isotones are defined as atoms or nuclides that possess the same number of neutrons $(N)$ but different numbers of protons $(Z)$,which consequently leads to different mass numbers $(A = Z + N)$.
Since they have different atomic numbers,they exhibit different chemical properties and occupy different positions in the periodic table.
Since they have different atomic numbers,they exhibit different chemical properties and occupy different positions in the periodic table.
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346
ChemistryEasyMCQMHT CET · 2022
Which of the following is $NOT$ a characteristic of chemisorption?
A
It is specific.
B
Heat released is in the range of $40-200 \ kJ/mol$.
C
Formation of multimolecular layer of adsorbate.
D
It is irreversible.
Solution
(C) Chemisorption (chemical adsorption) involves the formation of strong chemical bonds between the adsorbate and the adsorbent.
$1$. It is highly specific in nature.
$2$. The enthalpy of adsorption is high,typically in the range of $40-200 \ kJ/mol$.
$3$. It is an irreversible process.
$4$. Chemisorption results in the formation of a unimolecular layer,not a multimolecular layer.
Therefore,the formation of a multimolecular layer is a characteristic of physisorption,not chemisorption.
$1$. It is highly specific in nature.
$2$. The enthalpy of adsorption is high,typically in the range of $40-200 \ kJ/mol$.
$3$. It is an irreversible process.
$4$. Chemisorption results in the formation of a unimolecular layer,not a multimolecular layer.
Therefore,the formation of a multimolecular layer is a characteristic of physisorption,not chemisorption.
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347
ChemistryEasyMCQMHT CET · 2022
Which among the following is an example of sorption?
A
$A$. Oxygen gas is passed over finely divided nickel
B
$B$. Charcoal is added to methylene blue solution
C
$C$. Chalk is dipped in ink
D
$D$. Hydrogen gas is passed over platinum
Solution
(C) Sorption is a phenomenon where both adsorption and absorption occur simultaneously.
When chalk is dipped in ink,the ink particles (solute) get adsorbed on the surface of the chalk,while the solvent (water) is absorbed into the bulk of the chalk.
Therefore,this process represents both adsorption and absorption,which is termed as sorption.
When chalk is dipped in ink,the ink particles (solute) get adsorbed on the surface of the chalk,while the solvent (water) is absorbed into the bulk of the chalk.
Therefore,this process represents both adsorption and absorption,which is termed as sorption.
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348
ChemistryMediumMCQMHT CET · 2022
What is the effect of pressure on the extent of adsorption?
A
Increases as pressure increases continuously
B
Increases at the start as pressure increases and then remains constant
C
Decreases as pressure increases
D
No change in the extent of adsorption if pressure increases or decreases
Solution
(B) According to the Freundlich adsorption isotherm,the extent of adsorption $(x/m)$ increases with an increase in pressure $(P)$ at low pressures.
As the pressure increases further,the surface of the adsorbent becomes saturated,and the rate of adsorption decreases until it reaches a point where it becomes independent of pressure (saturation pressure).
Therefore,the extent of adsorption increases at the start and then remains constant at higher pressures.
As the pressure increases further,the surface of the adsorbent becomes saturated,and the rate of adsorption decreases until it reaches a point where it becomes independent of pressure (saturation pressure).
Therefore,the extent of adsorption increases at the start and then remains constant at higher pressures.
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349
ChemistryEasyMCQMHT CET · 2022
Which of the following phenomena is exhibited by adsorption?
A
Exothermic
B
Endothermic
C
Bulk
D
None
Solution
(A) Adsorption is a surface phenomenon where the concentration of molecules increases at the surface rather than in the bulk of a solid or liquid.
During adsorption,there is a decrease in the residual forces of the surface,which leads to a decrease in surface energy.
This decrease in surface energy is released as heat,making adsorption an $ \text{exothermic} $ process.
During adsorption,there is a decrease in the residual forces of the surface,which leads to a decrease in surface energy.
This decrease in surface energy is released as heat,making adsorption an $ \text{exothermic} $ process.
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350
ChemistryEasyMCQMHT CET · 2022
Which of the following conversions is carried out by using a charcoal catalyst?
A
$SO_{2(g)} + Cl_{2(g)} \rightarrow SO_2Cl_{2(l)}$
B
$4HCl + O_2 \rightarrow 2Cl_2 + 2H_2O$
C
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$
D
$2ZnS_{(s)} + 3O_{2(g)} \rightarrow 2ZnO_{(s)} + 2SO_{2(g)}$
Solution
(A) The reaction $SO_{2(g)} + Cl_{2(g)} \rightarrow SO_2Cl_{2(l)}$ is catalyzed by activated charcoal.
This is a classic example of heterogeneous catalysis where charcoal acts as the catalyst.
This is a classic example of heterogeneous catalysis where charcoal acts as the catalyst.
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