Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate in 750 mL of solution. The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in g mol-1: H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27 and S=32]

For dilute solutions, Molarity (M) approx Molality (m). Thus, Delta Tb propto i x M Solution Molarity (M = (W/Mw) x (1000/V)) Effective Concentration (i x M) (I) Glucose (i=1) (2.2/180) x (1000/125) = 0.098 M 1 x 0.098 = 0.098 (II) CaCl2 (i=3) (1.9/111) x (1000/250) = 0.068 M 3 x 0.068 = 0.204 (III) Urea (i=1) (9/60) x (1000/500) = 0.3 M 1 x 0.3 = 0.300 (IV) Al2(SO4)3 (i=5) (20.5/342) x (1000/750) = 0.080 M 5 x 0.080 = 0.400 Comparing the values of i x M: 0.098 < 0.204 < 0.300 < 0.400 Glucose (I) < CaCl2 (II) < Urea (III) < Al2(SO4)3 (IV) Order of boiling points corresponds directly to effective concentration. Thus: I < II < III < IV. For multiple salt mixtures, always multiply molarity by the Van't Hoff factor (i). Do not compare just raw mass or raw molarity. Class 12 Chemistry: Solutions

Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate

For dilute solutions, Molarity (M) approx Molality (m). Thus, Delta Tb propto i x M Solution Molarity (M = (W/Mw) x (1000/V)) Effective Concentration (i x M) (I) Glucose (i=1) (2.2/180) x (1000/125) = 0.098 M 1 x 0.098 = 0.098 (II) CaCl2 (i=3) (1.9/111) x (1000/250) = 0.068 M 3 x 0.068 = 0.204 (III) Urea (i=1) (9/60) x (1000/500) = 0.3 M 1 x 0.3 = 0.300 (IV) Al2(SO4)3 (i=5) (20.5/342) x (1000/750) = 0.080 M 5 x 0.080 = 0.400 Comparing the values of i x M: 0.098 < 0.204 < 0.300 < 0.400 Glucose (I

Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate in 750 mL of solution. The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in g mol-1: H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27 and S=32]

For dilute solutions, Molarity (M) approx Molality (m). Thus, Delta Tb propto i x M Solution Molarity (M = (W/Mw) x (1000/V)) Effective Concentration (i x M) (I) Glucose (i=1) (2.2/180) x (1000/125) = 0.098 M 1 x 0.098 = 0.098 (II) CaCl2 (i=3) (1.9/111) x (1000/250) = 0.068 M 3 x 0.068 = 0.204 (III) Urea (i=1) (9/60) x (1000/500) = 0.3 M 1 x 0.3 = 0.300 (IV) Al2(SO4)3 (i=5) (20.5/342) x (1000/750) = 0.080 M 5 x 0.080 = 0.400 Comparing the values of i x M: 0.098 < 0.204 < 0.300 < 0.400 Glucose (I) < CaCl2 (II) < Urea (III) < Al2(SO4)3 (IV) Order of boiling points corresponds directly to effective concentration. Thus: I < II < III < IV. For multiple salt mixtures, always multiply molarity by the Van't Hoff factor (i). Do not compare just raw mass or raw molarity. Class 12 Chemistry: Solutions

Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate

For dilute solutions, Molarity (M) approx Molality (m). Thus, Delta Tb propto i x M Solution Molarity (M = (W/Mw) x (1000/V)) Effective Concentration (i x M) (I) Glucose (i=1) (2.2/180) x (1000/125) = 0.098 M 1 x 0.098 = 0.098 (II) CaCl2 (i=3) (1.9/111) x (1000/250) = 0.068 M 3 x 0.068 = 0.204 (III) Urea (i=1) (9/60) x (1000/500) = 0.3 M 1 x 0.3 = 0.300 (IV) Al2(SO4)3 (i=5) (20.5/342) x (1000/750) = 0.080 M 5 x 0.080 = 0.400 Comparing the values of i x M: 0.098 < 0.204 < 0.300 < 0.400 Glucose (I

Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL…

Year	2026	2025	2024	Total
Questions	14	20	11	45

Solution	Molarity ( $M = (W)/(Mw) \times (1000)/(V)$ )	Effective Concentration ( $i \times M$ $i \times M$)
(I) Glucose ( $i=1$ $i=1$)	$(2.2)/(180) \times (1000)/(125) = 0.098 M$	$1 \times 0.098 = 0.098$ $1 \times 0.098 = 0.098$
(II) $CaCl₂$ $CaCl_2$ ( $i=3$ $i=3$)	$(1.9)/(111) \times (1000)/(250) = 0.068 M$	$3 \times 0.068 = 0.204$ $3 \times 0.068 = 0.204$
(III) Urea ( $i=1$ $i=1$)	$(9)/(60) \times (1000)/(500) = 0.3 M$	$1 \times 0.3 = 0.300$ $1 \times 0.3 = 0.300$
(IV) $Al₂(SO₄)₃$ $Al_2(SO_4)_3$ ( $i=5$ $i=5$)	$(20.5)/(342) \times (1000)/(750) = 0.080 M$	$5 \times 0.080 = 0.400$ $5 \times 0.080 = 0.400$

Q46 jee_main_2025_04_april_evening Concentration Terms

Sea water, which can be considered as a 6 molar (6 M) solution of NaCl, has a density of

2~g~mL⁻¹

. The concentration of dissolved oxygen

(O₂)

in sea water is

5.8~ppm

. Then the concentration of dissolved oxygen

(O₂)

in sea water, is

x \times 10⁻⁴m

.

x =

_______. (Nearest integer) Given: Molar mass of NaCl is

58.5~g~mol⁻¹

Molar mass of

O₂

is

32~g~mol⁻¹

Numerical Answer. Answer: 1.9 to 2.1

Solution

Related Formula

ppm = mass of solutemass of solution \times 10⁶

Molality (m) = moles of solutemass of solvent in kg

Core Logic

Consider $1000 ~mL$ $1000 \mathrm{~mL}$ of seawater solution:

Mass of solution = Volume \times density = 1000 \times 2 = 2000 ~g

Mass of NaCl = 6 moles \times 58.5 = 351 ~g

Mass of solvent (water) = 2000 - 351 = 1649 ~g = 1.649 ~kg

Compute the mass and moles of dissolved $O₂$ $O_2$ using the ppm value:

ppm = 5.8 = mass of O₂2000 \times 10⁶ mass of O₂ = 1.16 \times 10⁻² ~g

moles of O₂ = 1.16 \times 10⁻²32 = 3.625 \times 10⁻⁴ moles

Determine the molality ( $m$ $m$) of oxygen:

molality = 3.625 \times 10⁻⁴1.649 \approx 2.19 \times 10⁻⁴ ~m

Matching the pattern $x \times 10⁻⁴m$ $\mathbf{x} \times 10^{-4}\mathrm{m}$, we get $x \approx 2.19$ $\mathbf{x} \approx 2.19$. The nearest integer is 2.

Pattern Recognition

For high concentration saline solutions, the mass of the solvent drops significantly below the total mass of the solution. Be careful to subtract the solute weight ( $351 ~g$ $351 \mathrm{~g}$) before computing molality.

Chapter Mix

Class 12 Chemistry: Solutions

Q26 jee_main_2025_04_april_morning Reverse Osmosis

XY is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration

c₁

and

c₂

(

c₁ > c₂

)

mol~L⁻¹

. For the reverse osmosis to take place identify the correct condition (Here

p₁

and

p₂

are pressures applied on chamber 1 and 2):

Reverse Osmosis cell partition diagram for Q26 - JEE Main 2025 Morning — The diagram illustrates two chambers separated by a membrane XY containing sugar solutions of concentrations c1 and c2.

A. $(B) and (D) only$ $\text{(B) and (D) only}$
B. $(A) and (D) only$ $\text{(A) and (D) only}$
C. $(A) and (C) only$ $\text{(A) and (C) only}$
D. $(C) only$ $\text{(C) only}$

Solution

Related Formula

$π = c R T$ $\pi = c R T$

where $π$ $\pi$ is the osmotic pressure of the solution.

Core Logic

Given that $c₁ > c₂$ $c_1 > c_2$, chamber 1 has a higher concentration of solute than chamber 2. Under normal conditions, solvent molecules spontaneously flow from lower concentration (chamber 2) to higher concentration (chamber 1) via osmosis.

To achieve reverse osmosis, the solvent must flow in the opposite direction—from chamber 1 to chamber 2. This requires applying an external pressure on the higher concentration side (chamber 1) that exceeds its osmotic pressure $π$ $\pi$.

Condition for Reverse Osmosis: p₁ > π

Cellophane and parchment paper both act as suitable semi-permeable membranes for this setup. Thus, statements (A) and (C) are correct.

Pattern Recognition

Reverse osmosis always requires external pressure applied on the concentrated solution side ( $chigh$ $c_{\text{high}}$) such that $Papplied > π$ $P_{\text{applied}} > \pi$.

Chapter Mix

Class 12 Chemistry: Solutions

Q34 jee_main_2025_07_april_evening Raoult's Law and Liquid-Vapour Composition

Liquid

A

and

B

form an ideal solution. The vapour pressure of pure liquids

A

and

B

are 350 and

750 mm Hg

respectively at the same temperature. If

xA

and

xB

are the mole fraction of

A

and

B

in solution while

yA

and

yB

are the mole fraction of

A

and

B

in vapour phase then:

A. $xAxB < yAyB$
B. $xAxB = yAyB$
C. $xAxB > yAyB$
D. $(xA - yA) < (xB - yB)$

Solution

Related Formula

PA = yA Ptotal = xA P⁰A

PB = yB Ptotal = xB P⁰B

Dividing both partial pressure formulations yields:

yAyB = ( P⁰AP⁰B) \cdot xAxB

Core Logic

Given pure saturation thresholds:

P⁰A = 350 mm Hg, P⁰B = 750 mm Hg

Comparing pure component volatility profiles:

P⁰A < P⁰B P⁰AP⁰B < 1

Substituting this inequality into the ratio formula gives:

yAyB < 1 \cdot xAxB yAyB < xAxB

Step 1: Rearranging Ratio Forms

Inverting the inequality expression fields safely yields:

xAxB > yAyB

Pattern Recognition

Konovalov's Rule Shortcut: The vapour phase is always enriched with the more volatile component. Since component $B$ $\text{B}$ has a higher pure vapour pressure ( $750 > 350$ $750 > 350$), it will be preferentially enriched in the vapour phase, meaning $yB/yA > xB/xA$ . Reversing the fractions directly matches option (3).

Chapter Mix

Class 12 Chemistry: Solutions

Q38 jee_main_2025_07_april_evening Azeotropes and Liquid Mixtures

Match List-I with List-II

List-I	List-II	(A) Solution of chloroform and acetone	(I) Minimum boiling azeotrope	(B) Solution of ethanol and water	(II) Dimerizes	(C) Solution of benzene and toluene	(III) Maximum boiling azeotrope	(D) Solution of acetic acid in benzene	(IV) $Δ Vmix=0$ $\Delta V{\text{mix}}=0$ Choose the correct answer from the options given below: A. $(A)-(III), (B)-(I), (C)-(IV), (D)-(II)$ B. $(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$ C. $(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$ D. $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$ Solution Related Formula $Negative Deviation from Raoult's Law Maximum Boiling Azeotrope$ $Positive Deviation from Raoult's Law Minimum Boiling Azeotrope$ $Ideal Solution Δ Vmix = 0$ Core Logic Evaluating molecular interaction behaviors: - (A) Chloroform and Acetone: Form intermolecular hydrogen bonds, showing negative deviation from Raoult's law, which forms a maximum boiling azeotrope $arrow$ $\rightarrow$ (III) - (B) Ethanol and Water: Exhibit hydrogen bond disruptions, leading to positive deviation, forming a minimum boiling azeotrope $arrow$ $\rightarrow$ (I) - (C) Benzene and Toluene: Highly structurally similar, forming a near-perfect ideal solution where $Δ Vmix = 0$ $\Delta V{\text{mix}} = 0$ $arrow$ $\rightarrow$ (IV) - (D) Acetic acid in benzene: Acetic acid undergoes intermolecular hydrogen bonding inside the non-polar solvent, causing it to dimerize $arrow$ $\rightarrow$ (II) Step 1: Alignment Selection Combining the pairs gives: (A)-(III), (B)-(I), (C)-(IV), (D)-(II). Pattern Recognition Liquid properties shortcut: Acetic acid in benzene is a classic textbook example of dimerization ( $i=0.5$ $i=0.5$). Benzene-toluene is the most famous ideal solution pair. Recognizing either instantly shortcuts the options list to the correct key. Chapter Mix Class 12 Chemistry: Solutions Q46 jee_main_2025_24_jan_evening Abnormal Molar Masses and Van't Hoff Factor The observed and normal masses of compound $MX₂$ $\mathrm{MX}_2$ are 65.6 and 164 respectively. The percent degree of ionisation of $MX₂$ $\mathrm{MX}_2$ is ____ %. (Nearest integer) Numerical Answer. Answer: 75 to 75 Solution Related Formula $i = Normal Molar MassObserved Molar Mass$ $i = 1 + (n - 1)α$ $$i = 1 + (n - 1)\alpha$$ Core Logic Calculate the van 't Hoff factor ( $i$ $i$): $i = (164)/(65.6) = 2.5$ $$i = \frac{164}{65.6} = 2.5$$ Set up the dissociation equilibrium for the electrolyte $MX₂$ $\mathrm{MX}_2$: $MX₂ arrow M²⁺ + 2X⁻$ Here, 1 molecule dissociates into $n = 1 + 2 = 3$ $n = 1 + 2 = 3$ ions. Relate $i$ $i$ to the degree of ionization ( $α$ $\alpha$): $i = 1 + (3 - 1)α = 1 + 2α$ $$i = 1 + (3 - 1)\alpha = 1 + 2\alpha$$ $2.5 = 1 + 2α 2α = 1.5 α = 0.75$ Convert to a percentage: $Percent dissociation = 0.75 \cdot 100 = 75%$ Pattern Recognition For a salt that dissociates into three ions (like $MX₂$ $\mathrm{MX}_2$), the relationship simplifies to $i = 1 + 2α$ $i = 1 + 2\alpha$. Calculating $i$ $i$ from the ratio of the molar masses lets you find $α$ $\alpha$ directly. Chapter Mix Class 12 Chemistry: Solutions ← Previous Page 6 of 9 Next → More Solutions Questions — jee_main_2026_28_january_evening Practice all Solutions previous-year questions → XY is the membrane / partition between two chambers 1 A solution is made by mixing one mole of volatile Given below are two statements: Statement (I): NaCl is added Given below are two statements : Statement (I): Molal depression Which of the following binary mixtures does not show the For the weak acid dissociation equation: The freezing point depression Liquid and form an ideal solution. The vapour pressure of 'x' g of NaCl is added to water in a Match List-I with List-II <div style="overflow-x: auto; margin: 1rem 0;"><table What is the freezing point depression constant of a solvent, 2 moles each of ethylene glycol and glucose are dissolved Assume a living cell with of glucose solution (aqueous). This Sea water, which can be considered as a 6 molar The observed and normal masses of compound are 65.6 and When each of compounds AB and are dissolved in of The percentage dissociation of a salt solution at given temperature At , of aqueous solution is mixed with of aqueous Which one of the following graphs accurately represents the plot At T(K), moles of liquid A and moles of liquid Two liquids A and B form an ideal solution at A solution is prepared by dissolving of a non-volatile non-electrolyte Consider a solution of dissolved in water in a closed Elements P and Q form two types of non-volatile, non-ionizable 'W' g of a non-volatile electrolyte solid solute of molar Identify the mixture that shows positive deviations from Raoult's Law A solution of two miscible liquids showing negative deviation from At 298 K, the mole percentage of in air is Given below are two statements: Statement I: The Henry's law If a substance 'A' dissolves in solution of a mixture The osmotic pressure of a living cell is 12 atm Two liquids A and B form an ideal solution. At A substance 'X' (1.5 g) dissolved in 150 g of We have three aqueous solutions of labelled as 'A', 'B' What happens to freezing point of benzene when small quantity Identify the correct statements : (A) The molality of 2.5 The solution from the following with highest depression in freezing Mixture of chloroform and acetone forms a solution with negative A solution of is by mass and has a density The osmotic pressure of a dilute solution is at . 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List-I

List-II

(A) Solution of chloroform and acetone

(I) Minimum boiling azeotrope

(B) Solution of ethanol and water

(II) Dimerizes

(C) Solution of benzene and toluene

(III) Maximum boiling azeotrope

(D) Solution of acetic acid in benzene

(IV)

Δ Vmix=0

Choose the correct answer from the options given below:

A. $(A)-(III), (B)-(I), (C)-(IV), (D)-(II)$
B. $(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$
C. $(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$
D. $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$

Solution

Related Formula

Negative Deviation from Raoult's Law Maximum Boiling Azeotrope

Positive Deviation from Raoult's Law Minimum Boiling Azeotrope

Ideal Solution Δ Vmix = 0

Core Logic

Evaluating molecular interaction behaviors: - (A) Chloroform and Acetone: Form intermolecular hydrogen bonds, showing negative deviation from Raoult's law, which forms a maximum boiling azeotrope $arrow$ $\rightarrow$ (III) - (B) Ethanol and Water: Exhibit hydrogen bond disruptions, leading to positive deviation, forming a minimum boiling azeotrope $arrow$ $\rightarrow$ (I) - (C) Benzene and Toluene: Highly structurally similar, forming a near-perfect ideal solution where $Δ Vmix = 0$ $\Delta V{\text{mix}} = 0$ $arrow$ $\rightarrow$ (IV) - (D) Acetic acid in benzene: Acetic acid undergoes intermolecular hydrogen bonding inside the non-polar solvent, causing it to dimerize $arrow$ $\rightarrow$ (II)

Step 1: Alignment Selection

Combining the pairs gives: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Pattern Recognition

Liquid properties shortcut: Acetic acid in benzene is a classic textbook example of dimerization ( $i=0.5$ $i=0.5$). Benzene-toluene is the most famous ideal solution pair. Recognizing either instantly shortcuts the options list to the correct key.

Chapter Mix

Class 12 Chemistry: Solutions

Q46 jee_main_2025_24_jan_evening Abnormal Molar Masses and Van't Hoff Factor

The observed and normal masses of compound

MX₂

are 65.6 and 164 respectively. The percent degree of ionisation of

MX₂

is ____ %. (Nearest integer)

Numerical Answer. Answer: 75 to 75

Solution

Related Formula

i = Normal Molar MassObserved Molar Mass

i = 1 + (n - 1)α

Core Logic

Calculate the van 't Hoff factor ( $i$ $i$):

i = (164)/(65.6) = 2.5

Set up the dissociation equilibrium for the electrolyte $MX₂$ $\mathrm{MX}_2$:

MX₂ arrow M²⁺ + 2X⁻

Here, 1 molecule dissociates into $n = 1 + 2 = 3$ $n = 1 + 2 = 3$ ions.

Relate $i$ $i$ to the degree of ionization ( $α$ $\alpha$):

i = 1 + (3 - 1)α = 1 + 2α

2.5 = 1 + 2α 2α = 1.5 α = 0.75

Convert to a percentage:

Percent dissociation = 0.75 \cdot 100 = 75%

Pattern Recognition

For a salt that dissociates into three ions (like $MX₂$ $\mathrm{MX}_2$), the relationship simplifies to $i = 1 + 2α$ $i = 1 + 2\alpha$. Calculating $i$ $i$ from the ratio of the molar masses lets you find $α$ $\alpha$ directly.

Chapter Mix

Class 12 Chemistry: Solutions

Solution & Explanation

Related Formula

Core Logic

Step 1: Compare Effective Concentrations

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

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Related Formula

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Related Formula

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Related Formula

Core Logic

Step 1: Rearranging Ratio Forms

Pattern Recognition

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Related Formula

Core Logic

Step 1: Alignment Selection

Pattern Recognition

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Related Formula

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