Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circC, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below:

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer duration. Class 12 Chemistry: Solutions

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circ

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer durat

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circC, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below:

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer duration. Class 12 Chemistry: Solutions

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circ

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer durat

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to…

Year	2026	2025	2024	Total
Questions	14	20	11	45

Q49 jee_main_2025_02_april_evening Elevation of Boiling Point and Molar Mass Determination

When

1~g

each of compounds AB and

AB₂

are dissolved in

15~g

of water separately, they increased the boiling point of water by

2.7~K

and

1.5~K

respectively. The atomic mass of A (in amu) is

\times 10⁻¹

(Nearest integer) (Given : Molal boiling point elevation constant is

0.5~ K~kg~mol⁻¹

)

Numerical Answer. Answer: 25 to 25

Solution

Related Formula

Δ Tb = Kb \cdot m = Kb \cdot ( wsoluteMsolute \cdot 1000wsolvent )

Core Logic

Both dissolved compounds are non-electrolytes (van 't Hoff factor $i = 1$ $i = 1$). We calculate the molar masses of $AB$ $\mathrm{AB}$ and $AB₂$ $\mathrm{AB_2}$ individually, then solve for the individual atomic masses of elements A and B.

Step 1: Determine Molar Mass of AB

Given $Δ Tb = 2.7~K$ , $wsolute = 1~g$ $w_{\text{solute}} = 1~\mathrm{g}$, $wsolvent = 15~g$ $w_{\text{solvent}} = 15~\mathrm{g}$, and $Kb = 0.5~ K~kg~mol⁻¹$ :

2.7 = 0.5 \times 1MAB \times (1000)/(15)

MAB = (0.5 \times 1000)/(15 \times 2.7) = (500)/(40.5) \approx 12.3457~ g~mol⁻¹

Step 2: Determine Molar Mass of AB_2

Given $Δ Tb = 1.5~K$ , $wsolute = 1~g$ $w_{\text{solute}} = 1~\mathrm{g}$, $wsolvent = 15~g$ $w_{\text{solvent}} = 15~\mathrm{g}$, and $Kb = 0.5$ $K_{\mathrm{b}} = 0.5$:

1.5 = 0.5 \times 1MAB₂ \times (1000)/(15)

MAB₂ = (0.5 \times 1000)/(15 \times 1.5) = (500)/(22.5) \approx 22.2222~ g~mol⁻¹

Step 3: Solve for Atomic Mass of A

Let the atomic masses of elements A and B be $a$ $a$ and $b$ $b$ respectively:

a + b = 12.3457 --- (1)

a + 2b = 22.2222 --- (2)

Subtracting equation (1) from (2):

b = 22.2222 - 12.3457 = 9.8765~amu

Substituting $b$ $b$ back into equation (1):

a = 12.3457 - 9.8765 = 2.4692~amu

Expressing $a$ $a$ in the requested format ( $\times 10⁻¹$ $\times 10^{-1}$):

a = 24.692 \times 10⁻¹ \approx 25 \times 10⁻¹

Pattern Recognition

Mathematical consistency checks: Since $AB₂$ $\mathrm{AB_2}$ has more atoms than $AB$ $\mathrm{AB}$ of identical constituent mass, its molar mass should be higher, leading to a smaller elevation of boiling point for the same mass dissolved, which perfectly matches our $2.7~K arrow 1.5~K$ change.

Chapter Mix

Class 12 Chemistry: Solutions

Q jee_main_2025_02_april_morning Henry's Law Constant and Temperature Dependance

Which of the following graph correctly represents the plots of

KH

at 1 bar gases in water versus temperature?

A.
B.
C.
D.

Solution

Related Formula

Henry's Law formula connects partial pressure to solubility component:

p = KH \cdot x

Core Logic

As temperature increases, gas dissolution is typically exothermic, meaning solubility initially drops, causing the Henry's constant $KH$ $K_{\mathrm{H}}$ to curve upward dynamically before varying at extreme points. For standard non-reactive noble/molecular gases at regular ranges, the magnitude order follows:

KH(He) > KH(N₂) > KH(CH₄)

Step 1: Selection

Graph (4) illustrates the correct relative order and curved profile properly across the given temperature frame.

Pattern Recognition

Higher $KH$ $K_{\mathrm{H}}$ value implies lower solubility of that gas at a given pressure. Helium is notoriously insoluble in water compared to organic or polarizable molecules like methane, hence its plot line must live at the top.

Chapter Mix

Class 12 Chemistry: Solutions

Q32 jee_main_2025_02_april_morning Raoult's Law and Vapour Pressure

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is

200 ~mmHg

and that of the solution is

500 ~mmHg

. The vapour pressure of pure B and the least volatile component of the solution, respectively, are :

A. $(1) 1400 mmHg, A$ $(1)\ 1400\ \mathrm{mmHg},\ \text{A}$
B. $(2) 1400 mmHg, B$ $(2)\ 1400\ \mathrm{mmHg},\ \text{B}$
C. $(3) 600 mmHg, B$ $(3)\ 600\ \mathrm{mmHg},\ \text{B}$
D. $(4) 600 mmHg, A$ $(4)\ 600\ \mathrm{mmHg},\ \text{A}$

Solution

Related Formula

Raoult's Law for a multi-component solution mixtures:

PS = PA⁰ \cdot XA + PB⁰ \cdot XB

Core Logic

Let's determine the mole fractions first based on the input molar amounts:

XA = (1)/(1+3) = (1)/(4), XB = (3)/(1+3) = (3)/(4)

Substitute the known properties into Raoult's equation block:

500 = 200 \times (1)/(4) + PB⁰ \times (3)/(4)

500 = 50 + PB⁰ \times (3)/(4)

450 = PB⁰ \times (3)/(4) PB⁰ = 600 ~mmHg

Comparing pure state components pressures: $PA⁰ = 200 ~mmHg$ $P_{\mathrm{A}}^0 = 200 \mathrm{~mmHg}$ and $PB⁰ = 600 ~mmHg$ $P_{\mathrm{B}}^0 = 600 \mathrm{~mmHg}$. Lower vapor pressure indicates stronger intermolecular cohesion, making A the least volatile component.

Step 1: Finalization

Thus, the vapour pressure of pure B is $600~mmHg$ $600\mathrm{~mmHg}$ and the least volatile chemical is A.

Pattern Recognition

Volatility is directly proportional to pure vapor pressure ( $P⁰$ $P^0$). Don't mix up 'least volatile' with 'lowest mole fraction'—always evaluate based solely on the isolated values of $P⁰$ $P^0$.

Chapter Mix

Class 12 Chemistry: Solutions

Q49 jee_main_2025_07_april_morning Van't Hoff Factor

The percentage dissociation of a salt

(MX₃)

solution at given temperature (van't Hoff factor

i = 2

) is ______ %. (Nearest integer)

Numerical Answer. Answer: 33 to 33

Solution

Related Formula

i = 1 + (n - 1)α

Core Logic

For the salt $MX₃$ $\mathrm{MX}_3$:

MX₃ arrow M³⁺ + 3X^-

Number of ions formed per formula unit, $n = 1 + 3 = 4$ $n = 1 + 3 = 4$.

Given $i = 2$ $i = 2$, let's find the degree of dissociation ( $α$ $\alpha$):

i = 1 + (4 - 1)α

2 = 1 + 3α 3α = 1 α = (1)/(3) \approx 0.3333

Percentage dissociation:

% dissociation = α \times 100 = 33.33 % \approx 33 %

Pattern Recognition

Shortcut: For $MX₃$ $MX_3$ dissociating into $4$ $4$ ions, $α = (i - 1) / 3$ $\alpha = (i - 1) / 3$. Since $i = 2$ $i = 2$, $α = 1/3 = 33%$ $\alpha = 1/3 = 33\%$ directly.

Evaluation Rubric / Model Answer

Simple formula application verifying the degree of dissociation to obtain a precise $33$ $33$ percent.

Chapter Mix

Class 12 Chemistry: Solutions

Q33 jee_main_2025_08_april_evening Azeotropic Mixtures

Which of the following binary mixtures does not show the behaviour of minimum boiling azeotropes?

A. $H₂O + CH₃COC₂H₅$
B. $C₆H₅OH + C₆H₅NH₂$
C. $CS₂ + CH₃COCH₃$
D. $CH₃OH + CHCl₃$ $\text{CH}_3\text{OH} + \text{CHCl}_3$

Solution

Core Logic

Let's connect deviation behaviors to azeotropic styles:

Minimum Boiling Azeotropes: Formed by liquid binary solutions that display a strong positive deviation from Raoult's Law. In these mixtures, inter-molecular forces between components ( $A-B$ $A-B$) are weaker than pure self-interactions ( $A-A$ $A-A$ or $B-B$ $B-B$).
Maximum Boiling Azeotropes: Formed by liquid binary mixtures showing a notable negative deviation from Raoult's Law. Here, new inter-molecular interactions ( $A-B$ $A-B$) become significantly stronger.

Analyzing Phenol ( $C₆H₅OH$ $C_6H_5OH$) + Aniline ( $C₆H₅NH₂$ $C_6H_5NH_2$): The phenolic $-OH$ $-\text{OH}$ proton forms strong intermolecular hydrogen bonds with the lone pair of the $-NH₂$ $-\text{NH}_2$ group of aniline. These new forces exceed the initial individual fluid bonds, lowering vapor pressure below ideal expectations (negative deviation) and creating a maximum boiling azeotrope.

Pattern Recognition

Phenol + Aniline, and Chloroform + Acetone are classic textbook models of strong negative deviation from Raoult's Law. Negative deviation explicitly pairs with maximum boiling azeotropes, standing out instantly against minimum boiling selections.

Chapter Mix

Class 12 Chemistry: Solutions

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Solutions Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Determine Molar Mass of AB

Step 2: Determine Molar Mass of AB_2

Step 3: Solve for Atomic Mass of A

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Selection

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finalization

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Core Logic

Pattern Recognition

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