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

Q jee_main_2025_24_jan_morning Depression in Freezing Point

Consider the given plots of vapour pressure (VP) vs temperature (T/K) Which amongst the following options is correct graphical representation showing

Δ Tf

, depression in the freezing point of solvent in a solution?

A.
B.
C.
D.

Solution

Core Logic

Let the initial pressure of the reactant $N₂O₅$ $\mathrm{N}_2\mathrm{O}_{5}$ be $P₀$ $P_0$.

Setting up the stoichiometric reaction table:

$arraylcccc & N₂O5(g) & arrow & 2NO2(g) & + & (1)/(2)O2(g) Initially (t = 0): & P₀ & & 0 & & 0 At time t: & P₀ - x & & 2x & & (x)/(2) array$

The total pressure of the gaseous mixture at any time $t$ $t$ is given by:

Ptotal = (P₀ - x) + 2x + (x)/(2) = P₀ + (3x)/(2)

When $50%$ $50\%$ of the reaction is completed, the change in the reactant's pressure is:

x = 0.5 P₀ = (P₀)/(2)

Substituting the value of $x$ $x$ into the total pressure expression:

Ptotal = P₀ + (3)/(2)((P₀)/(2)) = P₀ + (3P₀)/(4) = (7)/(4)P₀

Pattern Recognition

Track the change in the total pressure carefully using stoichiometric coefficients. For a $50%$ $50\%$ completion step, substitute the fractional equivalent ( $x = 0.5 P₀$ $x = 0.5 P_0$) directly into your total pressure expression.

Q42 jee_main_2025_28_jan_evening Osmosis and Osmotic Pressure

Assume a living cell with

0.9%

(ω/ω)

of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will:

A. Shrink since solution is $0.5% (ω/ω)$ $0.5\%\ (\omega/\omega)$
B. Shrink since solution is $0.45% (ω/ω)$ $0.45\%\ (\omega/\omega)$ as a result of association of glucose molecules (due to hydrogen bonding)
C. Swell up since solution is $1%$ $1\%$
D. Show no change in volume since solution is $0.9% (ω/ω)$ $0.9\%\ (\omega/\omega)$

Solution

Related Formula

Mass percentage from mole fraction calculation:

% w/w = (x₁ \cdot M₁)/(x₁ \cdot M₁ + x₂ \cdot M₂) \times 100

Core Logic

Inside the living cell, glucose concentration is $0.9% w/w$ $0.9\%\text{ w/w}$.

The surrounding solution has equal mole fractions of glucose and water ( $xglucose = 0.5$ $x_{\text{glucose}} = 0.5$, $xwater = 0.5$ $x_{\text{water}} = 0.5$).

Let's calculate the mass percentage of the outer solution:

Mass of glucose component = $0.5 \times 180 = 90 g$ $0.5 \times 180 = 90\mathrm{\ g}$
Mass of water component = $0.5 \times 18 = 9 g$ $0.5 \times 18 = 9\mathrm{\ g}$
Total solution mass = $90 + 9 = 99 g$ $90 + 9 = 99\mathrm{\ g}$

Step 1: Concentration Determination and Osmosis Profile

Outer mass percentage:

% w/w = (90)/(99) \times 100 \approx 90.9%

Because the external environment is highly concentrated (hypertonic) compared to the inner cell ( $0.9%$ $0.9\%$), water flows out of the cell via exosmosis, causing the cell to shrink.

Note: Because the reasoning establishes shrinkage but the quantitative figures in the options are highly mismatched, this question is officially designated as a Bonus question.

Pattern Recognition

An equal mole fraction solution of a high-molar-mass solute (glucose, $180 g/mol$ $180\mathrm{\ g/mol}$) and a low-molar-mass solvent (water, $18 g/mol$ $18\mathrm{\ g/mol}$) is always extremely concentrated by mass. Placing a standard living cell into such a hypertonic solution inevitably causes fluid loss and cellular shrinkage.

Chapter Mix

Class 12 Chemistry: Solutions

Q jee_main_2025_29_jan_morning Abnormal Molar Mass and van't Hoff Factor

1.24 ~g

of

AX₂

(molar mass

124 ~g ~mol⁻¹

) is dissolved in

1 ~kg

of water to form a solution with boiling point of

100.0156° C

, while

25.4 ~g

of

AY₂

(molar mass

250 ~g ~mol⁻¹

) in

2 ~kg

of water constitutes a solution with a boiling point of

100.0260° C

.

Kb(H₂O) = 0.52 K kg mol⁻¹

Which of the following is correct?

A. $AX2$ $\mathrm{AX}2$ and $AY₂$ $\mathrm{AY}_2$ (both) are completely unionised.
B. $AX₂$ $\mathrm{AX}_2$ and $AY₂$ $\mathrm{AY}_2$ (both) are fully ionised.
C. $AX₂$ $\mathrm{AX}_2$ is completely unionised while $AY₂$ $\mathrm{AY}_2$ is fully ionised.
D. $AX₂$ $\mathrm{AX}_2$ is fully ionised while $AY₂$ $\mathrm{AY}_2$ is completely unionised.

Solution

Formulas Used

Elevation in boiling point formula involving the van't Hoff factor ( $i$ $i$):

Δ Tb = i \cdot Kb \cdot m

Where:

$Δ Tb = Tb - Tb^°$ $\Delta T_b = T_b - T_b^\circ$ (Boiling point elevation)
$Kb = Ebullioscopic constant$ $K_b = \text{Ebullioscopic constant}$
$m = Molality of the solution ( moles of solutemass of solvent in kg)$

Core Logic

Step 1: Evaluate solution system $AX₂$ $\mathrm{AX}_2$

Δ Tb = 100.0156^ - 100.0000^ = 0.0156^

Molality m₁ = 1.24 g / 124 g mol⁻¹1 kg = 0.01 mol/kg

Using the elevation formula:

0.0156 = iAX₂ \cdot 0.52 \cdot 0.01

iAX₂ = (0.0156)/(0.0052) = 3

Since theoretical dissociation of $AX₂ arrow A²⁺ + 2X^-$ produces $3$ $3$ particles, $i = 3$ $i = 3$ implies that $AX₂$ $\mathrm{AX}_2$ is fully ionised.

---

Step 2: Evaluate solution system $AY₂$ $\mathrm{AY}_2$

Δ Tb = 100.0260^ - 100.0000^ = 0.0260^

Molality m₂ = 25.4 g / 250 g mol⁻¹2 kg = 0.0508 mol/kg

Using the elevation formula:

0.0260 = iAY₂ \cdot 0.52 \cdot 0.0508

iAY₂ = (0.0260)/(0.0264) \approx 1

Since $i \approx 1$ $i \approx 1$, it behaves as a non-electrolyte, meaning $AY₂$ $\mathrm{AY}_2$ is completely unionised.

Thus, $AX₂$ $\mathrm{AX}_2$ is fully ionised while $AY₂$ $\mathrm{AY}_2$ is completely unionised.

Pattern Recognition

A van't Hoff factor matching the complete stoichiometric ion count ( $i = 3$ $i = 3$ for $AX₂$ $\mathrm{AX}_2$) confirms complete ionisation, whereas a factor near unity ( $i = 1$ $i = 1$) indicates no dissociation into separate ions.

Correct Option: (D)

Q jee_main_2025_29_jan_morning van't Hoff Factor

If

A₂B

is

30%

ionised in an aqueous solution, then the value of van't Hoff factor (i) is ________

\times 10⁻¹

.

Numerical Answer. Answer: 16 to 16

Solution

Related Formula

i = 1 + (y - 1)α

Core Logic

For electrolyte $A₂B$ $\mathrm{A}_2\mathrm{B}$ undergoing dissociation:

A₂B arrow 2A^+ + B²⁻

Total count of ions produced per molecule $y = 3$ $y = 3$. Given degree of dissociation $α = 30% = 0.3$ $\alpha = 30\% = 0.3$.

Substituting into the formula:

i = 1 + (3 - 1) \cdot 0.3

i = 1 + 2 \cdot 0.3 = 1 + 0.6 = 1.6

Expressing the value in the requested format:

1.6 = 16 \times 10⁻¹

Thus, the integer value to enter is 16.

Pattern Recognition

Always calculate total stoichiometric species $y$ $y$ carefully before executing the linear factor combination to avoid basic arithmetic errors.

Chapter Mix

Class 12 Chemistry: Solutions

Q75 jee_main_2024_01_february_morning Abnormal Molar Masses

We have three aqueous solutions of

NaCl

labelled as 'A', 'B' and 'C' with concentration

0.1 ~M

,

0.01 ~M

&

0.001 ~M

, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the order.

A. $iA < iB < iC$ $i_A < i_B < i_C$
B. $iA < iC < iB$ $i_A < i_C < i_B$
C. $iA = iB = iC$ $i_A = i_B = i_C$
D. $iA > iB > iC$ $i_A > i_B > i_C$

Solution

Core Logic

For a strong electrolyte like $NaCl$ $NaCl$, the theoretical van 't Hoff factor $itheory$ $i_{theory}$ is 2 ( $Na^+$ $Na^+$ and $Cl^-$ $Cl^-$). However, in real solutions, ion-pairing (interionic attraction) occurs. At higher concentrations, the ions are closer together, leading to stronger interionic attractions that reduce the effective number of independent particles, thus lowering the observed $i$ $i$. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed $i$ $i$ approaches the theoretical value.

Step 1: Correlate Concentration with $i$ $i$

Higher concentration $\implies$ more ion-pairing $\implies$ lower $i$ $i$. Given concentrations: $A = 0.1 ~M$ $A = 0.1 \mathrm{~M}$ (highest concentration) $B = 0.01 ~M$ $B = 0.01 \mathrm{~M}$ $C = 0.001 ~M$ $C = 0.001 \mathrm{~M}$ (most dilute)

Therefore, the actual $i$ $i$ values follow the reverse order of concentration: $iA (0.1 ~M) < iB (0.01 ~M) < iC (0.001 ~M)$ .

Execution

Salt	Values of i (for different conc. of a Salt)
Salt	0.1 M	0.01 M	0.001 M
NaCl	1.87	1.94	1.97

$i$ $i$ approaches 2 as the solution becomes very dilute.

Pattern Recognition

For strong electrolytes, effective dissociation (and thus $i$ $i$) increases as dilution increases because ions interfere with each other less.

Chapter Mix

Class 12 Chemistry: Solutions

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

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Solution

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Core Logic

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Pattern Recognition

Chapter Mix

Solution

Formulas Used

Core Logic

Pattern Recognition

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Correlate Concentration with i$i$

Execution

Pattern Recognition

Chapter Mix

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Step 1: Correlate Concentration with $i$ $i$