'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1 to 25. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature? [Consider the solubility of NaCl remains unchanged over the temperature range]

Since solubility of NaCl remains unchanged, the number of dissolved moles of NaCl solute (nsolute) remains strictly constant. Thus, molarity M is strictly dependent on the volume of water (solvent) as temperature changes: Water exhibits unique anomalous density behavior near freezing: - From 1 to 4, the density of water **increases** to a maximum. This contraction means the volume (V) of water **decreases**. - From 4 to 25, the density of water **decreases** due to standard thermal expansion. Consequently, the volume (V) **increases**. Because volume is in the denominator of the molarity equation: - From 1 to 4: Volume decreases Molarity **increases**. - At 4: Volume is minimum Molarity reaches a **maximum**. - From 4 to 25: Volume increases Molarity **decreases**. This behavior is perfectly represented by **Plot (2)**, which features a distinct peak around 4. Water is at its densest (and occupies minimum volume) at exactly 3.98^ (4^). Any concentration unit based on volume (such as Molarity or Normality) will reach a corresponding maximum at this temperature. Class 12 Chemistry: Solutions

'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1 to 25. Which out of the following plots, is best suited for the change in the molarity (M) of the

Since solubility of NaCl remains unchanged, the number of dissolved moles of NaCl solute (nsolute) remains strictly constant. Thus, molarity M is strictly dependent on the volume of water (solvent) as temperature changes: Water exhibits unique anomalous density behavior near freezing: - From 1 to 4, the density of water **increases** to a maximum. This contraction means the volume (V) of water **decreases**. - From 4 to 25, the density of water **decreases** due to standard thermal expansion. Co

'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1 to 25. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature? [Consider the solubility of NaCl remains unchanged over the temperature range]

Since solubility of NaCl remains unchanged, the number of dissolved moles of NaCl solute (nsolute) remains strictly constant. Thus, molarity M is strictly dependent on the volume of water (solvent) as temperature changes: Water exhibits unique anomalous density behavior near freezing: - From 1 to 4, the density of water **increases** to a maximum. This contraction means the volume (V) of water **decreases**. - From 4 to 25, the density of water **decreases** due to standard thermal expansion. Consequently, the volume (V) **increases**. Because volume is in the denominator of the molarity equation: - From 1 to 4: Volume decreases Molarity **increases**. - At 4: Volume is minimum Molarity reaches a **maximum**. - From 4 to 25: Volume increases Molarity **decreases**. This behavior is perfectly represented by **Plot (2)**, which features a distinct peak around 4. Water is at its densest (and occupies minimum volume) at exactly 3.98^ (4^). Any concentration unit based on volume (such as Molarity or Normality) will reach a corresponding maximum at this temperature. Class 12 Chemistry: Solutions

'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1 to 25. Which out of the following plots, is best suited for the change in the molarity (M) of the

Since solubility of NaCl remains unchanged, the number of dissolved moles of NaCl solute (nsolute) remains strictly constant. Thus, molarity M is strictly dependent on the volume of water (solvent) as temperature changes: Water exhibits unique anomalous density behavior near freezing: - From 1 to 4, the density of water **increases** to a maximum. This contraction means the volume (V) of water **decreases**. - From 4 to 25, the density of water **decreases** due to standard thermal expansion. Co

'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1 to 25. Which out…

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) $Delta Vtextmix=0$ $\Delta V{\text{mix}}=0$ Choose the correct answer from the options given below: A. $text(A)-(III), (B)-(I), (C)-(IV), (D)-(II)$ B. $text(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$ C. $text(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$ D. $text(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$ Solution ### Related Formula $textNegative Deviation from Raoult's Law implies textMaximum Boiling Azeotrope$ $textPositive Deviation from Raoult's Law implies textMinimum Boiling Azeotrope$ $textIdeal Solution implies Delta Vtextmix = 0$ ### 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 $ightarrow$ $ ightarrow$ (III) - (B) Ethanol and Water: Exhibit hydrogen bond disruptions, leading to positive deviation, forming a minimum boiling azeotrope $ightarrow$ $ ightarrow$ (I) - (C) Benzene and Toluene: Highly structurally similar, forming a near-perfect ideal solution where $Delta Vtextmix = 0$ $\Delta V{\text{mix}} = 0$ $ightarrow$ $ ightarrow$ (IV) - (D) Acetic acid in benzene: Acetic acid undergoes intermolecular hydrogen bonding inside the non-polar solvent, causing it to dimerize $ightarrow$ $ ightarrow$ (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. ### Evaluation Rubric / Model Answer null ### 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 $mathrmMX_2$ $\mathrm{MX}_2$ are 65.6 and 164 respectively. The percent degree of ionisation of $mathrmMX_2$ $\mathrm{MX}_2$ is ____ %. (Nearest integer) Numerical Answer. Answer: 75 to 75 Solution ### Related Formula $i = fractextNormal Molar MasstextObserved Molar Mass$ $i = 1 + (n - 1)alpha$ $$i = 1 + (n - 1)\alpha$$ ### Core Logic 1. Calculate the van 't Hoff factor ( $i$ $i$): $i = frac16465.6 = 2.5$ $$i = \frac{164}{65.6} = 2.5$$ 2. Set up the dissociation equilibrium for the electrolyte $mathrmMX_2$ $\mathrm{MX}_2$: $mathrmMX_2 ightarrow mathrmM^2+ + 2mathrmX^-$ Here, 1 molecule dissociates into $n = 1 + 2 = 3$ $n = 1 + 2 = 3$ ions. 3. Relate $i$ $i$ to the degree of ionization ( $alpha$ $\alpha$): $i = 1 + (3 - 1)alpha = 1 + 2alpha$ $$i = 1 + (3 - 1)\alpha = 1 + 2\alpha$$ $2.5 = 1 + 2alpha implies 2alpha = 1.5 implies alpha = 0.75$ 4. Convert to a percentage: $textPercent dissociation = 0.75 cdot 100 = 75\%$ ### Pattern Recognition For a salt that dissociates into three ions (like $mathrmMX_2$ $\mathrm{MX}_2$), the relationship simplifies to $i = 1 + 2alpha$ $i = 1 + 2\alpha$. Calculating $i$ $i$ from the ratio of the molar masses lets you find $alpha$ $\alpha$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q42 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 $Delta T_mathrmf$ $\Delta T_{\mathrm{f}}$ , depression in the freezing point of solvent in a solution? A. Plot (1) Depression in Freezing Point B. Plot (2) Depression in Freezing Point C. Plot (3) Depression in Freezing Point D. Plot (4) Depression in Freezing Point Solution ### Related Formula $Delta T_f = T_f^0 - T_f$ $$\Delta T_f = T_f^0 - T_f$$ ### Core Logic Dissolving a non-volatile solute lower the vapor pressure of the solution relative to the pure solvent across all temperature thresholds. The freezing point is defined as the temperature at which the vapor pressure of the liquid phase matches that of its solid phase. Because the solution's vapor pressure curve lies below that of the pure liquid solvent, its intersection with the frozen solvent curve occurs at a lower temperature ( $T_f < T_f^0$ $T_f < T_f^0$). This shift creates the characteristic freezing point depression step: $Delta T_f = T_f^0 - T_f$ $\Delta T_f = T_f^0 - T_f$. Plot (3) correctly displays this thermodynamic behavior. Depression in Freezing Point solution plot for Q42 - JEE Main 2025 Morning ### Pattern Recognition The vapor pressure curve for the solution always runs lower than that of the pure solvent, shifting the freezing intersection to the left. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q42 jee_main_2025_28_jan_evening Osmosis and Osmotic Pressure Assume a living cell with $0.9\%$ $0.9\%$ $(omega/omega)$ $(\omega/\omega)$ 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 $0.5\%\ (omega/omega)$ $0.5\%\ (\omega/\omega)$ B. Shrink since solution is $0.45\%\ (omega/omega)$ $0.45\%\ (\omega/\omega)$ 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 $0.9\%\ (omega/omega)$ $0.9\%\ (\omega/\omega)$ Solution ### Related Formula Mass percentage from mole fraction calculation: $\%text w/w = fracx_1 cdot M_1x_1 cdot M_1 + x_2 cdot M_2 times 100$ ### Core Logic Inside the living cell, glucose concentration is $0.9\%text w/w$ $0.9\%\text{ w/w}$. The surrounding solution has equal mole fractions of glucose and water ( $x_textglucose = 0.5$ $x_{\text{glucose}} = 0.5$, $x_textwater = 0.5$ $x_{\text{water}} = 0.5$). Let's calculate the mass percentage of the outer solution: - Mass of glucose component = $0.5 times 180 = 90mathrm\ g$ $0.5 \times 180 = 90\mathrm{\ g}$ - Mass of water component = $0.5 times 18 = 9mathrm\ g$ $0.5 \times 18 = 9\mathrm{\ g}$ - Total solution mass = $90 + 9 = 99mathrm\ g$ $90 + 9 = 99\mathrm{\ g}$ ### Step 1: Concentration Determination and Osmosis Profile Outer mass percentage: $\%text w/w = frac9099 times 100 approx 90.9\%$ 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, $180mathrm\ g/mol$ $180\mathrm{\ g/mol}$) and a low-molar-mass solvent (water, $18mathrm\ g/mol$ $18\mathrm{\ g/mol}$) is always extremely concentrated by mass. Placing a standard living cell into such a hypertonic solution inevitably causes fluid loss and cellular shrinkage. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q jee_main_2025_29_jan_morning Abnormal Molar Mass and van't Hoff Factor $1.24 mathrm~g$ $1.24 \mathrm{~g}$ of $mathrmAX_2$ $\mathrm{AX}_{2}$ (molar mass $124 mathrm~g mathrm~mol^-1$ $124 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in $1 mathrm~kg$ $1 \mathrm{~kg}$ of water to form a solution with boiling point of $100.0156^circ mathrmC$ $100.0156^{\circ} \mathrm{C}$ , while $25.4 mathrm~g$ $25.4 \mathrm{~g}$ of $mathrmAY_2$ $\mathrm{AY}_{2}$ (molar mass $250 mathrm~g mathrm~mol^-1$ $250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in $2 mathrm~kg$ $2 \mathrm{~kg}$ of water constitutes a solution with a boiling point of $100.0260^circ mathrmC$ $100.0260^{\circ} \mathrm{C}$ . $mathrmK_b(H_2O) = 0.52 \, K \, kg \, mol^-1$ Which of the following is correct? A. $mathrmAX2$ $\mathrm{AX}2$ and $mathrmAY_2$ $\mathrm{AY}_2$ (both) are completely unionised. B. $mathrmAX_2$ $\mathrm{AX}_2$ and $mathrmAY_2$ $\mathrm{AY}_2$ (both) are fully ionised. C. $mathrmAX_2$ $\mathrm{AX}_2$ is completely unionised while $mathrmAY_2$ $\mathrm{AY}_2$ is fully ionised. D. $mathrmAX_2$ $\mathrm{AX}_2$ is fully ionised while $mathrmAY_2$ $\mathrm{AY}_2$ is completely unionised. Solution ### Formulas Used Elevation in boiling point formula involving the van't Hoff factor ( $i$ $i$): $Delta T_b = i cdot K_b cdot m$ $$\Delta T_b = i \cdot K_b \cdot m$$ Where: * $Delta T_b = T_b - T_b^circ$ $\Delta T_b = T_b - T_b^\circ$ (Boiling point elevation) * $K_b = textEbullioscopic constant$ $K_b = \text{Ebullioscopic constant}$ * $m = textMolality of the solution left(fractextmoles of solutetextmass of solvent in kgright)$ ### Core Logic Step 1: Evaluate solution system $mathrmAX_2$ $\mathrm{AX}_2$ $Delta T_b = 100.0156^circmathrmC - 100.0000^circmathrmC = 0.0156^circmathrmC$ $textMolality m_1 = frac1.24text g / 124text g mol^-11text kg = 0.01text mol/kg$ Using the elevation formula: $0.0156 = i_mathrmAX_2 cdot 0.52 cdot 0.01$ $i_mathrmAX_2 = frac0.01560.0052 = 3$ Since theoretical dissociation of $mathrmAX_2 rightarrow mathrmA^2+ + 2mathrmX^-$ produces $3$ $3$ particles, $i = 3$ $i = 3$ implies that $mathrmAX_2$ $\mathrm{AX}_2$ is fully ionised. --- Step 2: Evaluate solution system $mathrmAY_2$ $\mathrm{AY}_2$ $Delta T_b = 100.0260^circmathrmC - 100.0000^circmathrmC = 0.0260^circmathrmC$ $textMolality m_2 = frac25.4text g / 250text g mol^-12text kg = 0.0508text mol/kg$ Using the elevation formula: $0.0260 = i_mathrmAY_2 cdot 0.52 cdot 0.0508$ $i_mathrmAY_2 = frac0.02600.0264 approx 1$ Since $i approx 1$ $i \approx 1$, it behaves as a non-electrolyte, meaning $mathrmAY_2$ $\mathrm{AY}_2$ is completely unionised. Thus, $mathrmAX_2$ $\mathrm{AX}_2$ is fully ionised while $mathrmAY_2$ $\mathrm{AY}_2$ is completely unionised. ### Pattern Recognition A van't Hoff factor matching the complete stoichiometric ion count ( $i = 3$ $i = 3$ for $mathrmAX_2$ $\mathrm{AX}_2$) confirms complete ionisation, whereas a factor near unity ( $i = 1$ $i = 1$) indicates no dissociation into separate ions. Correct Option: (D) ← Previous Page 4 of 7 Next → More Solutions Questions — jee_main_2025_02_april_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 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 Consider the given plots of vapour pressure (VP) vs temperature 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 Elements P and Q form two types of non-volatile, non-ionizable Identify the mixture that shows positive deviations from Raoult's Law A solution of two miscible liquids showing negative deviation from If a substance 'A' dissolves in solution of a mixture We have three aqueous solutions of labelled as 'A', 'B' What happens to freezing point of benzene when small quantity The solution from the following with highest depression in freezing Molality of solution (density ) is ________ . 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

(III) Maximum boiling azeotrope

(D) Solution of acetic acid in benzene

(IV)

Delta Vtextmix=0

Choose the correct answer from the options given below:

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

Solution

### Related Formula

textNegative Deviation from Raoult's Law implies textMaximum Boiling Azeotrope

textPositive Deviation from Raoult's Law implies textMinimum Boiling Azeotrope

textIdeal Solution implies Delta Vtextmix = 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

ightarrow

(III) - (B) Ethanol and Water: Exhibit hydrogen bond disruptions, leading to positive deviation, forming a minimum boiling azeotrope

ightarrow

(I) - (C) Benzene and Toluene: Highly structurally similar, forming a near-perfect ideal solution where

Delta Vtextmix = 0

ightarrow

(IV) - (D) Acetic acid in benzene: Acetic acid undergoes intermolecular hydrogen bonding inside the non-polar solvent, causing it to dimerize

ightarrow

(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

). Benzene-toluene is the most famous ideal solution pair. Recognizing either instantly shortcuts the options list to the correct key. ### Evaluation Rubric / Model Answer null ### 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

mathrmMX_2

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

mathrmMX_2

is ____ %. (Nearest integer)

Numerical Answer. Answer: 75 to 75

Solution

### Related Formula

i = fractextNormal Molar MasstextObserved Molar Mass

i = 1 + (n - 1)alpha

### Core Logic 1. Calculate the van 't Hoff factor (

i

i = frac16465.6 = 2.5

2. Set up the dissociation equilibrium for the electrolyte

mathrmMX_2

mathrmMX_2 ightarrow mathrmM^2+ + 2mathrmX^-

Here, 1 molecule dissociates into

n = 1 + 2 = 3

ions. 3. Relate

i

to the degree of ionization (

alpha

i = 1 + (3 - 1)alpha = 1 + 2alpha

2.5 = 1 + 2alpha implies 2alpha = 1.5 implies alpha = 0.75

4. Convert to a percentage:

textPercent dissociation = 0.75 cdot 100 = 75\%

### Pattern Recognition For a salt that dissociates into three ions (like

mathrmMX_2

), the relationship simplifies to

i = 1 + 2alpha

. Calculating

i

from the ratio of the molar masses lets you find

alpha

directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

Q42 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

Delta T_mathrmf

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

A. Plot (1)
Depression in Freezing Point
B. Plot (2)
Depression in Freezing Point
C. Plot (3)
Depression in Freezing Point
D. Plot (4)
Depression in Freezing Point

Solution

### Related Formula

Delta T_f = T_f^0 - T_f

### Core Logic Dissolving a non-volatile solute lower the vapor pressure of the solution relative to the pure solvent across all temperature thresholds. The freezing point is defined as the temperature at which the vapor pressure of the liquid phase matches that of its solid phase. Because the solution's vapor pressure curve lies below that of the pure liquid solvent, its intersection with the frozen solvent curve occurs at a lower temperature (

T_f < T_f^0

). This shift creates the characteristic freezing point depression step:

Delta T_f = T_f^0 - T_f

. Plot (3) correctly displays this thermodynamic behavior.

Depression in Freezing Point solution plot for Q42 - JEE Main 2025 Morning

### Pattern Recognition The vapor pressure curve for the solution always runs lower than that of the pure solvent, shifting the freezing intersection to the left. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

Q42 jee_main_2025_28_jan_evening Osmosis and Osmotic Pressure

Assume a living cell with

0.9\%

(omega/omega)

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 $0.5\%\ (omega/omega)$ $0.5\%\ (\omega/\omega)$
B. Shrink since solution is $0.45\%\ (omega/omega)$ $0.45\%\ (\omega/\omega)$ 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 $0.9\%\ (omega/omega)$ $0.9\%\ (\omega/\omega)$

Solution

### Related Formula Mass percentage from mole fraction calculation:

\%text w/w = fracx_1 cdot M_1x_1 cdot M_1 + x_2 cdot M_2 times 100

### Core Logic Inside the living cell, glucose concentration is

0.9\%text w/w

. The surrounding solution has equal mole fractions of glucose and water (

x_textglucose = 0.5

x_textwater = 0.5

). Let's calculate the mass percentage of the outer solution: - Mass of glucose component =

0.5 times 180 = 90mathrm\ g

- Mass of water component =

0.5 times 18 = 9mathrm\ g

- Total solution mass =

90 + 9 = 99mathrm\ g

### Step 1: Concentration Determination and Osmosis Profile Outer mass percentage:

\%text w/w = frac9099 times 100 approx 90.9\%

Because the external environment is highly concentrated (hypertonic) compared to the inner cell (

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,

180mathrm\ g/mol

) and a low-molar-mass solvent (water,

18mathrm\ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

Q jee_main_2025_29_jan_morning Abnormal Molar Mass and van't Hoff Factor

1.24 mathrm~g

mathrmAX_2

(molar mass

124 mathrm~g mathrm~mol^-1

) is dissolved in

1 mathrm~kg

of water to form a solution with boiling point of

100.0156^circ mathrmC

, while

25.4 mathrm~g

mathrmAY_2

(molar mass

250 mathrm~g mathrm~mol^-1

) in

2 mathrm~kg

of water constitutes a solution with a boiling point of

100.0260^circ mathrmC

mathrmK_b(H_2O) = 0.52 \, K \, kg \, mol^-1

Which of the following is correct?

A. $mathrmAX2$ $\mathrm{AX}2$ and $mathrmAY_2$ $\mathrm{AY}_2$ (both) are completely unionised.
B. $mathrmAX_2$ $\mathrm{AX}_2$ and $mathrmAY_2$ $\mathrm{AY}_2$ (both) are fully ionised.
C. $mathrmAX_2$ $\mathrm{AX}_2$ is completely unionised while $mathrmAY_2$ $\mathrm{AY}_2$ is fully ionised.
D. $mathrmAX_2$ $\mathrm{AX}_2$ is fully ionised while $mathrmAY_2$ $\mathrm{AY}_2$ is completely unionised.

Solution

### Formulas Used Elevation in boiling point formula involving the van't Hoff factor (

i

Delta T_b = i cdot K_b cdot m

Where: *

Delta T_b = T_b - T_b^circ

(Boiling point elevation) *

K_b = textEbullioscopic constant

m = textMolality of the solution left(fractextmoles of solutetextmass of solvent in kgright)

### Core Logic **Step 1: Evaluate solution system

mathrmAX_2

Delta T_b = 100.0156^circmathrmC - 100.0000^circmathrmC = 0.0156^circmathrmC

textMolality m_1 = frac1.24text g / 124text g mol^-11text kg = 0.01text mol/kg

Using the elevation formula:

0.0156 = i_mathrmAX_2 cdot 0.52 cdot 0.01

i_mathrmAX_2 = frac0.01560.0052 = 3

Since theoretical dissociation of

mathrmAX_2 rightarrow mathrmA^2+ + 2mathrmX^-

produces

3

particles,

i = 3

implies that **

mathrmAX_2

is fully ionised**. --- **Step 2: Evaluate solution system

mathrmAY_2

Delta T_b = 100.0260^circmathrmC - 100.0000^circmathrmC = 0.0260^circmathrmC

textMolality m_2 = frac25.4text g / 250text g mol^-12text kg = 0.0508text mol/kg

Using the elevation formula:

0.0260 = i_mathrmAY_2 cdot 0.52 cdot 0.0508

i_mathrmAY_2 = frac0.02600.0264 approx 1

Since

i approx 1

, it behaves as a non-electrolyte, meaning **

mathrmAY_2

is completely unionised**. Thus, **

mathrmAX_2

is fully ionised while

mathrmAY_2

is completely unionised**. ### Pattern Recognition A van't Hoff factor matching the complete stoichiometric ion count (

i = 3

for

mathrmAX_2

) confirms complete ionisation, whereas a factor near unity (

i = 1

) indicates no dissociation into separate ions. **Correct Option:** **(D)**

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