Liquid A$\text{A}$ and B$\text{B}$ form an ideal solution. The vapour pressure of pure liquids A$\text{A}$ and B$\text{B}$ are 350 and 750 mm Hg$750\text{ mm Hg}$ respectively at the same temperature. If xA$\text{x}_{\text{A}}$ and xB$\text{x}_{\text{B}}$ are the mole fraction of A$\text{A}$ and B$\text{B}$ in solution while yA$\text{y}_{\text{A}}$ and yB$\text{y}_{\text{B}}$ are the mole fraction of A$\text{A}$ and B$\text{B}$ in vapour phase then:

Related Formula

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$\Delta T_{\mathrm{b}} = 2.7~\mathrm{K}$, 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$K_{\mathrm{b}} = 0.5~\mathrm{K~kg~mol^{-1}}$:

Step 2: Determine Molar Mass of AB_2

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

Step 3: Solve for Atomic Mass of A

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

Subtracting equation (1) from (2):

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

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

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$2.7~\mathrm{K} \rightarrow 1.5~\mathrm{K}$ change.

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Class 12 Chemistry: Solutions

Related Formula

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

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:

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.

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Class 12 Chemistry: Solutions

Related Formula

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

Core Logic

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

Substitute the known properties into Raoult's equation block:

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$P^0$.

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Class 12 Chemistry: Solutions

Related Formula

Core Logic

For the salt MX₃$\mathrm{MX}_3$:

• 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$\alpha$):

Percentage dissociation:

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.

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Class 12 Chemistry: Solutions

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.

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Class 12 Chemistry: Solutions