When 1~mathrmg$1~\mathrm{g}$ each of compounds AB and mathrmAB_2$\mathrm{AB}_2$ are dissolved in 15~mathrmg$15~\mathrm{g}$ of water separately, they increased the boiling point of water by 2.7~mathrmK$2.7~\mathrm{K}$ and 1.5~mathrmK$1.5~\mathrm{K}$ respectively. The atomic mass of A (in amu) is times 10^-1$\times 10^{-1}$ (Nearest integer) (Given : Molal boiling point elevation constant is 0.5~mathrmK~kg~mol^-1$0.5~\mathrm{K~kg~mol^{-1}}$)

### Related Formula i = 1 + (n - 1)alpha$$i = 1 + (n - 1)\alpha$$ ### Core Logic For the salt mathrmMX_3$\mathrm{MX}_3$: mathrmMX_3 rightarrow mathrmM^3+ + 3mathrmX^-$$\mathrm{MX}_3 \rightarrow \mathrm{M}^{3+} + 3\mathrm{X}^-$$ - 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 (alpha$\alpha$): i = 1 + (4 - 1)alpha$$i = 1 + (4 - 1)\alpha$$ 2 = 1 + 3alpha implies 3alpha = 1 implies alpha = frac13 approx 0.3333$$2 = 1 + 3\alpha \implies 3\alpha = 1 \implies \alpha = \frac{1}{3} \approx 0.3333$$ Percentage dissociation: \% text dissociation = alpha times 100 = 33.33 \% approx 33 \%$$\% \text{ dissociation} = \alpha \times 100 = 33.33 \% \approx 33 \%$$ ### Pattern Recognition Shortcut: For MX_3$MX_3$ dissociating into 4$4$ ions, alpha = (i - 1) / 3$\alpha = (i - 1) / 3$. Since i = 2$i = 2$, alpha = 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

### 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 (C_6H_5OH$C_6H_5OH$) + Aniline (C_6H_5NH_2$C_6H_5NH_2$)**: The phenolic -textOH$-\text{OH}$ proton forms strong intermolecular hydrogen bonds with the lone pair of the -textNH_2$-\text{NH}_2$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula Depression in freezing point colligative relation: Delta T_f = i cdot K_f cdot m$$\Delta T_f = i \cdot K_f \cdot m$$ Van 't Hoff factor for a weak acid dissociation: i = 1 + alpha$i = 1 + \alpha$ Dissociation constant formula: K_a = fracCalpha^21-alpha$$K_a = \frac{C\alpha^2}{1-\alpha}$$ ### Execution Step 1: Calculate the Van 't Hoff factor i$i$ using experimental freezing data: 0.20 = i times 1.8 times 0.1$$0.20 = i \times 1.8 \times 0.1$$ i = frac0.200.18 = frac2018 = frac109$$i = \frac{0.20}{0.18} = \frac{20}{18} = \frac{10}{9}$$ Step 2: Solve for degree of dissociation alpha$\alpha$: i = 1 + alpha implies frac109 = 1 + alpha$$i = 1 + \alpha \implies \frac{10}{9} = 1 + \alpha$$ alpha = frac109 - 1 = frac19$$\alpha = \frac{10}{9} - 1 = \frac{1}{9}$$ Step 3: Compute K_a$K_a$ substituting concentration C = 0.1 text M$C = 0.1 \text{ M}$ and alpha = frac19$\alpha = \frac{1}{9}$: K_a = frac0.1 times left(frac19right)^21 - frac19 = frac0.1 times frac181frac89 = frac0.181 times frac98 = frac0.172 = frac1720$$K_a = \frac{0.1 \times \left(\frac{1}{9}\right)^2}{1 - \frac{1}{9}} = \frac{0.1 \times \frac{1}{81}}{\frac{8}{9}} = \frac{0.1}{81} \times \frac{9}{8} = \frac{0.1}{72} = \frac{1}{720}$$ K_a approx 1.388 times 10^-3$$K_a \approx 1.388 \times 10^{-3}$$ ### Pattern Recognition When dealing with weak acids, always determine i$i$ first via colligative data, isolate alpha$\alpha$, and map directly to K_a = fracCalpha^21-alpha$K_a = \frac{C\alpha^2}{1-\alpha}$. Speed up computation by converting decimals into fractional fractions (0.2/0.18 = 10/9$0.2/0.18 = 10/9$) to maintain clean, mistake-free algebra. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Ionic Equilibrium

### Related Formula Delta T_f = i cdot K_f cdot m$$Delta T_f = i cdot K_f cdot m$$ ### Core Logic Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0^circC$0^{circ}C$, 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. ### Pattern Recognition This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer duration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula Freezing point depression relationship: Delta T_f = K_f cdot m$$\Delta T_f = K_f \cdot m$$ where m$m$ is the molality defined as: m = fractextmoles of solutetextmass of solvent in kg$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$ ### Step 1: Compute Molality Moles of non-volatile solute: textmoles = frac1\,mathrmg256\,mathrmg\,mol^-1$$\text{moles} = \frac{1\,\mathrm{g}}{256\,\mathrm{g\,mol}^{-1}}$$ Mass of solvent in kg: textmass = 50\,mathrmg = 50 times 10^-3\,mathrmkg$$\text{mass} = 50\,\mathrm{g} = 50 \times 10^{-3}\,\mathrm{kg}$$ Therefore, molality values map to: m = frac1256 times 50 times 10^-3 = frac100012800 = frac564\,mathrmmol\,kg^-1$$m = \frac{1}{256 \times 50 \times 10^{-3}} = \frac{1000}{12800} = \frac{5}{64}\,\mathrm{mol\,kg}^{-1}$$ ### Step 2: Calculate K_f$K_f$ Substituting values into the core formula: 0.40 = K_f cdot left(frac564right)$$0.40 = K_f \cdot \left(\frac{5}{64}\right)$$ K_f = frac0.40 times 645 = 0.08 times 64 = 5.12\,mathrmK\,kg\,mol^-1$$K_f = \frac{0.40 \times 64}{5} = 0.08 \times 64 = 5.12\,\mathrm{K\,kg\,mol}^{-1}$$ ### Pattern Recognition Sees: Direct calculation of cryogenic context constant (K_f$K_f$). Shortcut: Isolate K_f = fracDelta T_f cdot M cdot W_textsolvent1000 cdot w_textsolute$K_f = \frac{\Delta T_f \cdot M \cdot W_{\text{solvent}}}{1000 \cdot w_{\text{solute}}}$. Substituting instantly returns 5.12. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions