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 = fractextNormal Molar MasstextObserved Molar Mass$$i = \frac{\text{Normal Molar Mass}}{\text{Observed Molar Mass}}$$ 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^-$$\mathrm{MX}_{2} ightarrow \mathrm{M}^{2+} + 2\mathrm{X}^{-}$$ 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$$2.5 = 1 + 2\alpha \implies 2\alpha = 1.5 \implies \alpha = 0.75$$ 4. Convert to a percentage: textPercent dissociation = 0.75 cdot 100 = 75\%$$\text{Percent dissociation} = 0.75 \cdot 100 = 75\%$$ ### 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

### 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.

### 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

### 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$$\%\text{ w/w} = \frac{x_1 \cdot M_1}{x_1 \cdot M_1 + x_2 \cdot M_2} \times 100$$ ### 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\%$$\%\text{ w/w} = \frac{90}{99} \times 100 \approx 90.9\%$$ 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, 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

### 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)$m = \text{Molality of the solution } \left(\frac{\text{moles of solute}}{\text{mass of solvent in kg}}\right)$ ### Core Logic **Step 1: Evaluate solution system mathrmAX_2$\mathrm{AX}_2$** Delta T_b = 100.0156^circmathrmC - 100.0000^circmathrmC = 0.0156^circmathrmC$$\Delta T_b = 100.0156^\circ\mathrm{C} - 100.0000^\circ\mathrm{C} = 0.0156^\circ\mathrm{C}$$ textMolality m_1 = frac1.24text g / 124text g mol^-11text kg = 0.01text mol/kg$$\text{Molality } m_1 = \frac{1.24\text{ g} / 124\text{ g mol}^{-1}}{1\text{ kg}} = 0.01\text{ mol/kg}$$ Using the elevation formula: 0.0156 = i_mathrmAX_2 cdot 0.52 cdot 0.01$$0.0156 = i_{\mathrm{AX}_2} \cdot 0.52 \cdot 0.01$$ i_mathrmAX_2 = frac0.01560.0052 = 3$$i_{\mathrm{AX}_2} = \frac{0.0156}{0.0052} = 3$$ Since theoretical dissociation of mathrmAX_2 rightarrow mathrmA^2+ + 2mathrmX^-$\mathrm{AX}_2 \rightarrow \mathrm{A}^{2+} + 2\mathrm{X}^-$ 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$$\Delta T_b = 100.0260^\circ\mathrm{C} - 100.0000^\circ\mathrm{C} = 0.0260^\circ\mathrm{C}$$ textMolality m_2 = frac25.4text g / 250text g mol^-12text kg = 0.0508text mol/kg$$\text{Molality } m_2 = \frac{25.4\text{ g} / 250\text{ g mol}^{-1}}{2\text{ kg}} = 0.0508\text{ mol/kg}$$ Using the elevation formula: 0.0260 = i_mathrmAY_2 cdot 0.52 cdot 0.0508$$0.0260 = i_{\mathrm{AY}_2} \cdot 0.52 \cdot 0.0508$$ i_mathrmAY_2 = frac0.02600.0264 approx 1$$i_{\mathrm{AY}_2} = \frac{0.0260}{0.0264} \approx 1$$ 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)**