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 + (y - 1)alpha$$i = 1 + (y - 1)\alpha$$ ### Core Logic For electrolyte mathrmA_2mathrmB$\mathrm{A}_2\mathrm{B}$ undergoing dissociation: mathrmA_2B rightarrow 2mathrmA^+ + mathrmB^2-$$\mathrm{A_2B} \rightarrow 2\mathrm{A}^+ + \mathrm{B}^{2-}$$ Total count of ions produced per molecule y = 3$y = 3$. Given degree of dissociation alpha = 30\% = 0.3$\alpha = 30\% = 0.3$. Substituting into the formula: i = 1 + (3 - 1) cdot 0.3$$i = 1 + (3 - 1) \cdot 0.3$$ i = 1 + 2 cdot 0.3 = 1 + 0.6 = 1.6$$i = 1 + 2 \cdot 0.3 = 1 + 0.6 = 1.6$$ Expressing the value in the requested format: 1.6 = 16 times 10^-1$$1.6 = 16 \times 10^{-1}$$ 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

### Core Logic For a strong electrolyte like NaCl$NaCl$, the theoretical van 't Hoff factor i_theory$i_{theory}$ 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$\implies$ more ion-pairing implies$\implies$ lower i$i$. Given concentrations: A = 0.1 mathrm~M$A = 0.1 \mathrm{~M}$ (highest concentration) B = 0.01 mathrm~M$B = 0.01 \mathrm{~M}$ C = 0.001 mathrm~M$C = 0.001 \mathrm{~M}$ (most dilute) Therefore, the actual i$i$ values follow the reverse order of concentration: i_A (0.1 mathrm~M) < i_B (0.01 mathrm~M) < i_C (0.001 mathrm~M)$i_A (0.1 \mathrm{~M}) < i_B (0.01 \mathrm{~M}) < i_C (0.001 \mathrm{~M})$. ### Execution 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula m = fracM times 1000, (1000 times d) - (M times M_B)$$m = \frac{M \times 1000}, {(1000 \times d) - (M \times M_B)}$$ where, M$M$ = Molarity = 0.8text M$0.8\text{ M}$ d$d$ = Density = 1.06text g/cm^3$1.06\text{ g/cm}^3$ M_B$M_B$ = Molar mass of solute (H_2SO_4$H_2SO_4$) = 98text g/mol$98\text{ g/mol}$ ### Core Logic Substituting the given values into the equation: m = frac0.8 times 1000, (1000 times 1.06) - (0.8 times 98)$$m = \frac{0.8 \times 1000}, {(1000 \times 1.06) - (0.8 \times 98)}$$ Calculating the denominator parameters: textDenominator = 1060 - 78.4 = 981.6text g$$\text{Denominator} = 1060 - 78.4 = 981.6\text{ g}$$ ### Step 1: Final Resolution Solving for molality: m = frac800, 981.6 approx 0.815text m = 815 times 10^-3text m$$m = \frac{800}, {981.6} \approx 0.815\text{ m} = 815 \times 10^{-3}\text{ m}$$ Thus, the integer factor value is **815**. ### Pattern Recognition Ensure you explicitly subtract the mass of the solute from the total mass of the solution to correctly isolate the mass of the solvent needed for molality calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Core Logic A system demonstrating a negative deviation from Raoult's law implies tighter molecular attractions between components (A-B$A-B$ interactions are stronger than A-A$A-A$ or B-B$B-B$). This decreases the aggregate escaping tendency, yielding a decreased total vapour pressure. Consequently, a higher thermal energy threshold is required to reach the boiling threshold, causing an increased boiling point. ### Pattern Recognition Negative deviation rightarrow$\rightarrow$ Vapour Pressure drops rightarrow$\rightarrow$ Boiling Point rises inversely. ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula textMolarity (M) = frac\% text by mass times 10 times dM_w$$\text{Molarity } (M) = \frac{\% \text{ by mass} \times 10 \times d}{M_w}$$ ### Core Logic Let's assume we have 100 g of the solution. Mass of H_2SO_4$H_2SO_4$ in 100 g solution = 31.4text g$31.4\text{ g}$. Moles of H_2SO_4$H_2SO_4$ (n_textsolute$n_{\text{solute}}$) = frac31.498text mol$\frac{31.4}{98}\text{ mol}$. Volume of the solution (V$V$) can be found using density: V = fractextMass of solutiontextDensity = frac1001.25text mL$$V = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100}{1.25}\text{ mL}$$ ### Step 1: Calculating Molarity Molarity is defined as moles of solute per liter of solution: M = fracn_textsoluteV(textin mL) times 1000$$M = \frac{n_{\text{solute}}}{V(\text{in mL})} \times 1000$$ M = frac31.4 / 98100 / 1.25 times 1000$$M = \frac{31.4 / 98}{100 / 1.25} \times 1000$$ M = frac31.4 times 1.2598 times 100 times 1000$$M = \frac{31.4 \times 1.25}{98 \times 100} \times 1000$$ M = frac39.2598 times 10$$M = \frac{39.25}{98} \times 10$$ M = 0.4005 times 10$$M = 0.4005 \times 10$$ M = 4.005text M$$M = 4.005\text{ M}$$ Rounding off to the nearest integer gives 4. ### Pattern Recognition Whenever percentage by mass (w/w$w/w$) and density (d$d$ in g/mL) are given, use the direct formula: M = frac\%(w/w) times d times 10M_w$M = \frac{\%(w/w) \times d \times 10}{M_w}$. This saves enormous time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry