### 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

### Related Formula pi = C R T$\pi = C R T$ where pi$\pi$ is osmotic pressure, C$C$ is molar concentration, R$R$ is gas constant, and T$T$ is absolute temperature. ### Core Logic For a given dilute solution, the concentration (C$C$) and the gas constant (R$R$) are constant. Therefore, osmotic pressure is directly proportional to the absolute temperature. pi propto T$\pi \propto T$ fracpi_1T_1 = fracpi_2T_2$\frac{\pi_1}{T_1} = \frac{\pi_2}{T_2}$ ### Step 1: Calculation Given values: pi_1 = 7 times 10^5 text Pa = 70 times 10^4 text Nm^-2$\pi_1 = 7 \times 10^5 \text{ Pa} = 70 \times 10^4 \text{ Nm}^{-2}$ T_1 = 273 text K$T_1 = 273 \text{ K}$ T_2 = 283 text K$T_2 = 283 \text{ K}$ Rearranging for pi_2$\pi_2$: pi_2 = fracpi_1 cdot T_2T_1$\pi_2 = \frac{\pi_1 \cdot T_2}{T_1}$ pi_2 = frac7 times 10^5 times 283273$\pi_2 = \frac{7 \times 10^5 \times 283}{273}$ pi_2 = frac1981 times 10^5273$\pi_2 = \frac{1981 \times 10^5}{273}$ pi_2 = 7.2564 times 10^5 text Pa$\pi_2 = 7.2564 \times 10^5 \text{ Pa}$ Converting to the requested format (times 10^4 text Nm^-2$\times 10^4 \text{ Nm}^{-2}$): pi_2 = 72.564 times 10^4 text Nm^-2$\pi_2 = 72.564 \times 10^4 \text{ Nm}^{-2}$ Rounding off yields 72.56 (or 73 depending on required decimal places). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions