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

### Related Formula The net boiling point elevation for multiple non-volatile solutes is given by: Delta T_b = (i_1 m_1 + i_2 m_2) K_b$$\Delta T_b = (i_1 m_1 + i_2 m_2) K_b$$ ### Core Logic Both ethylene glycol and glucose are non-electrolytes, so their van 't Hoff factors are equal to unity (i_1 = i_2 = 1$i_1 = i_2 = 1$). textTotal moles of solute = 2 + 2 = 4text moles $$\text{Total moles of solute} = 2 + 2 = 4\text{ moles} $$ textMass of solvent (water) = 500text g = 0.5text kg $$ \text{Mass of solvent (water)} = 500\text{ g} = 0.5\text{ kg} $$ textTotal molality (m) = frac4text mol0.5text kg = 8text mol/kg $$ \text{Total molality (m)} = \frac{4\text{ mol}}{0.5\text{ kg}} = 8\text{ mol/kg} $$ ### Step 1: Compute Elevation and Final Temperature Delta T_b = 8 times 0.52 = 4.16text K $$ Delta T_b = 8 \times 0.52 = 4.16\text{ K} $$ textBoiling point of solution = T_b^circ + Delta T_b = 373.15text K + 4.16text K = 377.31text K approx 377.3text K $$ \text{Boiling point of solution} = T_b^{\circ} + \Delta T_b = 373.15\text{ K} + 4.16\text{ K} = 377.31\text{ K} \approx 377.3\text{ K} $$ ### Pattern Recognition Shortcut: Since both are molecular non-dissociating solutes, simply \sum their moles (2 + 2 = 4$2 + 2 = 4$). Diluting 4 moles in 0.5text kg$0.5\text{ kg}$ gives an effective concentration of 8text m$8\text{ m}$. Multiplying 8 times 0.52$8 \times 0.52$ gives a shift value of 4.16text K$4.16\text{ K}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula K_f = fracM_1 R T_f^2Delta H_fus = fracM_1 R T_fleft(fracDelta H_fusT_fright) = fracM_1 R T_fDelta S_fus$$K_f = \frac{M_1 R T_f^2}{\Delta H_{fus}} = \frac{M_1 R T_f}{\left(\frac{\Delta H_{fus}}{T_f}\right)} = \frac{M_1 R T_f}{\Delta S_{fus}}$$ ### Core Logic - **Statement I is correct:** Substituting Delta S_fus = fracDelta H_fusT_f$\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$ directly matches the given structural relationship formula. - **Statement II is incorrect:** Standard cryoscopic constants are: - For Benzene: K_f approx 5.12 mathrm~^circ C cdot kg cdot mol^-1$K_f \approx 5.12 \mathrm{~^\circ C \cdot kg \cdot mol^{-1}}$ - For Water: K_f approx 1.86 mathrm~^circ C cdot kg cdot mol^-1$K_f \approx 1.86 \mathrm{~^\circ C \cdot kg \cdot mol^{-1}}$ Therefore, K_f$K_f$ for benzene is greater than that of water, making Statement II false. ### Pattern Recognition Keep numerical benchmarks for common solvent colligative constants (K_b, K_f$K_b, K_f$ for water and benzene) memorized. Benzene has a far lower enthalpy of fusion and a higher freezing point, resulting in a significantly elevated K_f$K_f$ value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula textppm = fractextmass of solutetextmass of solution times 10^6$$\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$$ textMolality (m) = fractextmoles of solutetextmass of solvent in kg$$\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$ ### Core Logic 1. Consider 1000 mathrm~mL$1000 \mathrm{~mL}$ of seawater solution: textMass of solution = textVolume times textdensity = 1000 times 2 = 2000 mathrm~g$$\text{Mass of solution} = \text{Volume} \times \text{density} = 1000 \times 2 = 2000 \mathrm{~g}$$ textMass of NaCl = 6 text moles times 58.5 = 351 mathrm~g$$\text{Mass of NaCl} = 6 \text{ moles} \times 58.5 = 351 \mathrm{~g}$$ textMass of solvent (water) = 2000 - 351 = 1649 mathrm~g = 1.649 mathrm~kg$$\text{Mass of solvent (water)} = 2000 - 351 = 1649 \mathrm{~g} = 1.649 \mathrm{~kg}$$ 2. Compute the mass and moles of dissolved O_2$O_2$ using the ppm value: textppm = 5.8 = fractextmass of O_22000 times 10^6 implies textmass of O_2 = 1.16 times 10^-2 mathrm~g$$\text{ppm} = 5.8 = \frac{\text{mass of } O_2}{2000} \times 10^6 \implies \text{mass of } O_2 = 1.16 \times 10^{-2} \mathrm{~g}$$ textmoles of O_2 = frac1.16 times 10^-232 = 3.625 times 10^-4 text moles$$\text{moles of } O_2 = \frac{1.16 \times 10^{-2}}{32} = 3.625 \times 10^{-4} \text{ moles}$$ 3. Determine the molality (m$m$) of oxygen: textmolality = frac3.625 times 10^-41.649 approx 2.19 times 10^-4 mathrm~m$$\text{molality} = \frac{3.625 \times 10^{-4}}{1.649} \approx 2.19 \times 10^{-4} \mathrm{~m}$$ Matching the pattern mathbfx times 10^-4mathrmm$\mathbf{x} \times 10^{-4}\mathrm{m}$, we get mathbfx approx 2.19$\mathbf{x} \approx 2.19$. The nearest integer is **2**. ### Pattern Recognition For high concentration saline solutions, the mass of the solvent drops significantly below the total mass of the solution. Be careful to subtract the solute weight (351 mathrm~g$351 \mathrm{~g}$) before computing molality. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula pi = c R T$\pi = c R T$ where pi$\pi$ is the osmotic pressure of the solution. ### Core Logic Given that c_1 > c_2$c_1 > c_2$, chamber 1 has a higher concentration of solute than chamber 2. Under normal conditions, solvent molecules spontaneously flow from lower concentration (chamber 2) to higher concentration (chamber 1) via osmosis. To achieve **reverse osmosis**, the solvent must flow in the opposite direction—from chamber 1 to chamber 2. This requires applying an external pressure on the higher concentration side (chamber 1) that exceeds its osmotic pressure pi$\pi$. textCondition for Reverse Osmosis: p_1 > pi$$\text{Condition for Reverse Osmosis: } p_1 > \pi$$ Cellophane and parchment paper both act as suitable semi-permeable membranes for this setup. Thus, statements (A) and (C) are correct. ### Pattern Recognition Reverse osmosis always requires external pressure applied on the concentrated solution side (c_texthigh$c_{\text{high}}$) such that P_textapplied > pi$P_{\text{applied}} > \pi$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions