Solutions & Colligative Props

Cheatsheet Content

### 01 Mark Questions 1. **The molarity of the solution containing 5.0 g of NaOH in 250 mL solution is:** A. 0.1 M B. 0.5 M C. 1.0 M D. 2.0 M **Answer:** C. 0.5 M **Explanation:** * Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol * Moles of NaOH = Mass / Molar mass = 5.0 g / 40 g/mol = 0.125 mol * Volume of solution in Liters = 250 mL / 1000 mL/L = 0.250 L * Molarity (M) = Moles of solute / Volume of solution (L) = 0.125 mol / 0.250 L = 0.5 M 2. **In which case Raoult's law is not applicable?** A. 1 M NaCl B. 1 M urea C. 1 M glucose D. 1 M sucrose **Answer:** A. 1 M NaCl **Explanation:** Raoult's law is applicable to ideal solutions and non-volatile solutes. For electrolytes like NaCl, which dissociate into ions, the effective concentration of particles is higher than the stated molarity, leading to deviations from Raoult's law. Urea, glucose, and sucrose are non-electrolytes. 3. **At a constant temperature which of the following solution will have maximum vapour pressure?** (Molecular weight: NaCl : 58.5, H₂SO₄ = 98.0 g mol⁻¹) A. 1 molal NaCl (aq) B. 1 molar NaCl (aq) C. 1 molal H₂SO₄ (aq) D. 1 molar H₂SO₄ (aq) **Answer:** This question is flawed as "maximum vapour pressure" would correspond to the *pure solvent* or the solution with the *lowest concentration of non-volatile solute*. All options describe solutions with non-volatile solutes, which will lower the vapor pressure compared to pure water. Assuming the question implies "which solution will have the highest vapor pressure *among the given options*", it would be the one with the *least lowering* of vapor pressure. This requires a detailed comparison of effective molalities/molarities and van't Hoff factors. However, the exact phrasing "maximum vapour pressure" is problematic for solutions. 4. **To observe an elevation of boiling point of 0.05 °C, the amount of solute (molecular weight=100) to be added to 100 g of water (k_b = 0.5) is:** A. 2g B. 0.05 g C. 1g D. 0.75 g **Answer:** C. 1g **Explanation:** * $\Delta T_b = K_b \cdot m$ * $m = \Delta T_b / K_b = 0.05 \text{ °C} / 0.5 \text{ °C kg/mol} = 0.1 \text{ mol/kg}$ * Moles of solute = $m \times \text{mass of solvent (kg)} = 0.1 \text{ mol/kg} \times 0.1 \text{ kg} = 0.01 \text{ mol}$ * Mass of solute = Moles $\times$ Molecular weight = $0.01 \text{ mol} \times 100 \text{ g/mol} = 1 \text{ g}$ 5. **Which has least freezing point?** A. 1% sucrose B. 1% NaCl C. 1% CaCl₂ D. 1% Glucose **Answer:** C. 1% CaCl₂ **Explanation:** Freezing point depression is a colligative property, meaning it depends on the number of solute particles. * Sucrose and Glucose are non-electrolytes (i=1). * NaCl dissociates into 2 ions ($Na^+ + Cl^-$), so i=2. * CaCl₂ dissociates into 3 ions ($Ca^{2+} + 2Cl^-$), so i=3. For the same mass percentage, CaCl₂ produces the most particles, leading to the greatest freezing point depression and thus the least (lowest) freezing point. 6. **Which of the following is not colligative property?** A. Depression in freezing point B. Osmotic pressure C. Elevation in boiling point D. Increasing freezing point **Answer:** D. Increasing freezing point **Explanation:** Colligative properties are those that depend only on the number of solute particles in a solution, not on their identity. These include relative lowering of vapor pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. "Increasing freezing point" is not a colligative property; adding a non-volatile solute always *depresses* the freezing point. 7. **What is the molarity of 0.2 N Na₂CO₃ solution?** A. 0.1 M B. 0 M C. 0.4 M D. 0.2 M **Answer:** A. 0.1 M **Explanation:** * Normality (N) = Molarity (M) $\times$ n-factor * For Na₂CO₃, the n-factor (number of equivalents per mole) is 2, as it provides 2 $Na^+$ ions (or its charge is 2- for $CO_3^{2-}$). * Molarity (M) = Normality (N) / n-factor = 0.2 N / 2 = 0.1 M #### Assertion and Reason Questions (Q. Nos. 8 to 10) Choose the appropriate answer: A. Both A and R are true and R is correct explanation of A B. Both A and R are true and R is not correct explanation of A C. A is correct and R is incorrect D. A is incorrect and R is correct 8. **A: Azeotropic mixtures are formed only by non-ideal solutions and they may have boiling point either greater than both the components or less than both the components.** **R: The composition of the vapour phase is same as that of the liquid phase of an azeotropic mixture.** **Answer:** A. Both A and R are true and R is correct explanation of A **Explanation:** Azeotropes are indeed formed by non-ideal solutions, and their boiling point is either higher or lower than that of their components. The defining characteristic of an azeotrope is that the composition of the vapor phase is identical to that of the liquid phase, which prevents separation by fractional distillation. This identical composition is the reason for their constant boiling point behavior. 9. **A: The boiling point of 0.1 M urea solution is less than that of 0.1 M KCl solution.** **R: Elevation of boiling point is directly proportional to the number of species present in the solution.** **Answer:** A. Both A and R are true and R is correct explanation of A **Explanation:** * Urea is a non-electrolyte, so 0.1 M urea produces 0.1 M particles. * KCl is an electrolyte that dissociates into two ions ($K^+ + Cl^-$), so 0.1 M KCl produces approximately 0.2 M particles (assuming complete dissociation). * Since boiling point elevation is a colligative property (directly proportional to the number of particles as stated in R), 0.1 M KCl solution will have a greater elevation in boiling point than 0.1 M urea solution. * Therefore, the boiling point of 0.1 M urea solution will be *less* than that of 0.1 M KCl solution. Both A and R are true, and R explains A. 10. **A: The water pouch of instant cold pack for treating athletic injuries breaks when squeezed and NH₄NO₃ dissolves lowering the temperature.** **R: Addition of non-volatile solute into solvent results in depression of freezing point of solvent.** **Answer:** B. Both A and R are true and R is not correct explanation of A **Explanation:** * Assertion A describes an endothermic dissolution process (NH₄NO₃ dissolving in water absorbs heat, lowering temperature), which is correct. This is related to enthalpy of solution, not directly colligative properties. * Reason R describes freezing point depression, which is also a correct statement about colligative properties. * However, R does not explain A. The cold pack effect is due to the endothermic nature of NH₄NO₃ dissolution, not freezing point depression. Thus, both statements are true, but R does not explain A. ### 02 Marks Questions 11. **I. Give an example of solid solution in which solute is a gas.** **II. Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.** **Answer:** **I.** An example of a solid solution in which the solute is a gas is **hydrogen in palladium (H₂ in Pd)**. Palladium can absorb a significant volume of hydrogen gas, forming a solid solution. **II.** * Mass of aspirin (solute) = 6.5 g * Mass of acetonitrile (solvent) = 450 g * Total mass of solution = Mass of solute + Mass of solvent = 6.5 g + 450 g = 456.5 g * Mass percentage of aspirin = (Mass of aspirin / Total mass of solution) $\times$ 100 * Mass percentage = (6.5 g / 456.5 g) $\times$ 100 $\approx$ **1.42 %** 12. **Henry's law constant for the molality of Methane in Benzene at 298 K is 4.27 × 10⁵ mm Hg. Calculate the solubility of Methane in benzene at 298 K under 760 mm Hg.** **Answer:** * Henry's Law: $P_{gas} = K_H \cdot x_{gas}$ (where $x_{gas}$ is mole fraction of gas in solution) * Given: $K_H = 4.27 \times 10^5 \text{ mm Hg}$ * Partial pressure of methane ($P_{gas}$) = 760 mm Hg * Mole fraction of methane ($x_{gas}$) = $P_{gas} / K_H$ * $x_{gas} = 760 \text{ mm Hg} / (4.27 \times 10^5 \text{ mm Hg}) \approx 0.00178$ * The solubility of methane in benzene at 298 K under 760 mm Hg is approximately **0.00178 (as mole fraction)**. 13. **Give reason.** **I. Meat is preserved for a longer time by salting.** **II. Gases always tend to be less soluble in liquids as the temperature is raised.** **Answer:** **I. Meat is preserved for a longer time by salting** due to the principle of **osmosis**. When meat is heavily salted, the salt creates a hypertonic environment on the surface of the meat. Bacteria and other microorganisms present on the meat have a lower solute concentration inside their cells. Due to osmosis, water diffuses out of the microbial cells into the highly concentrated salt solution, causing the cells to dehydrate, shrivel, and die. This inhibits microbial growth and spoilage. **II. Gases always tend to be less soluble in liquids as the temperature is raised.** This is because the dissolution of gases in liquids is generally an **exothermic process** (releases heat). According to Le Chatelier's principle, if the temperature of the system is increased, the equilibrium will shift in the direction that absorbs heat, which is the reverse direction (gas coming out of solution). Additionally, at higher temperatures, gas molecules have higher kinetic energy and are more likely to escape from the liquid phase. 14. **The sample of drinking water was found to be severely contaminated with chloroform exposure to be a Carcinogen. The level of contamination was 15 ppm (by mass).** **I. Express this in percent by mass.** **II. Determine the molality of chloroform in the water sample.** **Answer:** **I. Express this in percent by mass:** * 15 ppm (parts per million) by mass means 15 grams of chloroform per $10^6$ grams of solution. * Percent by mass = (Mass of chloroform / Mass of solution) $\times$ 100 * Percent by mass = (15 / $10^6$) $\times$ 100 = $15 \times 10^{-4}$ % = **0.0015 %** **II. Determine the molality of chloroform in the water sample:** * Assume 1,000,000 g of solution. * Mass of chloroform = 15 g * Molar mass of chloroform (CHCl₃) = 12.01 (C) + 1.01 (H) + 3 $\times$ 35.45 (Cl) = 119.37 g/mol * Moles of chloroform = 15 g / 119.37 g/mol $\approx$ 0.1256 mol * Mass of water (solvent) $\approx$ Mass of solution - Mass of chloroform = 1,000,000 g - 15 g = 999,985 g $\approx$ 1000 kg (since chloroform is a very small fraction) * Molality (m) = Moles of solute / Mass of solvent (kg) * Molality = 0.1256 mol / 999.985 kg $\approx$ **$1.256 \times 10^{-4}$ mol/kg** (or $1.256 \times 10^{-4}$ m) 15. **Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.** **Answer:** * According to Raoult's Law, $P_{solution} = P^0_{solvent} \cdot x_{solvent}$ * $P^0_{water}$ = 17.535 mm Hg * Molar mass of glucose (C₆H₁₂O₆) = 6 $\times$ 12.01 + 12 $\times$ 1.01 + 6 $\times$ 16.00 = 180.18 g/mol * Moles of glucose = 25 g / 180.18 g/mol $\approx$ 0.1387 mol * Molar mass of water (H₂O) = 2 $\times$ 1.01 + 16.00 = 18.02 g/mol * Moles of water = 450 g / 18.02 g/mol $\approx$ 24.972 mol * Total moles = Moles of glucose + Moles of water = 0.1387 mol + 24.972 mol = 25.1107 mol * Mole fraction of water ($x_{water}$) = Moles of water / Total moles = 24.972 mol / 25.1107 mol $\approx$ 0.9945 * Vapor pressure of solution = $17.535 \text{ mm Hg} \times 0.9945 \approx$ **17.439 mm Hg** ### 03 Marks Questions 16. **Concentrate nitric acid used in laboratory work is 70 % nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL⁻¹.** **Answer:** * Assume 100 g of solution. * Mass of HNO₃ = 70 g (since it's 70% by mass) * Molar mass of HNO₃ = 1.01 (H) + 14.01 (N) + 3 $\times$ 16.00 (O) = 63.02 g/mol * Moles of HNO₃ = 70 g / 63.02 g/mol $\approx$ 1.1107 mol * Density of solution = 1.504 g/mL * Volume of solution = Mass of solution / Density = 100 g / 1.504 g/mL $\approx$ 66.49 mL * Volume of solution in Liters = 66.49 mL / 1000 mL/L = 0.06649 L * Molarity (M) = Moles of solute / Volume of solution (L) = 1.1107 mol / 0.06649 L $\approx$ **16.70 M** 17. **Calculate the osmotic pressure in pascals excited by a solution prepared by dissolving 1 g of polymer of molar mass 185, 000 in 450 mL of water at 37°C.** **Answer:** * Osmotic pressure ($\Pi$) = (n/V)RT = CRT * Mass of polymer = 1 g * Molar mass of polymer = 185,000 g/mol * Moles of polymer (n) = 1 g / 185,000 g/mol $\approx$ $5.405 \times 10^{-6}$ mol * Volume of solution (V) = 450 mL = 0.450 L (assuming volume of solution is approximately volume of solvent for dilute solutions) * Temperature (T) = 37°C + 273.15 = 310.15 K * Gas constant (R) = 8.314 J mol⁻¹ K⁻¹ (for pressure in Pascals) * $\Pi = (5.405 \times 10^{-6} \text{ mol} / 0.450 \text{ L}) \times 8.314 \text{ J mol⁻¹ K⁻¹} \times 310.15 \text{ K}$ * $\Pi \approx 0.00310 \text{ atm}$ (if R = 0.0821 L atm mol⁻¹ K⁻¹) * Using R = 8.314 J mol⁻¹ K⁻¹ (which gives pressure in Pa when V is in m³): * $V = 0.450 \text{ L} = 0.450 \times 10^{-3} \text{ m}^3$ * $\Pi = (5.405 \times 10^{-6} \text{ mol}) / (0.450 \times 10^{-3} \text{ m}^3) \times 8.314 \text{ J mol⁻¹ K⁻¹} \times 310.15 \text{ K}$ * $\Pi \approx$ **31.0 Pascals** 18. **A. Why an unripe mango placed in a concentrated salt solution to prepare pickle shrivels?** **B. What happened to freezing point of benzene when naphthalene is added?** **C. It is advisable to add ethylene glycol to water in car radiator while driving in hill station. Why?** **Answer:** **A. An unripe mango placed in a concentrated salt solution to prepare pickle shrivels** due to **osmosis**. The concentrated salt solution is hypertonic relative to the cells of the mango. Water molecules move from the higher water potential inside the mango cells to the lower water potential in the surrounding salt solution across the semi-permeable cell membranes. This loss of water causes the mango cells to lose turgor, leading to the shriveling of the mango. **B. When naphthalene is added to benzene, the freezing point of benzene will be depressed (lowered).** This is a colligative property known as **freezing point depression**. Naphthalene acts as a non-volatile solute in benzene (the solvent). The presence of solute particles interferes with the formation of the ordered crystal structure of the solvent, requiring a lower temperature for solidification to occur. **C. It is advisable to add ethylene glycol to water in a car radiator while driving in hill stations** because ethylene glycol acts as an **antifreeze**. In hill stations, temperatures can drop significantly below 0°C. If only water is used, it would freeze in the radiator, expand, and potentially damage the engine. Ethylene glycol, being a non-volatile solute, lowers the freezing point of water, preventing it from freezing even at very low temperatures. It also raises the boiling point, which can be beneficial in high-altitude conditions where water boils at a lower temperature. 19. **A. In order of boiling point of 4 equi-molar aqueous solution is C ### 04 Marks Questions 20. **A. State the condition of reverse osmosis.** **B. What is the molarity of K⁺ in aqua solution contains of K₂SO₄? (Molar mass = 174 g mol⁻¹)** **Answer:** **A. Condition of reverse osmosis:** Reverse osmosis occurs when a pressure greater than the osmotic pressure is applied to the solution side of a semi-permeable membrane. Normally, in osmosis, solvent molecules move from a region of lower solute concentration to a region of higher solute concentration. In reverse osmosis, the applied external pressure overcomes the osmotic pressure, forcing the solvent molecules (e.g., water) to move from the higher solute concentration side to the lower solute concentration side, leaving the solute behind. This process is used for desalination of seawater. **B. What is the molarity of K⁺ in aqueous solution contains of K₂SO₄? (Molar mass = 174 g mol⁻¹)** The question is incomplete as it does not provide the concentration (e.g., molarity or mass percentage) of K₂SO₄ solution. Assuming a hypothetical molarity for K₂SO₄ to demonstrate the calculation: * Let's assume the molarity of K₂SO₄ solution is 'X' M. * K₂SO₄ dissociates in water as: K₂SO₄ (aq) $\rightarrow$ 2K⁺ (aq) + SO₄²⁻ (aq) * From the stoichiometry, 1 mole of K₂SO₄ produces 2 moles of K⁺ ions. * Therefore, if the molarity of K₂SO₄ solution is X M, then the molarity of K⁺ ions will be **2X M**. (If a specific concentration of K₂SO₄ was given, e.g., 0.1 M, then the molarity of K⁺ would be 0.2 M). 21. **A 5% solution (by mass) of cane sugar in water has freezing point of 271 K and freezing point of pure water is 273.15 K. What is the freezing point of 5% solution (by mass) of glucose in water?** **Answer:** This problem can be solved using the concept of freezing point depression, $\Delta T_f = K_f \cdot m \cdot i$. Since both cane sugar and glucose are non-electrolytes, $i=1$. First, find $K_f$ for water using the cane sugar data. * For cane sugar solution: * $\Delta T_f = T^0_f - T_f = 273.15 \text{ K} - 271 \text{ K} = 2.15 \text{ K}$ * 5% solution by mass means 5 g cane sugar in 95 g water. * Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 12 $\times$ 12.01 + 22 $\times$ 1.01 + 11 $\times$ 16.00 = 342.34 g/mol * Moles of cane sugar = 5 g / 342.34 g/mol $\approx$ 0.014605 mol * Mass of water (solvent) = 95 g = 0.095 kg * Molality of cane sugar solution ($m_{cane sugar}$) = 0.014605 mol / 0.095 kg $\approx$ 0.1537 mol/kg * $K_f = \Delta T_f / m_{cane sugar} = 2.15 \text{ K} / 0.1537 \text{ mol/kg} \approx 13.99 \text{ K kg/mol}$ (This value of $K_f$ is unusually high for water, which is typically $\approx 1.86 \text{ K kg/mol}$. There might be an error in the provided $\Delta T_f$ for cane sugar, or it's a specific context. We will proceed with this calculated $K_f$). Now, calculate the freezing point depression for 5% glucose solution using this $K_f$. * For glucose solution: * 5% solution by mass means 5 g glucose in 95 g water. * Molar mass of glucose (C₆H₁₂O₆) = 180.18 g/mol * Moles of glucose = 5 g / 180.18 g/mol $\approx$ 0.02775 mol * Mass of water (solvent) = 95 g = 0.095 kg * Molality of glucose solution ($m_{glucose}$) = 0.02775 mol / 0.095 kg $\approx$ 0.2921 mol/kg * $\Delta T_f = K_f \cdot m_{glucose} = 13.99 \text{ K kg/mol} \times 0.2921 \text{ mol/kg} \approx 4.09 \text{ K}$ * Freezing point of glucose solution ($T_f$) = $T^0_f - \Delta T_f = 273.15 \text{ K} - 4.09 \text{ K} =$ **269.06 K** (Note: If we use the standard $K_f$ for water (1.86 K kg/mol), the $\Delta T_f$ for cane sugar would be $1.86 \times 0.1537 \approx 0.286$ K, leading to a freezing point of $273.15 - 0.286 = 272.864$ K, which contradicts the given 271 K. We must adhere to the consistency of the problem's given data.) 22. **KBr is 80% dissociated in aqueous solution of 0.5 m concentration. Calculate the temperature at which solution freezes. [given, $k_b$ for water = 1.86 K Kg mol⁻¹]** **Answer:** * Given molality (m) = 0.5 m * $K_f$ for water = 1.86 K kg/mol (Note: the question provides $k_b$ but uses $K_f$ value in the solution, assuming it meant $K_f$) * KBr dissociates as: KBr (aq) $\rightleftharpoons$ K⁺ (aq) + Br⁻ (aq) * If dissociation is 100%, $i=2$. * Percent dissociation ($\alpha$) = 80% = 0.80 * Van't Hoff factor (i) = $1 + (n-1)\alpha$, where n is the number of ions per formula unit (n=2 for KBr). * $i = 1 + (2-1) \times 0.80 = 1 + 0.80 = 1.80$ * Freezing point depression ($\Delta T_f$) = $i \cdot K_f \cdot m$ * $\Delta T_f = 1.80 \times 1.86 \text{ K kg/mol} \times 0.5 \text{ mol/kg} = 1.674 \text{ K}$ * Freezing point of pure water ($T^0_f$) = 0°C = 273.15 K * Freezing temperature of solution ($T_f$) = $T^0_f - \Delta T_f$ * $T_f = 273.15 \text{ K} - 1.674 \text{ K} =$ **271.476 K** (or -1.674 °C) ### Additional 04 Marks Questions (from second page) 23. **I. When kept in water, raisins swell in size. Name and explain the phenomenon involved with the help of diagram. Give three application of the phenomenon.** **III. What is the elevation in boiling point of solution of 13.44 g of CuCl₂ in 1 kg of water (using the given information, i.e, molecular weight of CuCl₂ = 134.4 and $k_b$ = 0.52 K m⁻¹)?** **Answer:** **I. Raisins swelling in water:** * **Phenomenon:** **Osmosis**. * **Explanation:** Raisins are dried grapes, and their cells contain a high concentration of sugars and other solutes. When placed in pure water (which has a very low solute concentration and thus a high water potential), water molecules move from the surrounding water into the raisin cells across their semi-permeable membranes. This inward movement of water increases the turgor pressure inside the cells, causing the raisins to swell. * **Diagram (Conceptual):** ``` [High Conc. Solute (Raisin)]