GSEB Class 12 Chemistry Solutions Chapter 2 Solutions

Gujarat Board Textbook Solutions Class 12 Chemistry Chapter 2 Solutions

GSEB Class 12 Chemistry Solutions InText Questions and Answers

 

Question 1.Calculate the mass percentage of benzene (C\(_{6}\)H\(_{6}\)) and carbon tetrachloride (CCl\(_{4}\)) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Answer:To determine the mass percentage of benzene, we use the formula: Mass percentage of benzene = \( \frac{W_{C_6H_6}}{W_{C_6H_6} + W_{CCl_4}} \times 100 \) Given, mass of benzene (\(W_{C_6H_6}\)) = 22 g Mass of carbon tetrachloride (\(W_{CCl_4}\)) = 122 g Therefore, mass percentage of benzene = \( \frac{22}{22 + 122} \times 100 = \frac{22}{144} \times 100 = 15.28\% \) The mass percentage of carbon tetrachloride can be calculated by subtracting the mass percentage of benzene from 100%: Mass percentage of CCl\(_{4}\) = \( 100 - \text{mass percentage of benzene} = 100 - 15.28 = 84.72\% \)
In simple words: Mass percentage tells us how much of a component is present in a solution relative to the total mass. We calculate it by dividing the mass of the component by the total mass of the solution and multiplying by 100.

🎯 Exam Tip: Remember that mass percentage calculations are straightforward. Ensure you correctly identify the mass of the solute and solvent, and pay attention to significant figures in your final answer. Mastering this basic concept is crucial for more complex solution chemistry problems.

 

Question 2.Calculate the mole fraction of benzene in the solution containing 30% by mass in carbon tetrachloride.

Answer:Let's assume the total mass of the solution is 100 g. If the solution contains 30% benzene by mass, then: Mass of C\(_{6}\)H\(_{6}\) = 30 g Mass of CCl\(_{4}\) = \( 100 - 30 = 70 \text{ g} \) Next, we calculate the number of moles for each component: Molar mass of C\(_{6}\)H\(_{6}\) = \( (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \text{ g/mol} \) Number of moles of C\(_{6}\)H\(_{6}\) (\(n_{C_6H_6}\)) = \( \frac{\text{Mass}}{\text{Molecular mass}} = \frac{30}{78} \approx 0.385 \text{ mol} \) Molar mass of CCl\(_{4}\) = \( 12 + (4 \times 35.5) = 12 + 142 = 154 \text{ g/mol} \) Number of moles of CCl\(_{4}\) (\(n_{CCl_4}\)) = \( \frac{\text{Mass}}{\text{Molecular mass}} = \frac{70}{154} \approx 0.455 \text{ mol} \) Now, calculate the mole fraction of benzene: Mole fraction of benzene (\(X_{C_6H_6}\)) = \( \frac{n_{C_6H_6}}{n_{C_6H_6} + n_{CCl_4}} = \frac{0.385}{0.385 + 0.455} = \frac{0.385}{0.840} \approx 0.458 \) The mole fraction of CCl\(_{4}\) can be found by subtracting the mole fraction of benzene from 1: Mole fraction of CCl\(_{4}\) = \( 1 - 0.458 = 0.542 \)
In simple words: The mole fraction of a substance in a solution is the ratio of its moles to the total moles of all components. First, determine the mass of each component, convert these masses to moles using molar masses, and then apply the mole fraction formula.

🎯 Exam Tip: When dealing with percentage by mass problems, always assume a convenient total mass (like 100 g) to simplify initial calculations. Remember to calculate the molar mass correctly for each compound, as it's a common source of error. Mole fraction is a unitless quantity, so don't include units in the final mole fraction value.

 

Question 3.Calculate the molarity of each of the following solutions:
(a) 30 g of Co(NO\(_{3}\))\(_{2}\).6H\(_{2}\)O in 4.3 L of solution.
(b) 30 ml of 0.5 M H\(_{2}\)SO\(_{4}\) diluted to 500 ml.

Answer:(a) For 30 g of Co(NO\(_{3}\))\(_{2}\).6H\(_{2}\)O in 4.3 L of solution: First, determine the molar mass (MB) of Co(NO\(_{3}\))\(_{2}\).6H\(_{2}\)O: Molar mass of Co = 58.9 g/mol Molar mass of N = 14 g/mol Molar mass of O = 16 g/mol Molar mass of H = 1 g/mol Molar mass (\(M_B\)) of Co(NO\(_{3}\))\(_{2}\).6H\(_{2}\)O = \( 58.9 + (2 \times (14 + (3 \times 16))) + (6 \times ((2 \times 1) + 16)) \) = \( 58.9 + (2 \times (14 + 48)) + (6 \times 18) \) = \( 58.9 + (2 \times 62) + 108 \) = \( 58.9 + 124 + 108 = 290.9 \text{ g/mol} \) (approx. 291 g/mol as given in OCR) Mass of solute (\(W_B\)) = 30 g Volume of solution (\(V\)) = 4.3 L Molarity = \( \frac{W_B}{M_B \times V \text{ (in L)}} = \frac{30}{291 \times 4.3} = \frac{30}{1251.3} \approx 0.023 \text{ M} \) (b) For 30 ml of 0.5 M H\(_{2}\)SO\(_{4}\) diluted to 500 ml: We use the dilution formula: \( M_1V_1 = M_2V_2 \) Given: Initial molarity (\(M_1\)) = 0.5 M Initial volume (\(V_1\)) = 30 ml Final volume (\(V_2\)) = 500 ml We need to find the final molarity (\(M_2\)). \( M_2 = \frac{M_1V_1}{V_2} = \frac{0.5 \times 30}{500} = \frac{15}{500} = 0.03 \text{ M} \)
In simple words: Molarity measures the concentration of a solution by finding the moles of solute per liter of solution. For dilutions, the total moles of solute remain constant, allowing us to use the \(M_1V_1 = M_2V_2\) formula to find the new concentration.

🎯 Exam Tip: Pay close attention to the units; volume should always be in liters for molarity calculations. When dealing with hydrated salts like Co(NO\(_{3}\))\(_{2}\).6H\(_{2}\)O, ensure you include the mass of the water of hydration in the molar mass calculation. For dilution problems, the \(M_1V_1=M_2V_2\) formula is your go-to, but ensure volumes are in consistent units (either both in ml or both in L).

 

Question 4.Calculate the mass of urea (NH\(_{2}\)CONH\(_{2}\)) required in making 2.5 kg of 0.25 molal aqueous solution?

Answer:A 0.25 molal aqueous solution means that: Moles of urea = 0.25 mole per kg of solvent Mass of solvent (water) = 1 kg = 1000 g First, calculate the molar mass of urea (NH\(_{2}\)CONH\(_{2}\)): Molar mass of N = 14 g/mol Molar mass of H = 1 g/mol Molar mass of C = 12 g/mol Molar mass of O = 16 g/mol Molar mass of urea = \( (2 \times 14) + (4 \times 1) + 12 + 16 = 28 + 4 + 12 + 16 = 60 \text{ g/mol} \) Now, calculate the mass of 0.25 moles of urea: Mass of urea for 1 kg water = \( 0.25 \text{ mol} \times 60 \text{ g/mol} = 15 \text{ g} \) This means 15 g of urea is present in 1000 g of water. So, the total mass of the solution when 15 g of urea is dissolved in 1000 g of water is: Mass of solution = \( 1000 \text{ g (water)} + 15 \text{ g (urea)} = 1015 \text{ g} = 1.015 \text{ kg} \) We need to prepare 2.5 kg of this solution. If 1.015 kg of solution contains 15 g of urea, then 2.5 kg of solution will require: Mass of urea = \( \frac{15 \text{ g urea}}{1.015 \text{ kg solution}} \times 2.5 \text{ kg solution} \) Mass of urea = \( \frac{15 \times 2.5}{1.015} = \frac{37.5}{1.015} \approx 36.94 \text{ g} \), which rounds to approximately 37 g.
In simple words: Molality relates moles of solute to kilograms of solvent. To find the total mass of solute needed for a desired amount of solution, first calculate the mass of solute per kilogram of solvent, then find the total mass of that small solution, and finally scale up to the target total mass.

🎯 Exam Tip: When working with molality, remember it's moles of solute per *kilogram of solvent*, not solution. Be careful with unit conversions, especially between grams and kilograms. A common mistake is to confuse the mass of solution with the mass of solvent. Always clearly distinguish between them in your calculations.

 

Question 5.Calculate (a) molality, (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL\(^{-1}\).

Answer:Let's assume the mass of the solution is 100 g. Given 20% (mass/mass) aqueous KI solution: Mass of KI (\(W_B\)) = 20 g Mass of water (\(W_A\)) = \( 100 - 20 = 80 \text{ g} \) First, calculate the molar mass of KI: Molar mass of K = 39 g/mol Molar mass of I = 127 g/mol Molar mass (\(M_B\)) of KI = \( 39 + 127 = 166 \text{ g/mol} \) (a) Molality (m): Molality is moles of solute per kg of solvent. Molality (m) = \( \frac{W_B \times 1000}{M_B \times W_A \text{ (in g)}} = \frac{20 \times 1000}{166 \times 80} = \frac{20000}{13280} \approx 1.506 \text{ mol kg}^{-1} \) (approx. 1.5 mol kg\(^{-1}\)) (b) Molarity (M): Density of solution = 1.202 g mL\(^{-1}\) Mass of solution = 100 g Volume of solution = \( \frac{\text{Mass of solution}}{\text{Density}} = \frac{100 \text{ g}}{1.202 \text{ g mL}^{-1}} \approx 83.19 \text{ mL} \) Convert volume to liters: \( 83.19 \text{ mL} = 0.08319 \text{ L} \) Molarity (M) = \( \frac{W_B}{M_B \times V \text{ (in L)}} = \frac{20 \text{ g}}{166 \text{ g/mol} \times 0.08319 \text{ L}} = \frac{20}{13.809} \approx 1.448 \text{ mol L}^{-1} \) (approx. 1.45 mol L\(^{-1}\)) Alternatively, using the formula: Molarity = \( \frac{W_B \times 1000 \times \text{Density of solution}}{M_B \times \text{Mass of solution}} \) (if considering 100g of solution and its volume) Molarity = \( \frac{20 \times 1000}{166 \times (\frac{100}{1.202})} = \frac{20000 \times 1.202}{166 \times 100} = \frac{24040}{16600} \approx 1.448 \text{ mol L}^{-1} \) (c) Mole fraction of KI (\(X_B\)): Number of moles of KI (\(n_B\)) = \( \frac{20 \text{ g}}{166 \text{ g/mol}} \approx 0.120 \text{ mol} \) Molar mass of water = 18 g/mol Number of moles of H\(_{2}\)O (\(n_A\)) = \( \frac{80 \text{ g}}{18 \text{ g/mol}} \approx 4.44 \text{ mol} \) Mole fraction of KI (\(X_B\)) = \( \frac{n_B}{n_A + n_B} = \frac{0.120}{4.44 + 0.120} = \frac{0.120}{4.56} \approx 0.026 \)
In simple words: This question asks for three different concentration measures: molality (moles per kg solvent), molarity (moles per liter solution), and mole fraction (moles of component divided by total moles). Each requires calculating the moles of solute and solvent, and molarity additionally needs the solution's volume which can be found using density.

🎯 Exam Tip: When given a percentage by mass, always assume a convenient total mass (e.g., 100 g) to easily find the mass of solute and solvent. Pay close attention to unit conversions for volume (mL to L) and mass (g to kg) when calculating molarity and molality. Remember that mole fraction is a ratio and thus unitless.

 

Question 6.H\(_{2}\)S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H\(_{2}\)S in water at STP is 0.195 m, calculate Henry's law constant?

Answer:Solubility of H\(_{2}\)S is 0.195 m, which means 0.195 moles of H\(_{2}\)S are present in 1 kg (1000 g) of water. Mass of water = 1000 g Molar mass of water (H\(_{2}\)O) = 18 g/mol Number of moles of water = \( \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.5 \text{ mol} \) Number of moles of H\(_{2}\)S = 0.195 mol Now, calculate the mole fraction of H\(_{2}\)S: Mole fraction of H\(_{2}\)S (\(X_{H_2S}\)) = \( \frac{\text{Moles of H}_2\text{S}}{\text{Moles of H}_2\text{S} + \text{Moles of H}_2\text{O}} \) \(X_{H_2S} = \frac{0.195}{0.195 + 55.5} = \frac{0.195}{55.695} \approx 0.0035 \) At STP (Standard Temperature and Pressure), the pressure is usually considered 1 bar for gases. According to Henry's Law, \( p = K_H X \) Where: \(p\) = partial pressure of the gas = 1 bar \(K_H\) = Henry's law constant \(X\) = mole fraction of the gas in the solution = 0.0035 Rearranging the formula to find \(K_H\): \( K_H = \frac{p}{X} = \frac{1 \text{ bar}}{0.0035} \approx 285.71 \text{ bar} \) (approx. 285.9 bar as in OCR due to rounding)
In simple words: Henry's law states that the partial pressure of a gas above a liquid is proportional to its mole fraction in the liquid. To find Henry's constant, first convert the given molality into the mole fraction of the gas, and then divide the gas's pressure (at STP) by this mole fraction.

🎯 Exam Tip: When a gas's solubility is given in molality, it implies moles per kg of solvent. Always convert this to mole fraction to apply Henry's Law. Remember the standard pressure at STP for gases (often 1 bar or 1 atm), as it's crucial for the calculation. Ensure accurate rounding to maintain precision in the final constant value.

 

Question 7.Henry's law constant for CO\(_{2}\) in water is 1.67 \(\times\) 10\(^{8}\) Pa at 298 K. Calculate the quantity of CO\(_{2}\) in 500 ml of soda water when packed under 2.5 atm CO\(_{2}\) pressure at 298 K?

Answer:Given: Henry's law constant (\(K_H\)) = 1.67 \(\times\) 10\(^{8}\) Pa Pressure of CO\(_{2}\) (\(p\)) = 2.5 atm Volume of water = 500 ml = 500 g (assuming density of water is 1 g/mL) Temperature = 298 K First, convert the pressure from atm to Pa: 1 atm = 101325 Pa (approximately 1 \(\times\) 10\(^{5}\) Pa) \(p = 2.5 \text{ atm} \times 101325 \text{ Pa/atm} = 253312.5 \text{ Pa} \) (approx. 2.5 \(\times\) 10\(^{5}\) Pa as in OCR) According to Henry's Law: \( p = K_H X \) Where \(X\) is the mole fraction of CO\(_{2}\). \( X_{CO_2} = \frac{p}{K_H} = \frac{2.5 \times 10^5 \text{ Pa}}{1.67 \times 10^8 \text{ Pa}} \approx 0.0015 \) Now, calculate the number of moles of water in 500 g: Molar mass of water (H\(_{2}\)O) = 18 g/mol Number of moles of water (\(n_{H_2O}\)) = \( \frac{500 \text{ g}}{18 \text{ g/mol}} \approx 27.78 \text{ mol} \) Let \(n_{CO_2}\) be the number of moles of CO\(_{2}\). The mole fraction of CO\(_{2}\) is given by: \( X_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \) Since \(X_{CO_2}\) is very small (0.0015), we can approximate \(n_{CO_2} + n_{H_2O} \approx n_{H_2O}\) for simplicity in the denominator: \( 0.0015 \approx \frac{n_{CO_2}}{27.78} \) \( n_{CO_2} = 0.0015 \times 27.78 \approx 0.04167 \text{ mol} \) (approx. 0.0417 mol as in OCR) Finally, calculate the quantity (mass) of CO\(_{2}\): Molar mass of CO\(_{2}\) = \( 12 + (2 \times 16) = 12 + 32 = 44 \text{ g/mol} \) Mass of CO\(_{2}\) = \( n_{CO_2} \times \text{Molar mass of CO}_2 \) Mass of CO\(_{2}\) = \( 0.0417 \text{ mol} \times 44 \text{ g/mol} \approx 1.8348 \text{ g} \) (approx. 1.836 g)
In simple words: To find the amount of CO\(_{2}\) dissolved in soda water, first use Henry's Law to calculate its mole fraction from the given pressure and constant. Then, convert the water volume to moles and use the mole fraction formula to find the moles of CO\(_{2}\), which can then be converted to mass.

🎯 Exam Tip: Ensure consistent units when using Henry's Law; if \(K_H\) is in Pa, convert pressure to Pa. When the mole fraction of the solute is very small, you can often simplify the denominator in the mole fraction calculation by assuming the moles of solute are negligible compared to the moles of solvent, but it's good practice to verify this assumption if time allows. Always convert moles back to mass if "quantity" is requested.

 

Question 8.The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase?

Answer:Given: Vapour pressure of pure liquid A (\(P^0_A\)) = 450 mm Hg Vapour pressure of pure liquid B (\(P^0_B\)) = 700 mm Hg Total vapour pressure (\(P_{total}\)) = 600 mm Hg According to Raoult's Law, the total vapour pressure of a mixture is given by: \( P_{total} = X_A P^0_A + X_B P^0_B \) We know that \( X_A + X_B = 1 \), so \( X_B = 1 - X_A \). Substitute \(X_B\) in the equation: \( P_{total} = X_A P^0_A + (1 - X_A) P^0_B \) \( 600 = X_A (450) + (1 - X_A) (700) \) \( 600 = 450 X_A + 700 - 700 X_A \) \( 600 - 700 = 450 X_A - 700 X_A \) \( -100 = -250 X_A \) \( X_A = \frac{-100}{-250} = \frac{100}{250} = 0.4 \) Now, calculate \(X_B\): \( X_B = 1 - X_A = 1 - 0.4 = 0.6 \) **Composition of the liquid mixture:** Mole fraction of A (\(X_A\)) = 0.4 Mole fraction of B (\(X_B\)) = 0.6 **Composition of the vapour phase:** The partial pressures of components A and B in the solution are: Partial pressure of A (\(P_A\)) = \( X_A P^0_A = 0.4 \times 450 = 180 \text{ mm Hg} \) Partial pressure of B (\(P_B\)) = \( X_B P^0_B = 0.6 \times 700 = 420 \text{ mm Hg} \) According to Dalton's Law of Partial Pressures, the mole fraction of a component in the vapour phase (\(Y_i\)) is given by: \( Y_i = \frac{P_i}{P_{total}} \) Mole fraction of A in vapour phase (\(Y_A\)) = \( \frac{P_A}{P_{total}} = \frac{180}{600} = 0.3 \) Mole fraction of B in vapour phase (\(Y_B\)) = \( \frac{P_B}{P_{total}} = \frac{420}{600} = 0.7 \)
In simple words: This problem involves using Raoult's Law to determine the liquid phase composition based on pure component vapor pressures and the total vapor pressure. Then, Dalton's Law of Partial Pressures is applied to find the vapor phase composition using the calculated partial pressures and the total vapor pressure.

🎯 Exam Tip: Clearly differentiate between mole fractions in the liquid phase (\(X_A, X_B\)) and the vapor phase (\(Y_A, Y_B\)). Ensure you apply Raoult's Law for the liquid phase and Dalton's Law for the vapor phase correctly. Remember the relationship \(X_A + X_B = 1\) (and \(Y_A + Y_B = 1\)) to simplify calculations.

 

Question 9.Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH\(_{2}\)CONH\(_{2}\)) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Answer:Given: Vapour pressure of pure water (\(P^0\)) = 23.8 mm Hg Mass of urea (\(W_{urea}\)) = 50 g Mass of water (\(W_{water}\)) = 850 g Temperature = 298 K First, calculate the number of moles of each component: Molar mass of urea (NH\(_{2}\)CONH\(_{2}\)) = \( (2 \times 14) + (4 \times 1) + 12 + 16 = 60 \text{ g/mol} \) Moles of urea (\(n_{urea}\)) = \( \frac{50 \text{ g}}{60 \text{ g/mol}} \approx 0.833 \text{ mol} \) Molar mass of water (H\(_{2}\)O) = 18 g/mol Moles of water (\(n_{water}\)) = \( \frac{850 \text{ g}}{18 \text{ g/mol}} \approx 47.22 \text{ mol} \) Next, calculate the mole fraction of water (\(X_{water}\)): \( X_{water} = \frac{n_{water}}{n_{water} + n_{urea}} = \frac{47.22}{47.22 + 0.833} = \frac{47.22}{48.053} \approx 0.9826 \) (approx. 0.9827 as in OCR) Now, calculate the vapour pressure of water for this solution (\(P\)) using Raoult's Law: \( P = P^0 \times X_{water} = 23.8 \text{ mm Hg} \times 0.9826 \approx 23.385 \text{ mm Hg} \) (approx. 23.39 mm Hg) Finally, calculate the relative lowering of vapour pressure: Relative lowering of vapour pressure = \( \frac{P^0 - P}{P^0} \) Alternatively, relative lowering of vapour pressure is equal to the mole fraction of the solute (\(X_{urea}\)): \( X_{urea} = 1 - X_{water} = 1 - 0.9826 = 0.0174 \) (approx. 0.0173)
In simple words: When a non-volatile solute like urea is added to water, the water's vapor pressure decreases. Raoult's law helps calculate this new vapor pressure by multiplying the pure solvent's vapor pressure by its mole fraction in the solution. The relative lowering is simply the fraction by which the vapor pressure has decreased, which is equal to the mole fraction of the solute.

🎯 Exam Tip: Remember that relative lowering of vapor pressure is a colligative property and is directly equal to the mole fraction of the non-volatile solute. Ensure accurate calculation of molar masses and moles for both solvent and solute. Precision in intermediate steps helps avoid rounding errors in the final answer.

 

Question 10.Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C?

Answer:Given: Boiling point of pure water (\(T^0_b\)) = 99.63°C Boiling point of solution (\(T_b\)) = 100°C Mass of solvent (water, \(W_A\)) = 500 g First, calculate the elevation in boiling point (\(\Delta T_b\)): \( \Delta T_b = T_b - T^0_b = 100°C - 99.63°C = 0.37°C \) or 0.37 K Molar mass of sucrose (\(C_{12}H_{22}O_{11}\)) = \( (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \text{ g/mol} \) Molal elevation constant for water (\(K_b\)) = 0.52 K kg mol\(^{-1}\) (standard value) We use the formula for elevation in boiling point: \( \Delta T_b = K_b \times m \) Where \(m\) is the molality. We also know that \( m = \frac{W_B \times 1000}{M_B \times W_A \text{ (in g)}} \) So, \( \Delta T_b = K_b \times \frac{W_B \times 1000}{M_B \times W_A} \) We need to find the mass of sucrose (\(W_B\)). Rearrange the formula: \( W_B = \frac{\Delta T_b \times M_B \times W_A}{K_b \times 1000} \) Substitute the given values: \( W_B = \frac{0.37 \text{ K} \times 342 \text{ g/mol} \times 500 \text{ g}}{0.52 \text{ K kg mol}^{-1} \times 1000} \) \( W_B = \frac{63270}{520} \approx 121.67 \text{ g} \) (approx. 122 g)
In simple words: Adding a solute raises the boiling point of a solvent. To find how much solute is needed, calculate the required boiling point elevation, then use the boiling point elevation formula (which relates elevation to the molality and \(K_b\) of the solvent) to determine the mass of solute.

🎯 Exam Tip: Ensure that the temperature change (\(\Delta T_b\)) is in Kelvin if \(K_b\) is in K kg mol\(^{-1}\), though a change in °C is numerically equivalent to a change in K. Remember to use the correct molar mass for the solute (sucrose) and the correct mass of the *solvent* (water) in grams or kilograms, depending on the formula variant used.

 

Question 11.Calculate the mass of ascorbic acid (Vitamin C, C\(_{6}\)H\(_{8}\)O\(_{6}\)) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K\(_{f}\) = 3.9 K kg mol\(^{-1}\).

Answer:Given: Depression in melting point (\(\Delta T_f\)) = 1.5°C or 1.5 K Mass of solvent (acetic acid, \(W_A\)) = 75 g Molal depression constant for acetic acid (\(K_f\)) = 3.9 K kg mol\(^{-1}\) First, calculate the molar mass of ascorbic acid (C\(_{6}\)H\(_{8}\)O\(_{6}\)): Molar mass of C = 12 g/mol Molar mass of H = 1 g/mol Molar mass of O = 16 g/mol Molar mass (\(M_B\)) of ascorbic acid = \( (6 \times 12) + (8 \times 1) + (6 \times 16) = 72 + 8 + 96 = 176 \text{ g/mol} \) We use the formula for depression in freezing point: \( \Delta T_f = K_f \times m \) Where \(m\) is the molality. And \( m = \frac{W_B \times 1000}{M_B \times W_A \text{ (in g)}} \) So, \( \Delta T_f = K_f \times \frac{W_B \times 1000}{M_B \times W_A} \) We need to find the mass of ascorbic acid (\(W_B\)). Rearrange the formula: \( W_B = \frac{\Delta T_f \times M_B \times W_A}{K_f \times 1000} \) Substitute the given values: \( W_B = \frac{1.5 \text{ K} \times 176 \text{ g/mol} \times 75 \text{ g}}{3.9 \text{ K kg mol}^{-1} \times 1000} \) \( W_B = \frac{19800}{3900} \approx 5.0769 \text{ g} \) (approx. 5.08 g)
In simple words: To lower the melting point of a solvent, a specific amount of solute is needed. This amount is found by using the freezing point depression formula, which links the desired temperature drop to the solvent's cryoscopic constant, the solute's molar mass, and the mass of the solvent.

🎯 Exam Tip: Ensure that all units are consistent (e.g., K for temperature difference, g for mass, g/mol for molar mass). Double-check the molar mass calculation for complex organic compounds like ascorbic acid. Pay attention to whether the problem asks for the mass of solute or solvent, and rearrange the formula accordingly.

 

Question 12.Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 ml of water at 37°C.

Answer:Given: Mass of polymer (\(W_B\)) = 1.0 g Molar mass of polymer (\(M_B\)) = 185,000 g/mol Volume of solution (\(V\)) = 450 ml = 0.450 L Temperature (\(T\)) = 37°C = \( 37 + 273 = 310 \text{ K} \) Gas constant (\(R\)) = 0.083 L bar K\(^{-1}\) mol\(^{-1}\) (We need to convert to Pa eventually, so it's better to use \(R\) in J mol\(^{-1}\) K\(^{-1}\) or convert bar to Pa later). Let's use \(R = 0.083 \text{ L bar K}^{-1}\text{ mol}^{-1}\) first, and then convert bar to Pa. The formula for osmotic pressure (\(\pi\)) is: \( \pi V = nRT \) or \( \pi = CRT \) where \(C\) is molarity (\(n/V\)) \( n = \frac{W_B}{M_B} \) So, \( \pi = \frac{W_B}{M_B} \frac{RT}{V} \) Substitute the values: \( \pi = \frac{1.0 \text{ g}}{185000 \text{ g/mol}} \times \frac{0.083 \text{ L bar K}^{-1}\text{ mol}^{-1} \times 310 \text{ K}}{0.450 \text{ L}} \) \( \pi = \frac{1.0 \times 0.083 \times 310}{185000 \times 0.450} \) \( \pi = \frac{25.73}{83250} \approx 0.000309 \text{ bar} \) (approx. 3.09 \(\times\) 10\(^{-4}\) bar) Now, convert the osmotic pressure from bar to Pascal (Pa): 1 bar = \( 10^5 \) Pa \( \pi = 0.000309 \text{ bar} \times 10^5 \text{ Pa/bar} \approx 30.9 \text{ Pa} \)
In simple words: Osmotic pressure is a colligative property that depends on the concentration of solute particles. To calculate it, use the formula \(\pi = CRT\) (or \(\pi V = nRT\)), where \(C\) is molarity, \(R\) is the gas constant, and \(T\) is temperature in Kelvin. Ensure units are consistent for calculation.

🎯 Exam Tip: Always convert temperature to Kelvin for osmotic pressure calculations. Choose the correct gas constant \(R\) value based on the desired pressure units (e.g., L bar K\(^{-1}\) mol\(^{-1}\) for bar, or J K\(^{-1}\) mol\(^{-1}\) for Pa after unit conversion). Pay close attention to unit conversions, especially when moving between different pressure units like bar and Pascal.

GSEB Class 12 Chemistry Solutions Text Book Questions And Answers

 

Question 1.Define the term solution. How many types of solutions are formed? Write briefly about each type with an example?

Answer:A **solution** is a homogeneous mixture composed of two or more substances. In a solution, the component present in the larger amount is designated as the **solvent**, while the component present in the smaller quantity is called the **solute**. Since both the solvent and solute can exist in gaseous, liquid, or solid states, various types of binary solutions can be formed. Here are the nine possible types:

Solute	Solvent	Example
Gas	Gas	Air (mixture of N\(_{2}\), O\(_{2}\), etc.)
Liquid	Gas	Humidity in air (water vapor in air)
Solid	Gas	Vapours of iodine in air
Gas	Liquid	Aerated water (CO\(_{2}\) in water)
Liquid	Liquid	Alcohol in water
Solid	Liquid	Sugar in water
Gas	Solid	H\(_{2}\) in palladium
Liquid	Solid	Amalgam of mercury with sodium
Solid	Solid	Copper in gold (Alloys)

In simple words: A solution is a uniform mixture of at least two substances, one (solvent) dissolving the other (solute). Solutions can be formed from any combination of gases, liquids, and solids, leading to nine different types with varied examples like air (gas-gas) or sugar water (solid-liquid).

🎯 Exam Tip: Understand the definition of a homogeneous mixture and the roles of solute and solvent. Being able to recall and provide at least one example for each of the nine types of solutions (gas-gas, gas-liquid, etc.) is key to demonstrating a comprehensive understanding of the topic.

 

Question 2.Suppose a solid solution is formed between two substances. One of whose particles are very large and the other particles are very small. What kind of solid solution is this likely to be?

Answer:This type of solid solution is likely a **solid in solid** solution, specifically an **alloy**. An example is copper dissolved in gold. When particles of one substance are significantly smaller than the other, they can often fit into the interstitial spaces of the larger substance's crystal lattice, forming an interstitial alloy, or substitute if the size difference allows.
In simple words: When a solid solution forms with very different particle sizes, like copper in gold, it's typically an alloy where one solid is mixed into another solid.

🎯 Exam Tip: Identify the phases of the solute and solvent first. For solid-in-solid solutions, consider alloys as the primary example. Understanding how particle size differences influence the type of solid solution (e.g., substitutional vs. interstitial) is beneficial for more detailed questions.

 

Question 3.Define the following terms:
1. Mole fraction
2. Molality
3. Molarity
4. Mass percentage

Answer:1. **Mole fraction (X):** The mole fraction of any component within a solution is defined as the ratio of the number of moles of that specific component to the total number of moles of all components present in the solution. For a binary solution containing components A and B: Mole fraction of A, \(X_A = \frac{\text{Number of moles of component A}}{\text{Total number of moles of A and B}} = \frac{n_A}{n_A + n_B} \) where \(n_A\) and \(n_B\) are the numbers of moles of components A and B, respectively. 2. **Molality (m):** Molality represents the number of moles of the solute that are present per kilogram (or 1000 g) of the solvent. The standard unit for molality is moles per kg (mol kg\(^{-1}\)) or m (molal). Molality = \( \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{W_B \times 1000}{M_B \times W_A \text{ in gram}} \) 3. **Molarity (M):** Molarity quantifies the number of moles of the solute present per litre of the solution. The unit of molarity is moles per litre (mol L\(^{-1}\)) or M (molar). Molarity = \( \frac{\text{Number of moles of solute}}{\text{Volume in litres of the solution}} = \frac{W_B \times 1000}{M_B \times V \text{ in ml}} \) 4. **Mass percentage (\(w/W\)):** Mass percentage refers to the mass of a component present in 100 mass units of the solution. It is a unitless measure. Mass percentage of solute = \( \frac{\text{Mass of solute} \times 100}{\text{(Mass of solute + Mass of solvent)}} = \frac{W_B \times 100}{(W_A + W_B)} \)
In simple words: These terms describe solution concentration: mole fraction shows component moles relative to total moles, molality is moles of solute per kilogram of solvent, molarity is moles of solute per liter of solution, and mass percentage is the mass of a component per 100 units of total solution mass.

🎯 Exam Tip: Understand the key distinction between molality (per kg *solvent*) and molarity (per L *solution*), as this is a frequent point of confusion. Always remember the units for each concentration term. Be precise with the definitions and ensure you can write out the corresponding mathematical formulas for each.

 

Question 4.Concentrated nitric acid used in laboratory works is 68% 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:Given: Mass percentage of HNO\(_{3}\) = 68% Density of solution = 1.504 g mL\(^{-1}\) Assume 1 litre (1000 mL) of the solution. Mass of 1 litre of solution = Density \(\times\) Volume Mass of solution = \( 1.504 \text{ g/mL} \times 1000 \text{ mL} = 1504 \text{ g} \) Since the solution is 68% nitric acid by mass: Mass of HNO\(_{3}\) in 1504 g of solution = \( \frac{68}{100} \times 1504 \text{ g} = 1022.72 \text{ g} \) Next, calculate the molar mass of HNO\(_{3}\): Molar mass of H = 1 g/mol Molar mass of N = 14 g/mol Molar mass of O = 16 g/mol Molar mass of HNO\(_{3}\) = \( 1 + 14 + (3 \times 16) = 1 + 14 + 48 = 63 \text{ g/mol} \) Now, calculate the molarity: Molarity = \( \frac{\text{Mass of solute in g}}{\text{Molecular mass of solute}} \times \frac{1}{\text{Volume of solution in L}} \) Molarity = \( \frac{1022.72 \text{ g}}{63 \text{ g/mol}} \times \frac{1}{1 \text{ L}} = 16.2336 \text{ M} \) (approx. 16.23 M)
In simple words: To find the molarity of a percentage-by-mass solution with a given density, first calculate the total mass of a specific volume (like 1 liter) of solution. Then, find the mass of the solute using the percentage, convert this mass to moles, and finally divide by the volume in liters.

🎯 Exam Tip: Always begin by assuming a convenient volume (e.g., 1 L or 1000 mL) for molarity problems where density and mass percentage are given. This allows direct calculation of the mass of solution and subsequently the mass of solute. Molar mass calculation for the acid is a critical step, so ensure its accuracy.

 

Question 5.A solution of glucose in water is labelled as 10% w/W, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g ml\(^{-1}\), then what shall be the molarity of the solution?

Answer:Given: Solution of glucose in water is 10% w/W (weight/weight). Density of solution = 1.2 g mL\(^{-1}\) Assume total mass of solution = 100 g. Mass of glucose (\(W_B\)) = 10 g Mass of water (\(W_A\)) = \( 100 - 10 = 90 \text{ g} \) First, calculate molar masses: Molar mass of glucose (C\(_{6}\)H\(_{12}\)O\(_{6}\)) = \( (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \text{ g/mol} \) Molar mass of water (H\(_{2}\)O) = 18 g/mol **1. Molality (m):** Molality = \( \frac{\text{Moles of glucose}}{\text{Mass of water in kg}} \) Moles of glucose (\(n_B\)) = \( \frac{10 \text{ g}}{180 \text{ g/mol}} = 0.0556 \text{ mol} \) Mass of water in kg = \( \frac{90 \text{ g}}{1000 \text{ g/kg}} = 0.090 \text{ kg} \) Molality (m) = \( \frac{0.0556 \text{ mol}}{0.090 \text{ kg}} \approx 0.617 \text{ mol kg}^{-1} \) **2. Mole fraction of each component:** Moles of glucose (\(n_B\)) = 0.0556 mol Moles of water (\(n_A\)) = \( \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ mol} \) Total moles = \( n_A + n_B = 5 + 0.0556 = 5.0556 \text{ mol} \) Mole fraction of glucose (\(X_B\)) = \( \frac{n_B}{n_A + n_B} = \frac{0.0556}{5.0556} \approx 0.011 \) Mole fraction of water (\(X_A\)) = \( \frac{n_A}{n_A + n_B} = \frac{5}{5.0556} \approx 0.989 \) (Alternatively, \(X_A = 1 - X_B = 1 - 0.011 = 0.989\)) **3. Molarity (M):** Molarity = \( \frac{\text{Moles of glucose}}{\text{Volume of solution in L}} \) Mass of solution = 100 g Density of solution = 1.2 g mL\(^{-1}\) Volume of solution = \( \frac{\text{Mass of solution}}{\text{Density}} = \frac{100 \text{ g}}{1.2 \text{ g mL}^{-1}} \approx 83.33 \text{ mL} \) Volume in L = \( \frac{83.33 \text{ mL}}{1000 \text{ mL/L}} = 0.08333 \text{ L} \) Molarity (M) = \( \frac{0.0556 \text{ mol}}{0.08333 \text{ L}} \approx 0.667 \text{ mol L}^{-1} \)
In simple words: This problem asks for three concentration units (molality, mole fraction, and molarity) for a glucose solution. Start by assuming a total mass, then calculate moles of each component. Use these values along with the solution's density to find the molality (moles per kg solvent), mole fractions (moles of component over total moles), and molarity (moles per liter solution).

🎯 Exam Tip: When given a percentage by mass, assuming a 100 g solution simplifies mass calculations for both solute and solvent. Remember to convert grams of solvent to kilograms for molality. For molarity, convert total solution mass to volume using density, and then convert volume to liters.

 

Question 6.A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution?

Answer:Let's first calculate the mass of solute in each initial solution. **From the first solution:** Mass of 25% solution = 300 g Mass of solute in 300 g of 25% solution = \( \frac{25}{100} \times 300 \text{ g} = 75 \text{ g} \) **From the second solution:** Mass of 40% solution = 400 g Mass of solute in 400 g of 40% solution = \( \frac{40}{100} \times 400 \text{ g} = 160 \text{ g} \) Now, calculate the total mass of solute and total mass of the resulting solution. Total mass of solute = Mass of solute from first solution + Mass of solute from second solution Total mass of solute = \( 75 \text{ g} + 160 \text{ g} = 235 \text{ g} \) Total mass of solution = Mass of first solution + Mass of second solution Total mass of solution = \( 300 \text{ g} + 400 \text{ g} = 700 \text{ g} \) Finally, calculate the mass percentage of the resulting solution: Mass percentage of solute = \( \frac{\text{Total mass of solute}}{\text{Total mass of solution}} \times 100 \) Mass percentage of solute = \( \frac{235 \text{ g}}{700 \text{ g}} \times 100 \approx 33.57\% \) (approx. 33.5% as in OCR) The mass percentage of solvent in the resulting solution = \( 100\% - 33.57\% = 66.43\% \) (approx. 66.5%)
In simple words: To find the mass percentage of a mixed solution, first calculate the total mass of solute from each individual solution. Then, sum these solute masses and divide by the total mass of all solutions combined, multiplying by 100 to get the percentage.

🎯 Exam Tip: This type of problem requires careful tracking of both solute and solution masses for each component before mixing. Summing the solute masses and total solution masses separately is crucial before calculating the final mass percentage. Avoid premature rounding to maintain accuracy.

 

Question 7.An antifreeze solution is prepared from 222.6 g of ethylene glycol (C\(_{2}\)H\(_{6}\)O\(_{2}\)) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g ml\(^{-1}\) then what shall be the molarity of the solution?

Answer:Given: Mass of ethylene glycol (\(W_B\)) = 222.6 g Mass of water (\(W_A\)) = 200 g Density of solution = 1.072 g mL\(^{-1}\) First, calculate the molar mass of ethylene glycol (C\(_{2}\)H\(_{6}\)O\(_{2}\)): Molar mass of C = 12 g/mol Molar mass of H = 1 g/mol Molar mass of O = 16 g/mol Molar mass (\(M_B\)) of ethylene glycol = \( (2 \times 12) + (6 \times 1) + (2 \times 16) = 24 + 6 + 32 = 62 \text{ g/mol} \) **1. Molality (m):** Molality = \( \frac{\text{Moles of ethylene glycol}}{\text{Mass of water in kg}} \) Moles of ethylene glycol (\(n_B\)) = \( \frac{222.6 \text{ g}}{62 \text{ g/mol}} = 3.59 \text{ mol} \) Mass of water in kg = \( \frac{200 \text{ g}}{1000 \text{ g/kg}} = 0.200 \text{ kg} \) Molality (m) = \( \frac{3.59 \text{ mol}}{0.200 \text{ kg}} = 17.95 \text{ mol kg}^{-1} \) **2. Molarity (M):** Molarity = \( \frac{\text{Moles of ethylene glycol}}{\text{Volume of solution in L}} \) Total mass of solution = Mass of ethylene glycol + Mass of water Total mass of solution = \( 222.6 \text{ g} + 200 \text{ g} = 422.6 \text{ g} \) Volume of solution = \( \frac{\text{Total mass of solution}}{\text{Density}} = \frac{422.6 \text{ g}}{1.072 \text{ g mL}^{-1}} \approx 394.22 \text{ mL} \) Volume in L = \( \frac{394.22 \text{ mL}}{1000 \text{ mL/L}} = 0.39422 \text{ L} \) Molarity (M) = \( \frac{3.59 \text{ mol}}{0.39422 \text{ L}} \approx 9.106 \text{ mol L}^{-1} \) (approx. 9.1 M)
In simple words: This problem asks for both molality and molarity of an antifreeze solution. Molality is calculated by finding moles of solute per kg of solvent. Molarity requires the total volume of the solution, which is found by combining solute and solvent masses and dividing by the solution's density.

🎯 Exam Tip: Clearly distinguish between molality (moles of solute per kg of *solvent*) and molarity (moles of solute per L of *solution*). Remember that to calculate molarity when density is provided, you must first calculate the total mass of the solution and then use the density to find the total volume. Pay attention to unit conversions (g to kg, mL to L).

 

Question 8.A sample of drinking water was found to be severely contaminated with chloroform (CHCl\(_{3}\)) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
1. Express this in percent by mass
2. Determine the molality of chloroform in the water sample.

Answer:Given: Contamination level of chloroform = 15 ppm (by mass) **1. Express in percent by mass:** Parts per million (ppm) is defined as: \( \text{ppm} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6 \) Percent by mass is defined as: \( \text{Percent by mass} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \) So, \( \text{Percent by mass} = \frac{\text{ppm}}{10^6} \times 100 \) Percent by mass = \( \frac{15}{10^6} \times 100 = 15 \times 10^{-4} \% \) **2. Determine the molality of chloroform:** 15 ppm by mass means 15 g of CHCl\(_{3}\) is present in \( 10^6 \) g of solution. Mass of chloroform (\(W_B\)) = 15 g Mass of solution = \( 10^6 \) g Mass of solvent (water, \(W_A\)) = Mass of solution - Mass of solute = \( 10^6 \text{ g} - 15 \text{ g} \approx 10^6 \text{ g} \) (since 15 g is negligible compared to \( 10^6 \) g) First, calculate the molar mass of chloroform (CHCl\(_{3}\)): Molar mass of C = 12 g/mol Molar mass of H = 1 g/mol Molar mass of Cl = 35.5 g/mol Molar mass (\(M_B\)) of CHCl\(_{3}\) = \( 12 + 1 + (3 \times 35.5) = 13 + 106.5 = 119.5 \text{ g/mol} \) Now, calculate the molality (m): Molality = \( \frac{\text{Moles of chloroform}}{\text{Mass of water in kg}} \) Moles of chloroform = \( \frac{W_B}{M_B} = \frac{15 \text{ g}}{119.5 \text{ g/mol}} \approx 0.1255 \text{ mol} \) Mass of water in kg = \( \frac{10^6 \text{ g}}{1000 \text{ g/kg}} = 1000 \text{ kg} \) Molality (m) = \( \frac{0.1255 \text{ mol}}{1000 \text{ kg}} = 1.255 \times 10^{-4} \text{ mol kg}^{-1} \) (approx. 1.26 \(\times\) 10\(^{-4}\) m)
In simple words: To convert parts per million (ppm) to mass percentage, divide ppm by \(10^4\). To find molality, convert the given ppm concentration to masses of solute and solvent, calculate the solute's moles, and then divide by the solvent's mass in kilograms.

🎯 Exam Tip: When dealing with very low concentrations (like ppm), the mass of the solvent can often be approximated as the mass of the solution itself. Remember the conversion factor between ppm and percent by mass. Accurate molar mass calculation is crucial for molality.

 

Question 9.What role does the molecular interaction play in a solution of alcohol and water?

Answer:In a solution of alcohol and water, strong molecular interactions play a crucial role. Both alcohol (e.g., ethanol) and water molecules are polar and capable of forming extensive hydrogen bonds with each other. The hydroxyl (-OH) groups in alcohol molecules can form hydrogen bonds with water molecules, and water molecules can form hydrogen bonds with each other and with alcohol molecules. These strong intermolecular hydrogen bonding interactions between alcohol and water molecules are comparable to or even stronger than the hydrogen bonds within pure alcohol or pure water. This extensive network of hydrogen bonding allows them to mix completely in all proportions, making them perfectly miscible.
In simple words: Alcohol and water mix completely because their molecules form strong hydrogen bonds with each other. These new interactions are similar to or stronger than the bonds within their pure forms, allowing them to dissolve fully into one another.

🎯 Exam Tip: For questions about miscibility, always consider the types of intermolecular forces present, especially hydrogen bonding. The strength of solute-solvent interactions relative to solute-solute and solvent-solvent interactions determines solubility. Emphasize the role of hydrogen bonding for alcohol and water miscibility.

 

Question 10.Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer:Gases tend to be less soluble in liquids as the temperature is raised because the dissolution of gases in liquids is an exothermic process, meaning it releases heat. According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in temperature, it will adjust itself to counteract the change. For gas dissolution: \( \text{Gas} + \text{Liquid} \rightleftharpoons \text{Solution} + \text{Heat} \) When the temperature is increased, the equilibrium shifts in the direction that absorbs heat, which is the reverse direction (towards the undissolved gas). This causes the dissolved gas to escape from the solution. Furthermore, as temperature increases, the kinetic energy of the gaseous molecules also increases. These higher-energy gas molecules are more likely to overcome the attractive forces holding them in the liquid phase and escape into the gas phase, thereby reducing their solubility.
In simple words: Gases become less soluble in liquids at higher temperatures because dissolving gases is an exothermic process. Increased heat shifts the equilibrium, causing dissolved gas molecules (which now have more kinetic energy) to escape the liquid and return to the gaseous state.

🎯 Exam Tip: Always link the effect of temperature on gas solubility to the exothermic nature of the dissolution process and Le Chatelier's principle. Also, mention the increased kinetic energy of gas molecules at higher temperatures as a contributing factor for them escaping the solution.

 

Question 11.State Henry's law and mention some important applications?

Answer:**Henry's Law:** Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the surface of the liquid. Mathematically, it can be expressed as \( p = K_H X \), where \(p\) is the partial pressure of the gas, \(K_H\) is Henry's law constant, and \(X\) is the mole fraction of the gas in the solution. **Applications:** 1. **Production of carbonated beverages:** To increase the solubility of CO\(_{2}\) in soft drinks and soda water, the bottles are sealed under high pressure. When the bottle is opened, the pressure above the liquid decreases, and CO\(_{2}\) bubbles out rapidly due to decreased solubility. 2. **Deep-sea diving (Decompression Sickness or 'Bends'):** Divers breathe compressed air (mixture of N\(_{2}\) and O\(_{2}\)) at high pressures underwater. Due to increased pressure, the solubility of atmospheric gases (especially N\(_{2}\)) in blood and body tissues increases. When divers ascend to the surface, the external pressure decreases, causing the dissolved gases to become less soluble and form bubbles in the blood, leading to 'bends'. To avoid this, tanks are filled with air diluted with helium (11.7% He, 56.2% N\(_{2}\), 32.1% O\(_{2}\)) because helium is less soluble in blood. 3. **Functioning of lungs (Respiration):** In the lungs, the partial pressure of oxygen is high, leading to its dissolution in the blood (forming oxyhaemoglobin). Conversely, in the tissues, the partial pressure of oxygen is lower, causing oxyhaemoglobin to release oxygen for cellular activity. Similarly, CO\(_{2}\) is taken up from the tissues (where its partial pressure is high) by the blood and released in the lungs (where its partial pressure is low).
In simple words: Henry's law explains that more gas dissolves in a liquid when the gas pressure above it is higher. This principle is used in making fizzy drinks, preventing 'the bends' in deep-sea divers by adjusting gas mixtures, and for efficient gas exchange in our lungs during breathing.

🎯 Exam Tip: When defining Henry's Law, include both the direct proportionality and the constant temperature condition. For applications, describe each example clearly, explaining how Henry's Law applies (e.g., increased pressure for carbonation, decreased pressure for bends, partial pressure differences in lungs).

 

Question 12.The partial pressure of ethane over a solution containing 6.56 \(\times\) 10\(^{-3}\) g of ethane is 1 bar. If the solution contains 5.00 \(\times\) 10\(^{-2}\) g of ethane, then what shall be the partial pressure of the gas?

Answer:According to Henry's Law, the solubility of a gas (\(m\)) is directly proportional to its partial pressure (\(p\)): \( m = K_H p \) Where \(K_H\) is Henry's law constant. This means \( K_H = \frac{m}{p} \). Since \(K_H\) is constant for a given gas at a specific temperature, we can write: \( \frac{m_1}{p_1} = \frac{m_2}{p_2} \) **First case:** Mass of ethane (\(m_1\)) = 6.56 \(\times\) 10\(^{-3}\) g Partial pressure (\(p_1\)) = 1 bar From this, we can find Henry's constant for ethane: \( K_H = \frac{6.56 \times 10^{-3} \text{ g}}{1 \text{ bar}} = 6.56 \times 10^{-3} \text{ g bar}^{-1} \) **Second case:** Mass of ethane (\(m_2\)) = 5.00 \(\times\) 10\(^{-2}\) g We need to find the new partial pressure (\(p_2\)). Using the formula \( p_2 = \frac{m_2}{K_H} \): \( p_2 = \frac{5.00 \times 10^{-2} \text{ g}}{6.56 \times 10^{-3} \text{ g bar}^{-1}} \) \( p_2 = \frac{0.050}{0.00656} \approx 7.6219 \text{ bar} \) (approx. 7.62 bar)
In simple words: Henry's law states that the amount of gas dissolved is directly proportional to its pressure. If we know the dissolved amount at one pressure, we can use this proportionality to find the pressure needed for a different dissolved amount, as Henry's constant remains the same.

🎯 Exam Tip: For problems involving changes in gas solubility and pressure, apply Henry's Law as a ratio: \(\frac{m_1}{p_1} = \frac{m_2}{p_2}\). Ensure consistent units for mass and pressure. This approach avoids needing to explicitly calculate \(K_H\) if not asked, though it's a good intermediate check.

 

Question 13.What is meant by positive and negative deviations from Raoult's law and how is the sign of \(\Delta_{mixing}H\) related to positive and negative deviations from Raoult's law?

Answer:A **non-ideal solution** is one that does not strictly adhere to Raoult's law across all concentrations and temperatures. Unlike ideal solutions where \(P_A = X_A P^0_A\) and \(P_B = X_B P^0_B\), non-ideal solutions exhibit deviations. These deviations are categorized into two types: **1. Non-ideal solutions showing positive deviation:** In these solutions, the partial vapour pressure of each component and the total vapour pressure of the solution are greater than what would be predicted by Raoult's law. That is, \( P_A > X_A P^0_A \), \( P_B > X_B P^0_B \) and \( P_S > X_A P^0_A + X_B P^0_B \). **Cause of positive deviation:** This occurs when the intermolecular attractive forces between the solute and solvent (A-B interactions) are weaker than the attractive forces between molecules of the pure components (A-A and B-B interactions). Because of these weaker A-B attractions, molecules escape more easily from the solution into the vapour phase. Consequently, the partial pressures of both components over the solution are higher than expected. * **Sign of \(\Delta_{mixing}H\):** Due to the decrease in attractive forces when mixing, energy is absorbed, making the process **endothermic**. Therefore, \(\Delta_{mixing}H > 0\) (positive). * **Sign of \(\Delta_{mixing}V\):** The weaker attractions also mean particles are held less tightly, leading to an **increase in volume** upon mixing. Therefore, \(\Delta_{mixing}V > 0\) (positive). **
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक आदर्श विलयन के लिए वाष्प दाब-मोल अंश संबंध को दर्शाता है। वक्र \(P^0_A\) और \(P^0_B\) शुद्ध घटकों A और B के वाष्प दाब हैं। सीधी रेखाएं Raoult के नियम के अनुसार घटकों A और B के आंशिक वाष्प दाब को उनके मोल अंश के फलन के रूप में दर्शाती हैं। वक्र P कुल वाष्प दाब को दर्शाता है। धनात्मक विचलन में, वास्तविक वक्र इन सीधी रेखाओं के ऊपर स्थित होते हैं, जो अपेक्षित से अधिक वाष्प दाब दिखाते हैं।** **Examples of solutions showing positive deviation:** * Acetone and CS\(_{2}\) * Acetone and alcohol (e.g., ethanol) * Acetone and benzene * Water and alcohol (e.g., ethanol) * Ether and acetone For instance, in a solution of ethanol and cyclohexane, pure ethanol molecules form strong hydrogen bonds. When cyclohexane is added, its molecules position themselves between ethanol molecules, disrupting the hydrogen bond network. This weakens the intermolecular attractions, leading to positive deviation. **2. Non-ideal solutions showing negative deviation:** In these solutions, the partial vapour pressure of each component and the total vapour pressure of the solution are less than what would be predicted by Raoult's law. That is, \( P_A < X_A P^0_A \), \( P_B < X_B P^0_B \) and \( P_S < X_A P^0_A + X_B P^0_B \). **Cause of negative deviation:** This occurs when the intermolecular attractive forces between the solute and solvent (A-B interactions) are stronger than the attractive forces between molecules of the pure components (A-A and B-B interactions). Because of these stronger A-B attractions, molecules have a reduced tendency to escape from the solution into the vapour phase. Consequently, the partial pressures of both components over the solution are lower than expected. * **Sign of \(\Delta_{mixing}H\):** Due to the increase in attractive forces when mixing, energy is released, making the process **exothermic**. Therefore, \(\Delta_{mixing}H < 0\) (negative). * **Sign of \(\Delta_{mixing}V\):** The stronger attractions also mean particles are held more tightly, leading to a **decrease in volume** upon mixing. Therefore, \(\Delta_{mixing}V < 0\) (negative). **
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक आदर्श विलयन के लिए वाष्प दाब-मोल अंश संबंध को दर्शाता है, जिसमें वक्र P कुल वाष्प दाब को प्रदर्शित करता है। ऋणात्मक विचलन में, घटकों A और B के आंशिक वाष्प दाब के वास्तविक वक्र (डॉट्स) Raoult के नियम द्वारा अपेक्षित सीधी रेखाओं के नीचे स्थित होते हैं, जो अपेक्षित से कम वाष्प दाब दिखाते हैं।** **Examples of solutions showing negative deviation:** * Chloroform and ethers * Benzene and chloroform * Water and nitric acid * Acetone and aniline * Chloroform and nitric acid For example, in a solution of chloroform and acetone, hydrogen bonding occurs between the hydrogen of chloroform and the oxygen of acetone, creating stronger intermolecular attractions than in their pure states. This increased attraction reduces the escaping tendency of molecules, leading to negative deviation.
In simple words: Positive deviation from Raoult's law means the solution's vapor pressure is higher than expected due to weaker solute-solvent interactions, making mixing endothermic (\(\Delta H > 0\)) and increasing volume (\(\Delta V > 0\)). Negative deviation means vapor pressure is lower than expected due to stronger solute-solvent interactions, making mixing exothermic (\(\Delta H < 0\)) and decreasing volume (\(\Delta V < 0\)).

🎯 Exam Tip: For this question, clearly define both types of deviations, explain their causes in terms of intermolecular forces, and state the sign of \(\Delta_{mixing}H\) and \(\Delta_{mixing}V\) for each. Providing suitable examples for each type of deviation is essential for a complete answer. Remember to explain the underlying energetic (endothermic/exothermic) and volume changes.

 

Question 14.An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer:Given: Percentage of non-volatile solute = 2% by mass Pressure exerted by the solution (\(P_A\)) = 1.004 bar Normal boiling point of the solvent (water) means \(P^0_A\) (pure water) = 1.013 bar (at 100°C) Assume total mass of solution = 100 g. Mass of solute (\(W_B\)) = 2 g Mass of solvent (water, \(W_A\)) = \( 100 - 2 = 98 \text{ g} \) Molar mass of solvent (water, \(M_A\)) = 18 g/mol According to Raoult's Law for relative lowering of vapor pressure: \( \frac{P^0_A - P_A}{P^0_A} = X_B = \frac{n_B}{n_A + n_B} \) For a dilute solution, \( n_A + n_B \approx n_A \). So, \( \frac{P^0_A - P_A}{P^0_A} \approx \frac{n_B}{n_A} = \frac{W_B/M_B}{W_A/M_A} = \frac{W_B M_A}{M_B W_A} \) Rearrange to solve for \(M_B\): \( M_B = \frac{W_B M_A P^0_A}{W_A (P^0_A - P_A)} \) Substitute the values: \( W_B = 2 \text{ g} \) \( M_A = 18 \text{ g/mol} \) \( P^0_A = 1.013 \text{ bar} \) \( W_A = 98 \text{ g} \) \( P_A = 1.004 \text{ bar} \) \( M_B = \frac{2 \times 18 \times 1.013}{98 \times (1.013 - 1.004)} \) \( M_B = \frac{36.468}{98 \times 0.009} \) \( M_B = \frac{36.468}{0.882} \approx 41.346 \text{ g mol}^{-1} \) (approx. 41.3 g mol\(^{-1}\))
In simple words: This problem uses the relative lowering of vapor pressure to find the molar mass of an unknown solute. By applying Raoult's Law, which relates the change in vapor pressure to the mole fraction of the solute, we can calculate the solute's molar mass using the given solution's vapor pressure, pure solvent's vapor pressure, and masses.

🎯 Exam Tip: Remember the full form of the relative lowering of vapor pressure equation, and know how to simplify it for dilute solutions (\(n_B \ll n_A\)). Ensure you correctly identify the pure solvent's vapor pressure (\(P^0_A\)) at its normal boiling point. Accurate arithmetic, especially for the denominator, is vital.

 

Question 15.Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Answer:Given: Vapour pressure of pure heptane (\(P^0_{heptane}\)) = 105.2 kPa Vapour pressure of pure octane (\(P^0_{octane}\)) = 46.8 kPa Mass of heptane = 26.0 g Mass of octane = 35 g Temperature = 373 K First, calculate the number of moles of each component: Molar mass of heptane (C\(_{7}\)H\(_{16}\)) = \( (7 \times 12) + (16 \times 1) = 84 + 16 = 100 \text{ g/mol} \) Moles of heptane (\(n_{heptane}\)) = \( \frac{26.0 \text{ g}}{100 \text{ g/mol}} = 0.26 \text{ mol} \) Molar mass of octane (C\(_{8}\)H\(_{18}\)) = \( (8 \times 12) + (18 \times 1) = 96 + 18 = 114 \text{ g/mol} \) Moles of octane (\(n_{octane}\)) = \( \frac{35 \text{ g}}{114 \text{ g/mol}} \approx 0.307 \text{ mol} \) Total moles = \( n_{heptane} + n_{octane} = 0.26 + 0.307 = 0.567 \text{ mol} \) Now, calculate the mole fraction of each component: Mole fraction of heptane (\(X_{heptane}\)) = \( \frac{n_{heptane}}{\text{Total moles}} = \frac{0.26}{0.567} \approx 0.458 \) Mole fraction of octane (\(X_{octane}\)) = \( \frac{n_{octane}}{\text{Total moles}} = \frac{0.307}{0.567} \approx 0.542 \) (Alternatively, \(X_{octane} = 1 - X_{heptane} = 1 - 0.458 = 0.542\)) Since it's an ideal solution, Raoult's Law applies: Total vapour pressure (\(P_{total}\)) = \( X_{heptane} P^0_{heptane} + X_{octane} P^0_{octane} \) Partial vapour pressure due to heptane = \( 0.458 \times 105.2 \text{ kPa} \approx 48.18 \text{ kPa} \) Partial vapour pressure due to octane = \( 0.542 \times 46.8 \text{ kPa} \approx 25.36 \text{ kPa} \) Total vapour pressure = \( 48.18 \text{ kPa} + 25.36 \text{ kPa} = 73.54 \text{ kPa} \)
In simple words: For an ideal solution, the total vapor pressure is found using Raoult's Law. First, convert the given masses of heptane and octane into moles, then calculate their mole fractions. Finally, multiply each component's mole fraction by its pure vapor pressure and sum these partial pressures to get the total vapor pressure of the mixture.

🎯 Exam Tip: Always calculate the moles and then mole fractions accurately for each component. Remember that Raoult's Law is directly applicable to ideal solutions, allowing you to sum the partial pressures to find the total pressure. Ensure all pressure units are consistent throughout the calculation.

 

Question 16.The vapour pressure of water is 12.3 kPa at 300K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it?

Answer:The vapor pressure of pure water (\(P^0_{\text{H}_2\text{O}}\)) is given as 12.3 kPa. For a 1 molal solution of a non-volatile solute: Number of moles of water in 1 kg (1000 g) = \( \frac{1000}{18} = 55.55 \) moles. Number of moles of solute = 1 mole. Total moles = \( 55.55 + 1 = 56.55 \) moles. Mole fraction of water (\(X_{\text{water}}\)) = \( \frac{\text{Moles of water}}{\text{Total moles}} = \frac{55.55}{56.55} \approx 0.9823 \). Using Raoult's Law, the vapor pressure of the solution (\(P\)) is given by: \( P = X_{\text{water}} \times P^0_{\text{H}_2\text{O}} \) \( P = 0.9823 \times 12.3 \text{ kPa} \) \( P = 12.08 \text{ kPa} \) The vapor pressure of the 1 molal solution is approximately 12.08 kPa.
In simple words: When a non-volatile substance is dissolved in water, the water's vapor pressure decreases. By finding the mole fraction of water in the solution, Raoult's law allows us to calculate this reduced vapor pressure.

🎯 Exam Tip: Remember to use Raoult's Law (P = X_solventP⁰_solvent) for calculating vapor pressure of solutions containing non-volatile solutes. Ensure correct calculation of mole fraction.

 

Question 17.Calculate the mass of a non-volatile solute (molar mass 40 g mol\(^{-1}\)) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Answer:Let \(W_B\) be the mass of the solute to be determined. Mass of solvent (octane), \(W_A = 114 \text{ g}\). Molar mass of solute, \(M_B = 40 \text{ g mol}^{-1}\). Molar mass of octane, \(M_A = 114 \text{ g mol}^{-1}\). Let the initial vapor pressure of pure octane be \(P^0_A\). The final vapor pressure of the solution, \(P_A\), is 80% of \(P^0_A\), so \(P_A = 0.8 P^0_A\). The relative lowering of vapor pressure is given by: \( \frac{P^0_A - P_A}{P^0_A} = \frac{W_B \times M_A}{M_B \times W_A} \) Substitute the known values: \( \frac{P^0_A - 0.8 P^0_A}{P^0_A} = \frac{W_B \times 114}{40 \times 114} \) \( \frac{0.2 P^0_A}{P^0_A} = \frac{W_B}{40} \) \( 0.2 = \frac{W_B}{40} \)
\( W_B = 0.2 \times 40 \)
\( W_B = 8 \text{ g} \) Thus, 8 g of the non-volatile solute must be dissolved.
In simple words: To reduce the vapor pressure of octane to 80% of its original value by adding a solute, we use the relative lowering of vapor pressure formula. This calculation shows that 8 grams of the solute are needed for the given amount of octane.

🎯 Exam Tip: For problems involving relative lowering of vapor pressure, ensure you correctly set up the ratio \((P^0 - P) / P^0\) and substitute the molar masses and masses of solute and solvent accurately. Molar mass of the solvent needs to be identified from the solvent name (octane).

 

Question 18.A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: 1. Molar mass of the solute 2. Vapour pressure of water at 298 K.

Answer:Let \(W_B\) be the mass of solute, \(M_B\) be the molar mass of solute. Let \(W_A\) be the mass of solvent (water), \(M_A\) be the molar mass of water (\(18 \text{ g mol}^{-1}\)). Let \(P^0_A\) be the vapor pressure of pure water. **Initial situation:** \(W_B = 30 \text{ g}\) \(W_A = 90 \text{ g}\) Vapor pressure of solution, \(P = 2.8 \text{ kPa}\) Using Raoult's Law for relative lowering of vapor pressure: \( \frac{P^0_A - P}{P^0_A} = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A} \) (for dilute solutions, or more accurately, using masses) \( \frac{P^0_A - P}{P^0_A} = \frac{W_B/M_B}{W_A/M_A} \) \( \frac{P^0_A - 2.8}{P^0_A} = \frac{30/M_B}{90/18} = \frac{30/M_B}{5} = \frac{6}{M_B} \)
\( 1 - \frac{2.8}{P^0_A} = \frac{6}{M_B} \) --- (Equation 1) **After adding 18 g of water:** New \(W_A = 90 \text{ g} + 18 \text{ g} = 108 \text{ g}\) New vapor pressure of solution, \(P' = 2.9 \text{ kPa}\) \( \frac{P^0_A - P'}{P^0_A} = \frac{W_B/M_B}{W_A'/M_A} \) \( \frac{P^0_A - 2.9}{P^0_A} = \frac{30/M_B}{108/18} = \frac{30/M_B}{6} = \frac{5}{M_B} \)
\( 1 - \frac{2.9}{P^0_A} = \frac{5}{M_B} \) --- (Equation 2) From Equation 1: \( \frac{2.8}{P^0_A} = 1 - \frac{6}{M_B} \) From Equation 2: \( \frac{2.9}{P^0_A} = 1 - \frac{5}{M_B} \) Divide Equation 1 by Equation 2: \( \frac{2.8/P^0_A}{2.9/P^0_A} = \frac{1 - 6/M_B}{1 - 5/M_B} \) \( \frac{2.8}{2.9} = \frac{(M_B - 6)/M_B}{(M_B - 5)/M_B} \) \( \frac{2.8}{2.9} = \frac{M_B - 6}{M_B - 5} \) \( 2.8(M_B - 5) = 2.9(M_B - 6) \) \( 2.8 M_B - 14 = 2.9 M_B - 17.4 \) \( 17.4 - 14 = 2.9 M_B - 2.8 M_B \)
\( 3.4 = 0.1 M_B \)
\( M_B = 34 \text{ g mol}^{-1} \) Now substitute \(M_B = 34\) into Equation 1 to find \(P^0_A\): \( 1 - \frac{2.8}{P^0_A} = \frac{6}{34} \) \( 1 - \frac{2.8}{P^0_A} = 0.1765 \) (approximately) \( \frac{2.8}{P^0_A} = 1 - 0.1765 = 0.8235 \)
\( P^0_A = \frac{2.8}{0.8235} \approx 3.4 \text{ kPa} \) Therefore: 1. Molar mass of the solute = \( 34 \text{ g mol}^{-1} \) 2. Vapor pressure of water at 298 K = \( 3.4 \text{ kPa} \)
In simple words: We have two scenarios of a solute dissolved in water, each with a different amount of water and a different vapor pressure. By applying Raoult's law to both scenarios and solving the resulting simultaneous equations, we can determine both the molar mass of the solute and the vapor pressure of pure water.

🎯 Exam Tip: This type of problem requires setting up two separate equations based on Raoult's law for relative lowering of vapor pressure and solving them simultaneously. Be careful with algebraic manipulation and ensure all quantities are correctly assigned.

 

Question 19.A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Answer:Given: Freezing point of pure water, \(T^0_f = 273.15 \text{ K}\) **For 5% cane sugar solution:** Freezing point of cane sugar solution, \(T_f = 271 \text{ K}\) Depression in freezing point, \( \Delta T_f = T^0_f - T_f = 273.15 - 271 = 2.15 \text{ K}\) Molar mass of cane sugar (\(C_{12}H_{22}O_{11}\)), \(M_B = 342 \text{ g mol}^{-1}\) 5% solution by mass means: Mass of solute (\(W_B\)) = 5 g Mass of solution = 100 g Mass of solvent (water, \(W_A\)) = \(100 - 5 = 95 \text{ g}\) Using the formula for depression in freezing point: \( \Delta T_f = K_f \times m \) where \(m\) is molality \( = \frac{W_B \times 1000}{M_B \times W_A} \) So, \( \Delta T_f = K_f \times \frac{W_B \times 1000}{M_B \times W_A} \) We can find \(K_f\) for water: \( K_f = \frac{\Delta T_f \times M_B \times W_A}{W_B \times 1000} \) \( K_f = \frac{2.15 \times 342 \times 95}{5 \times 1000} \) \( K_f = \frac{69802.5}{5000} = 13.96 \text{ K kg mol}^{-1} \) **For 5% glucose solution:** Molar mass of glucose (\(C_6H_{12}O_6\)), \(M_B = 180 \text{ g mol}^{-1}\) 5% solution by mass means: Mass of solute (\(W_B\)) = 5 g Mass of solvent (water, \(W_A\)) = \(100 - 5 = 95 \text{ g}\) Now calculate \(\Delta T_f\) for glucose solution using the determined \(K_f\): \( \Delta T_f = K_f \times \frac{W_B \times 1000}{M_B \times W_A} \) \( \Delta T_f = 13.96 \times \frac{5 \times 1000}{180 \times 95} \) \( \Delta T_f = 13.96 \times \frac{5000}{17100} \) \( \Delta T_f = 13.96 \times 0.2924 \) (approximately) \( \Delta T_f = 4.08 \text{ K} \) (approximately) Freezing point of glucose solution, \(T_f' = T^0_f - \Delta T_f \) \( T_f' = 273.15 - 4.08 \) \( T_f' = 269.07 \text{ K} \) The freezing point of the 5% glucose solution is 269.07 K.
In simple words: First, we use the known freezing point depression of a sugar solution to calculate the cryoscopic constant (K_f) of water. Then, using this calculated K_f and the same mass percentage for a glucose solution, we can determine its freezing point depression and thus its final freezing point.

🎯 Exam Tip: This problem involves two parts: first calculating the \(K_f\) using the given sugar solution data, and then using this \(K_f\) to find the freezing point of the glucose solution. Make sure to accurately calculate molality for both solutions and understand that \(K_f\) is a solvent-specific constant.

 

Question 20.At 300 K. 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Answer:Given: Temperature \(T = 300 \text{ K}\). Universal gas constant \(R\) value is needed. Assuming a suitable R value for pressure in bar and volume in Litres, \(R = 0.083 \text{ L bar K}^{-1} \text{ mol}^{-1}\). **Scenario 1:** Mass of glucose (\(W\)) = 36 g Volume of solution (\(V\)) = 1 L Osmotic pressure (\(\pi_1\)) = 4.98 bar Molar mass of glucose (\(M\)) = \(180 \text{ g mol}^{-1}\) Concentration (\(C_1\)) = Moles/Volume = \( \frac{W/M}{V} = \frac{36/180}{1} = \frac{0.2}{1} = 0.2 \text{ mol L}^{-1} \) Using Van't Hoff equation for osmotic pressure: \( \pi = CRT \) \( \pi_1 = C_1 RT \) \( 4.98 = 0.2 \times R \times 300 \) From this, we can verify R or proceed, but the question implies a change in osmotic pressure and asks for the new concentration. **Scenario 2:** New osmotic pressure (\(\pi_2\)) = 1.52 bar New concentration (\(C_2\)) = ? Since the temperature (\(T\)) and gas constant (\(R\)) are constant, we can use the ratio: \( \frac{\pi_1}{\pi_2} = \frac{C_1 RT}{C_2 RT} = \frac{C_1}{C_2} \) \( \frac{4.98}{1.52} = \frac{0.2}{C_2} \) \( C_2 = \frac{0.2 \times 1.52}{4.98} \) \( C_2 = \frac{0.304}{4.98} \) \( C_2 \approx 0.061 \text{ mol L}^{-1} \) The concentration of the solution for an osmotic pressure of 1.52 bars is approximately 0.061 mol L\(^{-1}\).
In simple words: Given the osmotic pressure and concentration of a glucose solution, we can find the relationship between osmotic pressure and concentration. Since temperature and the gas constant are fixed, a change in osmotic pressure directly implies a proportional change in concentration, allowing us to calculate the new concentration.

🎯 Exam Tip: When temperature and R are constant, osmotic pressure is directly proportional to concentration (\(\pi \propto C\)). This allows for a simple ratio-based calculation \(\frac{\pi_1}{C_1} = \frac{\pi_2}{C_2}\). Make sure to correctly identify and use the initial concentration.

 

Question 21.Suggest the most important type of intermolecular attractive interaction in the following pairs. 1. n-hexane and n-octane 2. I and CCl\(_{4}\) 3. NaClO\(_{4}\) and water 4. methanol and acetone 5. acetonitrile (CH\(_{3}\)CN) and acetone (C\(_{3}\)H\(_{6}\)O).

Answer:Let's analyze the intermolecular forces for each pair: 1. **n-hexane and n-octane:** Both are non-polar hydrocarbons. The primary intermolecular forces are London dispersion forces (also known as van der Waals forces). 2. **I and CCl\(_{4}\):** Iodine (\(I_2\)) is a non-polar molecule, and carbon tetrachloride (\(CCl_4\)) is also a non-polar molecule due to its symmetrical tetrahedral structure. Therefore, the main attractive forces are London dispersion forces. 3. **NaClO\(_{4}\) and water:** Sodium perchlorate (\(NaClO_4\)) is an ionic compound, dissociating into \(Na^+\) and \(ClO_4^-\) ions. Water (\(H_2O\)) is a polar molecule. The interaction between ions and polar water molecules is known as ion-dipole interaction. 4. **methanol and acetone:** Methanol (\(CH_3OH\)) is a polar molecule capable of hydrogen bonding. Acetone (\(CH_3COCH_3\)) is a polar molecule but cannot form hydrogen bonds with itself (it has no H directly bonded to O, N, or F). However, methanol can form hydrogen bonds with acetone's oxygen, and both are polar, so dipole-dipole interactions are also significant. The strongest interaction would be hydrogen bonding (between methanol's H and acetone's O) and dipole-dipole. 5. **acetonitrile (CH\(_{3}\)CN) and acetone (C\(_{3}\)H\(_{6}\)O):** Acetonitrile (\(CH_3CN\)) has a significant dipole moment due to the C≡N bond. Acetone (\(CH_3COCH_3\)) also has a significant dipole moment due to the C=O bond. Both are polar molecules. The primary intermolecular attractive interaction between them is dipole-dipole interaction.
In simple words: The type of attractive force depends on the polarity of the molecules. Non-polar substances interact via London dispersion forces, ionic compounds with polar solvents form ion-dipole interactions, and polar molecules mostly engage in dipole-dipole interactions, with hydrogen bonding being a special strong type of dipole-dipole interaction when applicable.

🎯 Exam Tip: To identify intermolecular forces, first determine if molecules are ionic, polar, or non-polar. Look for H-bonding (H bonded to F, O, or N). Then consider dipole-dipole (for polar molecules) and finally London dispersion forces (present in all molecules, dominant in non-polar ones).

 

Question 22.Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH\(_{3}\)OH, CH\(_{3}\)CN.

Answer:n-octane (\(C_8H_{18}\)) is a non-polar solvent. According to the "like dissolves like" principle, non-polar solvents dissolve non-polar solutes best. Let's analyze each solute: - **KCl:** Potassium chloride is an ionic compound. Ionic compounds are highly polar and interact strongly via ion-ion forces. They have very poor solubility in non-polar solvents like n-octane. - **CH\(_{3}\)OH (Methanol):** Methanol is a polar molecule and can form hydrogen bonds. Its strong polarity and H-bonding capacity make it poorly soluble in non-polar n-octane. - **CH\(_{3}\)CN (Acetonitrile):** Acetonitrile is a polar molecule but less polar than methanol (it does not have H-bonding capacity with itself to the extent of methanol). It has a dipole moment. Its solubility in n-octane will be higher than methanol but lower than non-polar substances. - **Cyclohexane (\(C_6H_{12}\)):** Cyclohexane is a non-polar hydrocarbon, similar in nature to n-octane. Due to similar intermolecular forces (London dispersion forces), it is highly miscible with n-octane. Therefore, the order of increasing solubility in n-octane is: **KCl < CH\(_{3}\)OH < CH\(_{3}\)CN < Cyclohexane** **Explanation:** 1. **KCl:** Being an ionic compound, it interacts strongly through ion-ion forces. n-octane is non-polar, so the energy required to break KCl's ionic lattice and form new ion-n-octane interactions is very high, leading to negligible solubility. 2. **CH\(_{3}\)OH (Methanol):** This is a highly polar compound capable of strong hydrogen bonding. These strong interactions are difficult to overcome by the weak London dispersion forces of n-octane, resulting in low solubility. 3. **CH\(_{3}\)CN (Acetonitrile):** Acetonitrile is a polar molecule with dipole-dipole interactions. While n-octane is non-polar, the dipole-dipole forces in acetonitrile are weaker than the hydrogen bonds in methanol, allowing for some degree of mixing with n-octane, but still limited due to the polarity mismatch. 4. **Cyclohexane:** It is a non-polar hydrocarbon. Since n-octane is also a non-polar hydrocarbon, both substances primarily exhibit London dispersion forces. These similar intermolecular forces allow them to mix freely, making cyclohexane highly soluble (miscible) in n-octane.
In simple words: Substances that are chemically similar (like non-polar n-octane and non-polar cyclohexane) dissolve well in each other because their intermolecular forces are compatible. Highly polar or ionic substances, like KCl and methanol, do not dissolve well in non-polar solvents because their strong internal forces cannot be overcome by the weak forces of the non-polar solvent.

🎯 Exam Tip: The principle "like dissolves like" is key here. Assess the polarity and type of intermolecular forces for both the solvent (n-octane, non-polar) and each solute. Ionic > Hydrogen bonding > Dipole-dipole > London dispersion in terms of strength and impact on solubility in non-polar solvents.

 

Question 23.Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? 1. Phenol 2. Toluene 3. Formic acid 4. Ethylene glycol 5. Chloroform 6. Pentanol

Answer:Water is a highly polar solvent capable of strong hydrogen bonding. Let's categorize the solubility of each compound in water: 1. **Phenol (\(C_6H_5OH\)):** Highly soluble in water. Phenol is a polar molecule and can form strong hydrogen bonds with water molecules through its -OH group. The benzene ring is non-polar, but the -OH group's strong polarity and H-bonding capability dominate the solubility for this size. 2. **Toluene (\(C_6H_5CH_3\)):** Insoluble in water. Toluene is predominantly a non-polar hydrocarbon. It lacks polar groups capable of forming strong interactions (like hydrogen bonds) with water molecules. 3. **Formic acid (\(HCOOH\)):** Highly soluble in water. Formic acid is a polar molecule with both a carbonyl group (\(C=O\)) and a hydroxyl group (\(-OH\)), allowing it to form extensive hydrogen bonds with water molecules. 4. **Ethylene glycol (\(HOCH_2CH_2OH\)):** Highly soluble in water. Ethylene glycol has two -OH groups, making it highly polar and capable of forming multiple strong hydrogen bonds with water molecules. 5. **Chloroform (\(CHCl_3\)):** Highly soluble in water. Chloroform is a polar molecule. While it doesn't have an -OH group for direct hydrogen bonding as a donor, its hydrogen atom is slightly acidic and can participate in hydrogen bonding as a donor with the oxygen of water, and its chlorine atoms can facilitate dipole-dipole interactions. More significantly, it forms strong hydrogen bonds where chloroform acts as a proton donor to water's oxygen. 6. **Pentanol (\(CH_3CH_2CH_2CH_2CH_2OH\)):** Partially soluble in water. Pentanol has a polar -OH group that can form hydrogen bonds with water. However, it also has a long non-polar hydrocarbon chain (\(C_5H_{11}\)). As the non-polar part increases in size, its hydrophobic nature starts to dominate, reducing overall solubility in water, making it only partially soluble. **Summary:** - **Insoluble:** Toluene - **Partially Soluble:** Pentanol - **Highly Soluble:** Phenol, Formic acid, Ethylene glycol, Chloroform
In simple words: Water dissolves substances best if they are polar or can form hydrogen bonds with it. Substances with long non-polar chains or entirely non-polar molecules will be insoluble or only partially soluble because their structures don't interact effectively with water.

🎯 Exam Tip: When determining solubility in water, assess the molecule's polarity and its ability to form hydrogen bonds. Small molecules with -OH, -NH, -COOH groups are typically highly soluble. As the non-polar hydrocarbon part of a molecule increases, solubility in water generally decreases, leading to partial or insolubility.

 

Question 24.If the density of some lake water is 1.25g mL\(^{-1}\) and contains 92 g of Na\(^+\) ions per kg of water, calculate the molality of Na ions in the lake?

Answer:Molality is defined as the number of moles of solute per kilogram of solvent. Given: Mass of Na\(^+\) ions = 92 g (solute) Mass of water = 1 kg = 1000 g (solvent) First, calculate the number of moles of Na\(^+\) ions. Molar mass of Na (\(Na^+\) ion) = \(23 \text{ g mol}^{-1}\) Number of moles of Na\(^+\) ions = \( \frac{\text{Mass of Na}^+}{\text{Molar mass of Na}^+} = \frac{92 \text{ g}}{23 \text{ g mol}^{-1}} = 4 \text{ moles} \) Now, calculate the molality: Molality (\(m\)) = \( \frac{\text{Number of moles of solute}}{\text{Mass of solvent (in kg)}} \) Molality = \( \frac{4 \text{ moles}}{1 \text{ kg}} = 4 \text{ m} \) The density of the lake water (1.25 g mL\(^{-1}\)) is additional information not required for calculating molality, as molality is based on the mass of the solvent.
In simple words: Molality is a measure of concentration based on the amount of solute per kilogram of solvent. Given 92 grams of sodium ions in 1 kilogram of water, we convert the grams of sodium into moles and then divide by the kilograms of water to find the molality.

🎯 Exam Tip: Understand the definitions of concentration terms like molality and molarity. Molality (moles of solute per kg of solvent) does not require solution density or volume, only masses. Molarity (moles of solute per L of solution) would require density if masses were given, but not for molality.

 

Question 25.Calculate the mass percentage of aspirin (C\(_{9}\)H\(_{8}\)O\(_{4}\)) in acetonitrile (CH\(_{3}\)CN) when 6.5 g of C\(_{9}\)H\(_{8}\)O\(_{4}\) is dissolved in 450 g of CH\(_{3}\)CN.

Answer:Mass percentage is calculated as the mass of the component divided by the total mass of the solution, multiplied by 100. Given: Mass of aspirin (solute) = 6.5 g Mass of acetonitrile (solvent) = 450 g Total mass of the solution = Mass of solute + Mass of solvent Total mass of the solution = \(6.5 \text{ g} + 450 \text{ g} = 456.5 \text{ g}\) Mass percentage of aspirin = \( \frac{\text{Mass of aspirin}}{\text{Total mass of solution}} \times 100 \) Mass percentage of aspirin = \( \frac{6.5 \text{ g}}{456.5 \text{ g}} \times 100 \) Mass percentage of aspirin = \( 0.0142387 \times 100 \) Mass percentage of aspirin \( \approx 1.424 \% \) The mass percentage of aspirin in the solution is approximately 1.424%.
In simple words: To find the mass percentage, we take the mass of the aspirin, divide it by the total mass of the solution (aspirin plus acetonitrile), and then multiply by 100.

🎯 Exam Tip: Always remember that mass percentage is (mass of component / total mass of solution) * 100. Ensure you add the masses of both solute and solvent to get the total mass of the solution before calculating.

 

Question 26.Nalorphene (C\(_{19}\)H\(_{21}\)NO\(_{3}\)), similar to morphine, is used to combat withdrawal system in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10\(^{-3}\) m aqueous solution required for the above dose?

Answer:Given: Dose of nalorphene required = 1.5 mg Molar mass of nalorphene (\(M_B\)) = 311 g mol\(^{-1}\) Molality of the aqueous solution (\(m\)) = 1.5 × 10\(^{-3}\) m Molality is defined as moles of solute per kg of solvent. A 1.5 × 10\(^{-3}\) m aqueous solution means that: Number of moles of nalorphene = 1.5 × 10\(^{-3}\) moles Mass of solvent (water) = 1 kg = 1000 g Now, calculate the mass of nalorphene corresponding to 1.5 × 10\(^{-3}\) moles: Mass of nalorphene (\(W_B\)) = Moles × Molar mass Mass of nalorphene = \(1.5 \times 10^{-3} \text{ mol} \times 311 \text{ g mol}^{-1} \) Mass of nalorphene = \(0.4665 \text{ g} = 466.5 \text{ mg}\) So, 466.5 mg of nalorphene is present in 1 kg (1000 g) of water. The total mass of solution containing 466.5 mg of nalorphene = Mass of solute + Mass of solvent Total mass of solution = \(0.4665 \text{ g} + 1000 \text{ g} = 1000.4665 \text{ g}\) We need to find the mass of this solution that contains the required dose of 1.5 mg of nalorphene. If 466.5 mg of nalorphene is in 1000.4665 g of solution, then 1.5 mg of nalorphene will be in: Mass of solution required = \( \frac{\text{Mass of solution (for 466.5 mg solute)}}{\text{Mass of solute (466.5 mg)}} \times \text{Required dose of solute} \) Mass of solution required = \( \frac{1000.4665 \text{ g}}{466.5 \text{ mg}} \times 1.5 \text{ mg} \) Mass of solution required = \( \frac{1000.4665}{466.5} \times 1.5 \) Mass of solution required \( \approx 2.1446 \times 1.5 \) Mass of solution required \( \approx 3.217 \text{ g} \) The mass of the 1.5 × 10\(^{-3}\) m aqueous solution required is approximately 3.22 g.
In simple words: First, we determine how much nalorphene is present in 1 kilogram of water for a 1.5 x 10^-3 molal solution. This allows us to find the total mass of solution containing that amount of nalorphene. Then, we use a proportion to calculate what mass of this solution contains the specific dose of 1.5 mg of nalorphene.

🎯 Exam Tip: This is a stoichiometric problem involving molality. Convert the given molality into mass of solute and solvent. Then use a ratio or proportion to find the mass of solution corresponding to the specific dose of the solute. Be careful with unit conversions (g to mg).

 

Question 27.Calculate the amount of benzoic acid (C\(_{6}\)H\(_{5}\)COOH) required for preparing 250 ml of 0.15 M solution in methanol?

Answer:Given: Volume of solution (\(V\)) = 250 ml = 0.250 L Molarity (\(M\)) = 0.15 M Solute is benzoic acid (\(C_6H_5COOH\)). First, calculate the molar mass of benzoic acid (\(C_6H_5COOH\)): C: \(7 \times 12.01 = 84.07\) H: \(6 \times 1.008 = 6.048\) O: \(2 \times 16.00 = 32.00\) Molar mass (\(M_B\)) = \(84.07 + 6.048 + 32.00 = 122.118 \text{ g mol}^{-1}\) (Using approximate integer atomic masses: C=12, H=1, O=16: \(7 \times 12 + 6 \times 1 + 2 \times 16 = 84 + 6 + 32 = 122 \text{ g mol}^{-1}\)) Molarity is defined as moles of solute per liter of solution. \( M = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} \) Moles of solute = Molarity × Volume of solution (L) Moles of benzoic acid = \(0.15 \text{ mol L}^{-1} \times 0.250 \text{ L} \) Moles of benzoic acid = \(0.0375 \text{ mol}\) Now, calculate the mass of benzoic acid required: Mass of solute (\(W_B\)) = Moles of solute × Molar mass of solute Mass of benzoic acid = \(0.0375 \text{ mol} \times 122 \text{ g mol}^{-1} \) Mass of benzoic acid = \(4.575 \text{ g}\) The amount of benzoic acid required is 4.575 g.
In simple words: To prepare a solution of a specific molarity and volume, we first calculate the total moles of solute needed. This is done by multiplying the molarity by the volume in liters. Then, we convert these moles into grams using the solute's molar mass.

🎯 Exam Tip: Molarity calculations require the volume of the *solution* in liters and the *molar mass* of the solute. Always convert volume from milliliters to liters before calculation. Accurately calculating the molar mass of the compound is crucial.

 

Question 28.The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly?

Answer:The depression in freezing point (\(\Delta T_f\)) is a colligative property, which depends on the number of solute particles in the solution, not on their identity. Stronger electrolytes dissociate more extensively, producing more ions and thus a greater number of solute particles. The acidic strength of these carboxylic acids follows the order: Acetic acid (CH\(_{3}\)COOH) < Trichloroacetic acid (CCl\(_{3}\)COOH) < Trifluoroacetic acid (CF\(_{3}\)COOH) **Explanation:** - **Trifluoroacetic acid (CF\(_{3}\)COOH)** is the strongest acid among the three because the highly electronegative fluorine atoms (F) exert a strong electron-withdrawing inductive effect (\(-I\) effect). This effect stabilizes the conjugate base (\(CF_3COO^-\)) by dispersing the negative charge, making the acid release its proton (\(H^+\)) more readily. Consequently, trifluoroacetic acid ionizes to the largest extent in water. - **Trichloroacetic acid (CCl\(_{3}\)COOH)** is also a strong acid due to the electron-withdrawing effect of the three chlorine atoms, but chlorine is less electronegative than fluorine, so its inductive effect is weaker than that of fluorine. Thus, it ionizes less than trifluoroacetic acid but more than acetic acid. - **Acetic acid (CH\(_{3}\)COOH)** is a relatively weak acid. The methyl group (\(CH_3\)) has a slight electron-donating inductive effect (\(+I\) effect), which destabilizes its conjugate base (\(CH_3COO^-\)) by intensifying the negative charge. This makes acetic acid ionize to the least extent among the three. Since trifluoroacetic acid ionizes most, it produces the largest number of ions (solute particles) per given amount of substance. Trichloroacetic acid produces an intermediate number, and acetic acid produces the fewest. A greater number of solute particles leads to a greater colligative effect, hence a greater depression in freezing point. Therefore, the depression in freezing point increases in the order: **Acetic acid < Trichloroacetic acid < Trifluoroacetic acid**.
In simple words: Freezing point depression is greater when there are more solute particles. Stronger acids ionize more in solution, creating more ions. Trifluoroacetic acid is the strongest, releasing the most ions, followed by trichloroacetic acid, and then acetic acid, which is the weakest. This explains why trifluoroacetic acid causes the largest freezing point depression.

🎯 Exam Tip: Colligative properties depend on the *number* of particles, not their nature. For acids, stronger acids dissociate more, producing more ions. Understand inductive effects (\(+I\) and \(-I\)) and their impact on acid strength to predict the extent of ionization and, consequently, the colligative property change.

 

Question 29.Henry's law constant for the molality of methane in benzene at 298 K is 4.27 × 10\(^{5}\) mm Hg. Calculate the solubility of methane in benzene at 298 K. under 760 mm Hg.

Answer:Henry's Law states that the partial pressure of the gas in the vapor phase (\(p\)) is proportional to the mole fraction of the gas (\(X\)) in the solution. \( p = K_H X \) Where \(K_H\) is Henry's Law constant. Given: Henry's Law constant (\(K_H\)) = \(4.27 \times 10^5 \text{ mm Hg}\) Partial pressure of methane (\(p\)) = 760 mm Hg (standard atmospheric pressure) We need to calculate the solubility, which is represented by the mole fraction (\(X\)) of methane in benzene. Rearranging Henry's Law equation: \( X = \frac{p}{K_H} \) \( X = \frac{760 \text{ mm Hg}}{4.27 \times 10^5 \text{ mm Hg}} \) \( X = 1.7798 \times 10^{-3} \) \( X \approx 1.78 \times 10^{-3} \) The solubility (mole fraction) of methane in benzene at 298 K under 760 mm Hg is approximately \(1.78 \times 10^{-3}\).
In simple words: Henry's Law tells us that the amount of gas dissolved in a liquid is directly proportional to the gas's pressure above the liquid. By dividing the given pressure of methane by Henry's law constant, we can find its solubility as a mole fraction in benzene.

🎯 Exam Tip: Ensure that the units of pressure and Henry's law constant are consistent (e.g., both in mm Hg or both in bar/atm). The solubility is typically expressed as mole fraction when using Henry's Law in this form.

 

Question 30.100 g of liquid A (molar mass 140 g mol\(^{-1}\)) was dissolved in 1000 g of liquid B (molar mass 180 g mol\(^{-1}\)). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr?

Answer:Given: Mass of liquid A (\(W_A\)) = 100 g Molar mass of liquid A (\(M_A\)) = 140 g mol\(^{-1}\) Mass of liquid B (\(W_B\)) = 1000 g Molar mass of liquid B (\(M_B\)) = 180 g mol\(^{-1}\) Vapor pressure of pure liquid B (\(P^0_B\)) = 500 torr Total vapor pressure of the solution (\(P_{total}\)) = 475 torr First, calculate the number of moles for each liquid: Moles of A (\(n_A\)) = \( \frac{W_A}{M_A} = \frac{100 \text{ g}}{140 \text{ g mol}^{-1}} \approx 0.7143 \text{ mol} \) Moles of B (\(n_B\)) = \( \frac{W_B}{M_B} = \frac{1000 \text{ g}}{180 \text{ g mol}^{-1}} \approx 5.5556 \text{ mol} \) Now, calculate the mole fraction for each liquid in the solution: Total moles = \(n_A + n_B = 0.7143 + 5.5556 = 6.2699 \text{ mol}\) Mole fraction of A (\(X_A\)) = \( \frac{n_A}{\text{Total moles}} = \frac{0.7143}{6.2699} \approx 0.1139 \) Mole fraction of B (\(X_B\)) = \( \frac{n_B}{\text{Total moles}} = \frac{5.5556}{6.2699} \approx 0.8861 \) (Alternatively, \(X_B = 1 - X_A = 1 - 0.1139 = 0.8861\)) Using Raoult's Law, the partial vapor pressure of B in the solution (\(P_B\)): \( P_B = X_B P^0_B = 0.8861 \times 500 \text{ torr} = 443.05 \text{ torr} \) The total vapor pressure of the solution is the sum of partial pressures of A and B: \( P_{total} = P_A + P_B \) We know \(P_{total}\) and \(P_B\), so we can find \(P_A\): \( P_A = P_{total} - P_B = 475 \text{ torr} - 443.05 \text{ torr} = 31.95 \text{ torr} \) Now, calculate the vapor pressure of pure liquid A (\(P^0_A\)) using Raoult's Law for A: \( P_A = X_A P^0_A \) \( 31.95 \text{ torr} = 0.1139 \times P^0_A \)
\( P^0_A = \frac{31.95}{0.1139} \approx 280.51 \text{ torr} \) Summary of results: - Vapor pressure of pure liquid A (\(P^0_A\)) = \(280.51 \text{ torr}\) - Vapor pressure of liquid A in the solution (\(P_A\)) = \(31.95 \text{ torr}\) - Vapor pressure of liquid B in the solution (\(P_B\)) = \(443.05 \text{ torr}\)
In simple words: We first determine the moles and mole fractions of each liquid. Using Raoult's Law and the pure vapor pressure of liquid B, we calculate its partial pressure in the solution. By subtracting this from the total given vapor pressure of the solution, we find the partial pressure of liquid A. Finally, we use liquid A's partial pressure and mole fraction to deduce the vapor pressure of pure liquid A.

🎯 Exam Tip: This problem integrates mole fraction calculations with Raoult's law and Dalton's law of partial pressures. Be methodical in calculating moles, mole fractions, partial pressures, and then finally the pure vapor pressure of the unknown component. Precision in intermediate calculations of mole fractions can improve final accuracy.

 

Question 31.The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are 3.30 × 10\(^{7}\) mm and 6.51 × 10\(^{7}\) mm respectively. Calculate the composition of these gases in water.

Answer:Given: Total pressure of air (\(P_{total}\)) = 10 atm Proportion of oxygen by volume = 20% Proportion of nitrogen by volume = 79% Henry's Law constant for O\(_{2}\) (\(K_H(O_2)\)) = \(3.30 \times 10^7 \text{ mm Hg}\) Henry's Law constant for N\(_{2}\) (\(K_H(N_2)\)) = \(6.51 \times 10^7 \text{ mm Hg}\) Temperature = 298 K First, convert total pressure to mm Hg: \(1 \text{ atm} = 760 \text{ mm Hg}\) \(P_{total} = 10 \text{ atm} \times 760 \text{ mm Hg/atm} = 7600 \text{ mm Hg}\) Now, calculate the partial pressure of oxygen (\(P_{O_2}\)) and nitrogen (\(P_{N_2}\)): Partial pressure is proportional to the volume fraction (for ideal gases). \(P_{O_2} = \text{Volume fraction of O}_2 \times P_{total} = \frac{20}{100} \times 10 \text{ atm} = 2 \text{ atm} \) \(P_{O_2} = 2 \text{ atm} \times 760 \text{ mm Hg/atm} = 1520 \text{ mm Hg}\) \(P_{N_2} = \text{Volume fraction of N}_2 \times P_{total} = \frac{79}{100} \times 10 \text{ atm} = 7.9 \text{ atm} \) \(P_{N_2} = 7.9 \text{ atm} \times 760 \text{ mm Hg/atm} = 6004 \text{ mm Hg}\) Using Henry's Law (\(X = \frac{p}{K_H}\)) to find mole fractions in water: Mole fraction of O\(_{2}\) in water (\(X_{O_2}\)): \( X_{O_2} = \frac{P_{O_2}}{K_H(O_2)} = \frac{1520 \text{ mm Hg}}{3.30 \times 10^7 \text{ mm Hg}} \) \( X_{O_2} \approx 4.606 \times 10^{-5} \) Mole fraction of N\(_{2}\) in water (\(X_{N_2}\)): \( X_{N_2} = \frac{P_{N_2}}{K_H(N_2)} = \frac{6004 \text{ mm Hg}}{6.51 \times 10^7 \text{ mm Hg}} \) \( X_{N_2} \approx 9.223 \times 10^{-5} \) The composition of these gases in water (expressed as mole fractions) is: - Mole fraction of oxygen (\(X_{O_2}\)) = \(4.61 \times 10^{-5}\) - Mole fraction of nitrogen (\(X_{N_2}\)) = \(9.22 \times 10^{-5}\)
In simple words: To find how much oxygen and nitrogen dissolve in water, we first calculate their individual pressures in the air. Then, using Henry's Law, which relates gas pressure to its solubility, we divide each gas's partial pressure by its respective Henry's constant to get its mole fraction in the water.

🎯 Exam Tip: Pay close attention to unit consistency (atm vs. mm Hg). Ensure partial pressures are calculated correctly from total pressure and volume fractions. Henry's law applies to each gas independently, relating its partial pressure to its mole fraction in the liquid phase.

 

Question 32.Determine the amount of CaCl\(_{2}\) (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm is 27°C.

Answer:Given: Van't Hoff factor (\(i\)) = 2.47 Volume of solution (\(V\)) = 2.5 L (assuming "2.5 litre of water" means the final solution volume is 2.5 L, which is common in osmotic pressure problems, though technically water is solvent) Osmotic pressure (\(\pi\)) = 0.75 atm Temperature (\(T\)) = 27°C = \(27 + 273.15 = 300.15 \text{ K}\) (or 300 K if approximate) Universal gas constant (\(R\)) = \(0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}\) (suitable for atm and L) The Van't Hoff equation for osmotic pressure is: \( \pi = i C R T \) Where \(C\) is the molar concentration (moles/volume). So, \( \pi = i \frac{n}{V} R T \) Rearranging to find the number of moles (\(n\)): \( n = \frac{\pi V}{i R T} \) \( n = \frac{0.75 \text{ atm} \times 2.5 \text{ L}}{2.47 \times 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \times 300 \text{ K}} \) \( n = \frac{1.875}{2.47 \times 24.63} \) \( n = \frac{1.875}{60.8361} \) \( n \approx 0.0308 \text{ mol} \) Now, calculate the molar mass of CaCl\(_{2}\): Ca: 40.08 g mol\(^{-1}\) Cl: \(2 \times 35.45 = 70.90 \text{ g mol}^{-1}\) Molar mass of CaCl\(_{2}\) = \(40.08 + 70.90 = 110.98 \text{ g mol}^{-1}\) (approximately 111 g mol\(^{-1}\)) Calculate the amount (mass) of CaCl\(_{2}\) dissolved: Mass = Moles × Molar mass Mass = \(0.0308 \text{ mol} \times 111 \text{ g mol}^{-1} \) Mass \( \approx 3.4188 \text{ g} \) The amount of CaCl\(_{2}\) dissolved is approximately 3.42 g.
In simple words: Osmotic pressure depends on the concentration of solute particles, factoring in the Van't Hoff factor for dissociation. By plugging the given osmotic pressure, volume, temperature, and Van't Hoff factor into the Van't Hoff equation, we calculate the total moles of CaCl₂. Then, we convert these moles into grams using its molar mass to find the required amount.

🎯 Exam Tip: Remember to include the Van't Hoff factor (\(i\)) for electrolyte solutions in colligative property calculations. Always convert temperature to Kelvin and choose the appropriate gas constant \(R\) based on the units of pressure (atm, bar, kPa).

 

Question 33.Determine the osmotic pressure of a solution prepared by dissolving 25 mgof K\(_{2}\)SO\(_{4}\) in 2 litre of water at 25°C, assuming that it is completely dissociated.

Answer:Given: Mass of K\(_{2}\)SO\(_{4}\) (\(W\)) = 25 mg = \(25 \times 10^{-3} \text{ g}\) = 0.025 g Volume of water (solution volume, \(V\)) = 2 L Temperature (\(T\)) = 25°C = \(25 + 273.15 = 298.15 \text{ K}\) (or 298 K) Assuming complete dissociation for K\(_{2}\)SO\(_{4}\): \(K_2SO_4 \implies 2K^+ + SO_4^{2-}\) This produces 3 ions for every molecule of K\(_{2}\)SO\(_{4}\). So, the Van't Hoff factor (\(i\)) = 3. First, calculate the molar mass of K\(_{2}\)SO\(_{4}\): K: \(2 \times 39.10 = 78.20\) S: \(1 \times 32.07 = 32.07\) O: \(4 \times 16.00 = 64.00\) Molar mass (\(M\)) = \(78.20 + 32.07 + 64.00 = 174.27 \text{ g mol}^{-1}\) (approximately 174 g mol\(^{-1}\)) Now, calculate the number of moles of K\(_{2}\)SO\(_{4}\): Moles (\(n\)) = \( \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.025 \text{ g}}{174 \text{ g mol}^{-1}} \approx 1.4368 \times 10^{-4} \text{ mol} \) Use the Van't Hoff equation for osmotic pressure: \( \pi = i \frac{n}{V} R T \) Using \(R = 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}\): \( \pi = 3 \times \frac{1.4368 \times 10^{-4} \text{ mol}}{2 \text{ L}} \times 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \times 298 \text{ K} \) \( \pi = 3 \times 7.184 \times 10^{-5} \text{ mol L}^{-1} \times 0.0821 \times 298 \text{ atm} \) \( \pi = 3 \times 0.00176 \) \( \pi \approx 0.00528 \text{ atm} \) Using \(R = 0.083 \text{ L bar K}^{-1} \text{ mol}^{-1}\) for pressure in bar: \( \pi = 3 \times \frac{1.4368 \times 10^{-4}}{2} \times 0.083 \times 298 \) \( \pi = 3 \times 7.184 \times 10^{-5} \times 0.083 \times 298 \) \( \pi \approx 0.00533 \text{ bar} \) The osmotic pressure of the solution is approximately 0.0053 atm (or 0.0053 bar). The original calculation in the prompt gives \(5.27 \times 10^{-3} \text{ atm}\), which is very close.
In simple words: To find the osmotic pressure of a salt solution, we first calculate the moles of the salt. Since K₂SO₄ dissociates into three ions, we multiply the moles by the Van't Hoff factor (3). Then, using the Van't Hoff equation, we calculate the osmotic pressure, which depends on the concentration, temperature, and gas constant.

🎯 Exam Tip: For ionic compounds, accurately determine the Van't Hoff factor (\(i\)) based on the number of ions formed upon dissociation. Convert mass to moles and ensure correct units for all constants and variables in the osmotic pressure equation.

GSEB Class 12 Chemistry Solutions Additional Important Questions and Answers

 

Question 1.Match the following

A	B
(i) Ideal solution	(a) \( \Delta H_{mixing} > 0 \)
(ii) Positive deviation	(b) Same osmotic pressure
(iii) Negative deviation	(c) CH\(_{3}\)OH + C\(_{2}\)H\(_{5}\)OH
(iv) Isotonic solution	(d) Benzene in benzoic acid
(v) Association (i<1)	(e) \( P_A < X_A P^0_A \) and \( P_B < X_B P^0_B \)

Answer: (i) c, (ii) a, (iii) e, (iv) b, (v) d
**Matching Explanation:** (i) **Ideal solution** matches **(c) CH\(_{3}\)OH + C\(_{2}\)H\(_{5}\)OH**. Ideal solutions are those where the solute-solvent interactions are similar to solute-solute and solvent-solvent interactions. Mixtures of structurally similar alcohols often behave ideally or nearly ideally. (ii) **Positive deviation** matches **(a) \( \Delta H_{mixing} > 0 \)**. Positive deviation from Raoult's law occurs when the solute-solvent interactions are weaker than the average of solute-solute and solvent-solvent interactions, leading to higher vapor pressure and an endothermic mixing process (\( \Delta H_{mixing} > 0 \)). (iii) **Negative deviation** matches **(e) \( P_A < X_A P^0_A \) and \( P_B < X_B P^0_B \)**. Negative deviation occurs when solute-solvent interactions are stronger than the average, leading to lower vapor pressure than predicted by Raoult's law (\(P_A < X_A P^0_A\) and \(P_B < X_B P^0_B\)). (iv) **Isotonic solution** matches **(b) Same osmotic pressure**. Isotonic solutions are solutions that have the same osmotic pressure across a semipermeable membrane. (v) **Association (i<1)** matches **(d) Benzene in benzoic acid**. When solute molecules associate (e.g., dimerize) in a solvent, the number of effective particles decreases, leading to a Van't Hoff factor (\(i\)) less than 1. Benzoic acid forms dimers in non-polar solvents like benzene through hydrogen bonding, resulting in association.
In simple words: This question tests understanding of different solution behaviors and properties. Ideal solutions follow Raoult's law perfectly. Positive deviations mean weaker intermolecular forces and heat absorption, while negative deviations mean stronger forces and heat release. Isotonic solutions have equal osmotic pressure, and association means solute particles combine, reducing their effective number.

🎯 Exam Tip: Thoroughly understand the characteristics of ideal solutions, positive deviations (weaker A-B interactions, endothermic mixing, V > 0), negative deviations (stronger A-B interactions, exothermic mixing, V < 0), and how colligative properties are affected by association or dissociation (Van't Hoff factor).

 

Question 2.Ethanol-water mixture obtained by the fermentation of sugars are rich in water. Can we get absolute alcohol from this mixture by fractional distillation? If not, explain why?

Answer:No, absolute alcohol cannot be obtained from an ethanol-water mixture by simple fractional distillation. **Explanation:** An ethanol-water mixture forms an **azeotrope**. An azeotrope is a mixture of two or more liquids that boils at a constant temperature and has a vapor phase composition identical to its liquid phase composition. This means that during distillation, the vapor produced has the same ratio of components as the liquid, preventing further separation by simple distillation. The ethanol-water mixture forms a **minimum boiling azeotrope** at approximately 95.6% ethanol and 4.4% water by mass, boiling at about 351.15 K (78.15 °C). This boiling point is lower than the boiling points of pure ethanol (78.3 °C) and pure water (100 °C). When an ethanol-water mixture is subjected to fractional distillation, the composition of the vapor phase will become increasingly enriched in ethanol until it reaches the azeotropic composition. Once the azeotropic composition is reached, the mixture will distill over at a constant temperature without any change in its composition. Therefore, it is impossible to obtain 100% pure (absolute) ethanol by fractional distillation of an aqueous ethanol solution. To obtain absolute alcohol, other methods like azeotropic distillation (using a third component like benzene or cyclohexane to form a new azeotrope with water) or molecular sieves are employed.
In simple words: No, you cannot get pure alcohol from an ethanol-water mix using normal distillation. This is because they form a special mixture called an azeotrope, which boils at a constant temperature and distills over with the same composition as the liquid. Once this point is reached, no further separation is possible by distillation.

🎯 Exam Tip: Understand the concept of azeotropes, especially minimum boiling azeotropes. Recognize that they limit separation by simple distillation. Remember the common example of the ethanol-water azeotrope.

A non-ideal solution is one which does not obey Raoult's law. For ideal solutions, (I) \(P_A \ne X_A P^0_A\) and \(P_B \ne X_B P^0_B\) (II) \( \Delta_{mixing} V \ne 0 \) (III) \( \Delta_{mixing} H \ne 0 \) Based on the nature of deviation from Raoult's law, non-ideal solutions are divided into two types. 1. non-ideal solutions with positive deviation and 2. non-ideal solutions with negative deviation.

1. Non-ideal solutions which show positive deviation:

In this case the partial vapour pressure of any component of the solution and the total vapour pressure of solution are greater than that expected from Raoult's law. i.e., \(P_A > X_A P^0_A\), \(P_B > X_B P^0_B\) and \(P_S > X_A P^0_A + X_B P^0_B\). In these solutions, \( \Delta_{mixing} H \) and \( \Delta_{mixing} V \) are positive.

Cause of positive deviation: Positive deviation is shown by liquid pairs for which A-B force of attraction is weaker than A-A and B-B forces of attractions. As a result, the escaping tendency of the molecules will be greater than that in the pure solutions. Thus, the partial pressure of both of them over the solution are greater than that predicted by Raoult's law and hence the solution exhibits a large vapour pressure than that expected.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक आदर्श विलयन के लिए वाष्प दाब-मोल अंश संबंध को दर्शाता है। इसमें P_A और P_B दो घटकों के आंशिक वाष्प दाब हैं जो घटक A और B के मोल अंश के साथ रैखिक रूप से बदलते हैं। कुल वाष्प दाब (P_S) P_A और P_B का योग है, जो एक सीधी रेखा बनाता है। यह राऊल्ट के नियम का पालन करने वाले आदर्श विलयन को दर्शाता है।

Due to the decrease in attractive forces, energy is increased. i.e., the process is endothermic. Thus \( \Delta_{mixing} H > 0 \) (+ve). Because of the decrease in attraction, the particles are loosely held and hence there is an increase in volume. Thus \( \Delta_{mixing} V > 0 \) (+ve). e.g. A solution of ethanol and cyclohexane. In pure ethanol molecules, there are intermolecular hydrogen bonds.

\( \dots H - O \dots H - O \dots \)
\( \quad \quad \quad C_2H_5 \quad \quad C_2H_5 \quad \quad C_2H_5 \)

When cyclohexane is added to ethanol, the cyclohexane molecules get placed in between the ethanol molecules, thereby breaking the hydrogen bonds. This reduces the intermolecular attraction. As a result, \( \Delta_{mixing} V \) and \( \Delta_{mixing} H \) are positive. Solutions such as - acetone and CS\(_{2}\) - acetone and alcohol - acetone and benzene - water and alcohol - ether and acetone etc. show positive deviation

2. Non-ideal solutions show negative deviations:

In this case the partial vapour pressure of any component of the solution and the total vapour pressure of the solution are less than that expected from Raoult's law. i.e; \(P_A < X_A P^0_A\), \(P_B < X_B P^0_B\) and \(P_S < X_A P^0_A + X_B P^0_B\). In these solutions \( \Delta_{mixing} H \) and \( \Delta_{mixing} V \) are negative.

Cause of negative Deviation: Negative deviation is shown by liquid pairs of which A-B force of attraction is stronger than A-A and B-B forces of attractions. Thus, in the solution the solute and solvent particles are held more tightly than those in the pure components. Due to this, the escaping tendency of the molecules of both the components will be less in the solution than in the pure components. Consequently, the partial pressures of both the components are less than that expected from Raoult's law and hence the solution exhibits a less vapour pressure than that expected.

In simple words: Non-ideal solutions deviate from Raoult's law. Positive deviation occurs when components interact weakly, leading to higher vapor pressure, increased volume, and energy absorption. Negative deviation happens when components interact strongly, causing lower vapor pressure, decreased volume, and energy release.

🎯 Exam Tip: For non-ideal solutions, distinguish between positive and negative deviations based on the strength of intermolecular forces (A-B vs. A-A and B-B), changes in enthalpy and volume of mixing, and their effect on vapor pressure relative to Raoult's law predictions. Cite common examples for each type of deviation.

 

Question 3.Vapour pressure – composition graph for a mixture of two substances at a particular temperature is given below.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक आदर्श विलयन के लिए वाष्प दाब-मोल अंश संबंध को दर्शाता है। इसमें दो वक्र (डॉट्स द्वारा) P_A और P_B दो घटकों के आंशिक वाष्प दाब को दिखाते हैं, जो उनके मोल अंश के साथ रैखिक रूप से बदलते हैं। ऊपर की ठोस रेखा कुल वाष्प दाब (P_S) को दर्शाती है, जो P_A और P_B का योग है और यह भी एक सीधी रेखा है। यह आरेख स्पष्ट रूप से दर्शाता है कि कैसे आदर्श विलयन राऊल्ट के नियम का पालन करते हैं।

(a) Sketch a version of the graph with boiling point as the Y-axis instead of vapour pressure showing the total pressure. (b) For one of the dotted lines \(P_A \propto X_A\) (\(P_A\) & \(X_A\) are vapour pressure and mole fraction of component A). To turn a proportionality into equality, \(P_A = kX_A\) where k is a constant. Deduce the value of k, when \(X_A = 1\)

Answer:(a) **Sketch of Boiling Point vs. Composition Graph:** For an ideal solution, the boiling point-composition graph would show a smooth curve, usually with boiling points varying between the boiling points of the pure components. The total pressure curve (as shown in the given diagram) goes from \(P^0_B\) to \(P^0_A\). When converting to a boiling point diagram, the boiling point generally follows an inverse trend to vapor pressure. If vapor pressure increases, boiling point decreases. Therefore, the boiling point curve would typically be concave downwards or upwards, depending on the relative boiling points of A and B, and would not necessarily be linear. For an ideal solution, the boiling point of the mixture will lie between the boiling points of the pure components.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक आदर्श विलयन के लिए क्वथनांक-मोल अंश संबंध को दर्शाता है। Y-अक्ष पर क्वथनांक (Boiling Point) और X-अक्ष पर मोल अंश (Mole Fraction) हैं। शुद्ध घटक A का क्वथनांक T_A है और शुद्ध घटक B का क्वथनांक T_B है। जैसे-जैसे घटक A का मोल अंश बढ़ता है, मिश्रण का क्वथनांक T_B से T_A की ओर एक चिकनी वक्र के रूप में परिवर्तित होता है, जो आदर्श विलयन में क्वथनांक के व्यवहार को दर्शाता है। (b) **Deducing the value of k:** Given the proportionality: \(P_A \propto X_A\) To turn this into an equality, we introduce a constant \(k\): \(P_A = k X_A\) When \(X_A = 1\), it means the solution consists solely of pure component A. In this case, the partial vapor pressure of A (\(P_A\)) becomes equal to the vapor pressure of pure liquid A (\(P^0_A\)). So, substituting \(X_A = 1\) and \(P_A = P^0_A\) into the equation: \(P^0_A = k \times 1\)
\(k = P^0_A\) Therefore, the constant \(k\) is equal to the vapor pressure of pure liquid A (\(P^0_A\)). This is Raoult's Law.
In simple words: For an ideal solution, the vapor pressure of a component is directly proportional to its mole fraction. When the mole fraction of component A is 1 (meaning it's pure A), its partial vapor pressure becomes its pure vapor pressure. Thus, the proportionality constant (k) is equal to the pure vapor pressure of component A.

🎯 Exam Tip: Understand that for ideal solutions, \(P_A = X_A P^0_A\), which is Raoult's Law. The constant of proportionality in this relationship is the vapor pressure of the pure component. Be able to sketch the boiling point-composition diagram by inverting the vapor pressure-composition relationship (lower vapor pressure means higher boiling point).

According to Raoult's law. it is obvious that the graph of the partial pressure of each component against its mole fraction should be a straight line. Thus a plot of \(P_A\) against \(X_A\) is a straight line passing through \(P^0_A\) When \(X_A = 1\) and a plot of \(P_B\) against \(X_B\) is also a straight line passing through \(P^0_B\) When \(X_B = 1\). These are represented by dotted lines in the figure. The total vapour pressure of the solution at any composition is equal to the sum of the partial pressures of the two components as shown by the solid hues in the figure.

The essential characteristics of an ideal solution are, 1. Obeys Raoult's law at all concentrations and temperatures. i.e., \(P_A = X_A P^0_A\) and \(P_B = X_B P^0_B\). 2. The total volume of an ideal solution is equal to the sum of the volumes of the individual components mixed. \( \Delta_{mixing} V = 0 \). 3. No. heat is liberated or absorbed in the formation of an ideal solution. i.e., \( \Delta_{mixing} H = 0 \).

The molecules of the ideal solution will exert same force on each other irrespective of their nature. Thus if two liquids A and B from an ideal solution, then the intermolecular forces between A-B, A-A and B-B will be equal. Some examples of ideal solutions are: - benzene and toluene - n-hexane and n-heptane - ethyl bromide and ethyl iodide etc.

 

Question 4.The teacher asked Ramu to prepare six solutions from the following compounds 1. C\(_{6}\)H\(_{6}\) 2. CCl\(_{4}\) 3. (CH\(_{3}\))\(_{2}\)CO 4. n-hexane 5. C\(_{2}\)H\(_{5}\)Br 6. C\(_{2}\)H\(_{5}\)-I 7. C\(_{6}\)H\(_{5}\)OH 8. HNO\(_{3}\)

Answer:The teacher asked Ramu to categorize these solutions into three types: ideal solutions, solutions showing positive deviation from Raoult's law, and solutions showing negative deviation from Raoult's law. Here are the six solutions formed from the given compounds and their categorization: 1. **C\(_{6}\)H\(_{6}\) (Benzene) + CCl\(_{4}\) (Carbon Tetrachloride):** * Both are non-polar molecules. The intermolecular forces (London dispersion forces) are very similar between benzene-benzene, CCl\(_{4}\)-CCl\(_{4}\), and benzene-CCl\(_{4}\). * **Category:** Ideal solution. 2. **(CH\(_{3}\))\(_{2}\)CO (Acetone) + CCl\(_{4}\) (Carbon Tetrachloride):** * Acetone is polar, CCl\(_{4}\) is non-polar. The A-B interactions (acetone-CCl\(_{4}\)) are weaker than the A-A (acetone-acetone dipole-dipole) and B-B (CCl\(_{4}\)-CCl\(_{4}\) dispersion) interactions. * **Category:** Solutions showing positive deviation from Raoult's law. 3. **C\(_{2}\)H\(_{5}\)Br (Ethyl Bromide) + C\(_{2}\)H\(_{5}\)-I (Ethyl Iodide):** * Both are alkyl halides, similar in structure and polarity. The intermolecular forces are mainly dipole-dipole and London dispersion forces, which are very similar between the components. * **Category:** Ideal solution. 4. **n-hexane + CH\(_{3}\)OH (Methanol):** * n-hexane is non-polar. Methanol is highly polar and forms strong hydrogen bonds. The A-B interactions (n-hexane-methanol) are significantly weaker than the A-A (methanol-methanol hydrogen bonds) interactions. * **Category:** Solutions showing positive deviation from Raoult's law. 5. **C\(_{6}\)H\(_{6}\) (Benzene) + CH\(_{3}\)OH (Methanol):** * Benzene is non-polar. Methanol is polar and forms strong hydrogen bonds. Similar to n-hexane + methanol, the intermolecular forces between benzene and methanol are weaker than those within pure methanol. * **Category:** Solutions showing positive deviation from Raoult's law. 6. **HNO\(_{3}\) (Nitric Acid) + H\(_{2}\)O (Water):** * Both nitric acid and water are highly polar and form hydrogen bonds. When mixed, strong new hydrogen bonds form between \(HNO_3\) and \(H_2O\). These A-B interactions are stronger than the average A-A and B-B interactions. * **Category:** Solutions showing negative deviation from Raoult's law.

Real solutions showing positive deviation from Raoult's law	Real solutions showing negative deviation from Raoult's law	Ideal solutions
(ii) Acetone + CCl\(_{4}\)	(vi) HNO\(_{3}\) + H\(_{2}\)O	(i) Benzene + CCl\(_{4}\)
(iv) n-hexane + Methanol		(iii) Ethyl Bromide + Ethyl Iodide
(v) Benzene + Methanol

In simple words: Ramu's task is to classify solutions based on how their vapor pressure deviates from ideal behavior. Ideal solutions have similar intermolecular forces. Positive deviation occurs when new interactions are weaker than the original ones. Negative deviation happens when new interactions are stronger than the original ones.

🎯 Exam Tip: To classify solutions, analyze the intermolecular forces present in the pure components and in the mixture. Similar forces typically lead to ideal solutions. Weaker A-B interactions compared to A-A and B-B cause positive deviation, while stronger A-B interactions cause negative deviation. Look for hydrogen bonding disruption (positive deviation) or formation (negative deviation).

 

Question 5.The graphical representation of vapour pressure Vs temperature for a liquid and a solution made by dissolving a non-volatile solute in it are given below.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ वाष्प दाब और तापमान के बीच संबंध को दर्शाता है। इसमें दो वक्र हैं: एक विलायक (सॉल्वेंट) के लिए और दूसरा उसमें अवाष्पशील विलेय घोलकर बने विलयन के लिए। सॉल्वेंट वक्र विलयन वक्र से ऊपर है, जो दर्शाता है कि किसी भी दिए गए तापमान पर सॉल्वेंट का वाष्प दाब विलयन से अधिक होता है। ये वक्र विभिन्न तापमानों (T₀, T_b, T_b') पर वाष्प दाब में परिवर्तन को दर्शाते हैं। क्षैतिज रेखा वायुमंडलीय दाब को दर्शाती है, और वक्रों के साथ इसका प्रतिच्छेदन क्वथनांक को इंगित करता है।

Analyse the graph and answer the following questions. 1. Which curve is for solvent? 2. Which curve represents solution? 3. What is meant by \(\Delta T_b\)? 4. Derive a formula for determining the molecular mass of solute? 5. Write the unit of \(K_b\)?

Answer:Based on the provided graph and principles of colligative properties: 1. **Which curve is for solvent?** Curve AB is for the solvent. At any given temperature, the vapor pressure of the pure solvent is higher than that of the solution containing a non-volatile solute. 2. **Which curve represents solution?** Curve CD represents the solution. The presence of a non-volatile solute lowers the vapor pressure of the solvent. 3. **What is meant by \(\Delta T_b\)?** \(\Delta T_b\) represents the **elevation in boiling point**. It is the difference between the boiling point of the solution (\(T_b\)) and the boiling point of the pure solvent (\(T^0_b\)). Mathematically, \( \Delta T_b = T_b - T^0_b \). The boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure. When a non-volatile solute is added, the vapor pressure is lowered, requiring a higher temperature to reach atmospheric pressure, thus elevating the boiling point. 4. **Derive a formula for determining the molecular mass of solute?** For dilute solutions, the elevation of boiling point (\(\Delta T_b\)) is directly proportional to the molal concentration (\(m\)) of the solute: \( \Delta T_b \propto m \) \( \Delta T_b = K_b \times m \) --- (Equation 1) Where \(K_b\) is the molal elevation constant (ebullioscopic constant) for the solvent. Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent: \( m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \) Let \(W_B\) be the mass of solute, \(M_B\) be the molar mass of solute, and \(W_A\) be the mass of solvent in grams. \( \text{moles of solute} = \frac{W_B}{M_B} \) \( \text{mass of solvent (in kg)} = \frac{W_A}{1000} \) So, \( m = \frac{W_B/M_B}{W_A/1000} = \frac{W_B \times 1000}{M_B \times W_A} \) --- (Equation 2) Substitute (Equation 2) into (Equation 1): \( \Delta T_b = K_b \times \frac{W_B \times 1000}{M_B \times W_A} \) To find the molecular mass of the solute (\(M_B\)), rearrange the formula: \( M_B = \frac{K_b \times W_B \times 1000}{\Delta T_b \times W_A} \) This formula allows us to calculate the molar mass of the solute from experimental measurements of boiling point elevation. 5. **Write the unit of \(K_b\)?** From \( \Delta T_b = K_b \times m \), we can write \( K_b = \frac{\Delta T_b}{m} \). The unit of \(\Delta T_b\) is Kelvin (K) or degree Celsius (°C). The unit of molality (\(m\)) is mol kg\(^{-1}\) (or 'm'). Therefore, the unit of \(K_b\) is **K kg mol\(^{-1}\)** or **°C kg mol\(^{-1}\)** or **K/m** or **°C/m**.
In simple words: The graph shows that adding a non-volatile substance lowers the vapor pressure of a liquid, making it boil at a higher temperature. This temperature increase is called the elevation in boiling point (\(\Delta T_b\)), which is directly related to the solute's concentration. We can use this relationship to find the solute's molecular mass.

🎯 Exam Tip: For colligative properties graphs, remember that the solvent always has a higher vapor pressure and a lower boiling point (and freezing point) compared to the solution. Be ready to derive the molecular mass formula from the colligative property definition and its relation to molality.

 

Question 5. The following sub-questions relate to the vapour pressure-temperature graph for a liquid and a solution:
(i) Which curve represents the solvent?
(ii) Which curve represents the solution?
(iii) What is the meaning of ΔT_b?
(iv) Derive a formula for calculating the molecular mass of the solute.
(v) State the unit of K_b?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक विलायक (सॉल्वेंट) और एक विलयन (सॉल्यूशन) के लिए वाष्प दाब और तापमान के बीच संबंध को दर्शाता है। इसमें दो वक्र दिखाए गए हैं जहाँ ऊपर वाला वक्र विलायक का प्रतिनिधित्व करता है और नीचे वाला वक्र विलयन का प्रतिनिधित्व करता है, जो दर्शाता है कि विलयन का वाष्प दाब विलायक से कम है।
Answer:
(i) The curve representing the solvent is AB.
(ii) The curve representing the solution is CD.
(iii) ΔT_b denotes the elevation in boiling point, which is the difference between the boiling point of the solution (T_b) and the boiling point of the pure solvent (T°_b), i.e., ΔT_b = T_b - T°_b.
(iv) **Derivation of Molecular Mass of Solute:**
The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the atmospheric pressure. When a non-volatile solute is introduced into a solvent, the vapor pressure decreases. Consequently, the temperature required for the solution's vapor pressure to match the external atmospheric pressure becomes higher than that of the pure solvent. This difference is known as the elevation in boiling point.
For dilute solutions, the elevation in boiling point (ΔT_b) is directly proportional to the molal concentration (m) of the solute:
\[ \Delta T_b \propto m \]
\[ \Delta T_b = K_b \cdot m \] Where K_b is the molal elevation constant (ebullioscopic constant).
Molality (m) is defined as:
\[ m = \frac{W_B \times 1000}{M_B \times W_A \text{ in gram}} \] Substituting this into the elevation in boiling point equation:
\[ \Delta T_b = K_b \times \frac{W_B \times 1000}{M_B \times W_A} \] Rearranging to solve for the molecular mass of the solute (M_B):
\[ M_B = \frac{K_b \times W_B \times 1000}{\Delta T_b \times W_A} \] This formula is highly useful for calculating the molecular mass of a solute when K_b, W_B, ΔT_b, and W_A are known. The elevation of boiling point is considered a colligative property, as it depends solely on the number of solute moles, not their chemical nature.
(v) The unit of K_b is K kg mol^-1 (or degrees/molality, K/m, or °C/m).
In simple words: The elevation of boiling point is directly proportional to the molality of the solution. This property, depending on the amount of solute, allows us to calculate the solute's molecular weight using a specific formula involving the boiling point constant and masses.

🎯 Exam Tip: Remember to clearly define all terms and provide the complete derivation for molecular mass from elevation in boiling point. Accuracy in units and proper substitution in formulas are crucial for full marks.

 

Question 6. The vapour pressure-temperature graph for a liquid and a solution is provided below. Analyse the graph and answer the following questions:
(i) Which curve represents the solvent?
(ii) Which curve represents a solution?
(iii) How much depression in the freezing point is observed?
(iv) Derive an expression for calculating the molecular mass of the solute.
(v) What is the value of K_f for water?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक विलायक (सॉल्वेंट) और एक विलयन (सॉल्यूशन) के लिए वाष्प दाब और तापमान के बीच संबंध को दर्शाता है। वक्र AB विलायक के वाष्प दाब को और वक्र CD विलयन के वाष्प दाब को दर्शाता है, जिसमें एक ठोस चरण का भी उल्लेख है। यह आरेख हिमांक अवनमन (depresion in freezing point) की अवधारणा को समझाता है, जहाँ एक विलयन का हिमांक शुद्ध विलायक से कम होता है।
Answer:
(i) Curve AB represents the solvent.
(ii) Curve CD represents the solution.
(iii) The depression in freezing point observed is ΔT_f = T°_f - T_f.
(iv) **Derivation of Molecular Mass of Solute:**
The freezing point of a liquid is defined as the temperature at which the liquid and solid phases of the substance possess the same vapor pressure. When a non-volatile solute is dissolved, the vapor pressure of the solution is lower than that of the pure solvent. Consequently, the solution's vapor pressure equals that of its pure solid only at a lower temperature.
This leads to the freezing point of the solution being lower than that of the pure solvent. The difference between the freezing points of the solvent (T°_f) and the solution (T_f) is termed the depression in freezing point (ΔT_f). Thus, ΔT_f = T°_f - T_f.
For dilute solutions, the depression in freezing point (ΔT_f) is directly proportional to the molal concentration (m) of the solute:
\[ \Delta T_f \propto m \]
\[ \Delta T_f = K_f \cdot m \] Here, K_f is the molal depression constant (cryoscopic constant) for the solvent. When m = 1 molal, ΔT_f = K_f.
Molality (m) is calculated as:
\[ m = \frac{W_B \times 1000}{M_B \times W_A} \] Substituting this into the freezing point depression equation:
\[ \Delta T_f = K_f \times \frac{W_B \times 1000}{M_B \times W_A} \] Rearranging the formula to determine the molecular mass of the solute (M_B):
\[ M_B = \frac{K_f \times W_B \times 1000}{\Delta T_f \times W_A} \] This formula is critical for calculating the molecular mass of a solute, provided K_f, W_B, ΔT_f, and W_A are known. Freezing point depression is a colligative property, meaning it depends on the number of solute moles rather than their intrinsic nature.
(v) The unit of K_f is K kg mol^-1 (or degrees/molality, K/m, or °C/m).
In simple words: Adding a solute lowers a liquid's freezing point, and this depression is proportional to the solute's concentration. This property helps determine the solute's molecular mass using a simple formula with the freezing point constant and component masses.

🎯 Exam Tip: Be precise with the distinction between boiling point elevation (ΔT_b, K_b) and freezing point depression (ΔT_f, K_f) in derivations. Highlight that both are colligative properties and the correct formula application is key.

 

Question 7. Does the following picture show the clearing of snow on roads? (a) Name a chemical useful for removing snow from the road. (b) Also explain the principle behind it.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक विलायक (सॉल्वेंट) और एक विलयन (सॉल्यूशन) के लिए वाष्प दाब और तापमान के बीच संबंध को दर्शाता है। वक्र AB विलायक के वाष्प दाब को और वक्र CD विलयन के वाष्प दाब को दर्शाता है, जिसमें एक ठोस चरण का भी उल्लेख है। यह आरेख हिमांक अवनमन (depresion in freezing point) की अवधारणा को समझाता है, जहाँ एक विलयन का हिमांक शुद्ध विलायक से कम होता है।
Answer:
The provided picture is a Vapour Pressure - Temperature graph, which illustrates freezing point depression, not the direct clearing of snow. However, the principle discussed is relevant to snow clearing.
(a) Chemicals useful for removing snow from roads include: NaCl (sodium chloride) or CaCl₂ (calcium chloride).
(b) The principle behind clearing ice from roads is the depression of the freezing point. When substances like NaCl or CaCl₂ are spread on roadways in winter, they dissolve in the surface water (even a thin film) to form a solution. This solution has a lower freezing point than pure water. For example, NaCl can melt ice down to -21°C, and CaCl₂ is effective down to -55°C. By lowering the freezing point, these chemicals prevent water from freezing at typical cold temperatures, or they cause existing ice to melt, making roads safer.
In simple words: Chemicals like salt are used to melt snow by lowering the freezing point of water. This creates a solution that stays liquid even at temperatures below water's normal freezing point, effectively clearing the roads.

🎯 Exam Tip: When describing the principle, clearly link the use of chemicals to the colligative property of freezing point depression. Mention specific examples like NaCl and CaCl₂ and their effective temperature ranges.

 

Question 8. When dried fruits and vegetables are placed in water, they slowly swell. Explain why? What would be the effect of temperature on this process?
Answer:
Dried fruits and vegetables swell when placed in water due to osmosis. Osmosis is the movement of solvent molecules (water in this case) from a region of higher solvent concentration (pure water) to a region of lower solvent concentration (inside the dried fruits/vegetables, which act as a concentrated solution) across a semi-permeable membrane. The cells of the fruits/vegetables act as semi-permeable membranes, allowing water to enter and cause swelling.
Increasing the temperature generally accelerates the process of osmosis. Higher temperatures increase the kinetic energy of water molecules, leading to faster movement across the membrane, and thus, more rapid swelling of the dried fruits and vegetables.
In simple words: Dried fruits absorb water and swell because of osmosis, where water moves into the more concentrated fruit cells. Higher temperatures speed up this water absorption.

🎯 Exam Tip: Define osmosis accurately and explain the role of the semi-permeable membrane. For temperature effects, mention increased kinetic energy and faster molecular movement.

 

Question 9. It is advised to add ethylene glycol to water in a car radiator while driving on a hill station. Explain?
Answer:
Adding ethylene glycol to water in a car radiator on a hill station is recommended because it forms an antifreeze solution. Ethylene glycol lowers the freezing point of water (freezing point depression) and also raises its boiling point. In cold hill station environments, the lower freezing point prevents the water in the radiator from freezing, which could otherwise damage the engine. Ethylene glycol also acts as a coolant due to its higher boiling point, preventing the engine from overheating, especially during strenuous driving uphill.
In simple words: Ethylene glycol is added to car radiators in cold places to stop the water from freezing and to raise its boiling point, protecting the engine from both freezing and overheating.

🎯 Exam Tip: Emphasize both aspects of antifreeze action-lowering the freezing point and raising the boiling point-and relate them to practical benefits in a car radiator.

 

Question 10. Sea water freezes at a lower temperature than distilled water but boils at a higher temperature than distilled water. Explain?
Answer:
Sea water contains dissolved solutes, primarily salts like NaCl, which are non-volatile. Distilled water, on the other hand, is pure and lacks these dissolved substances. The presence of non-volatile solutes in sea water leads to two colligative properties: freezing point depression and boiling point elevation. Therefore, sea water will freeze at a temperature below 0°C (lower than distilled water) and boil at a temperature above 100°C (higher than distilled water).
In simple words: Sea water freezes lower and boils higher than pure water because the salts dissolved in it change its freezing and boiling points.

🎯 Exam Tip: Clearly state that dissolved solutes are responsible for both freezing point depression and boiling point elevation, linking these colligative properties to the observed phenomena in sea water.

 

Question 11. The shells of two eggs are removed. One of the eggs is placed in pure water and the other is placed in a saturated solution of sodium chloride. What will be observed and explain the reason?
Answer:
When the shell of an egg is removed, the inner membrane acts as a semi-permeable membrane. If one egg is placed in pure water, it will swell. This is because water (higher solvent concentration) will move into the egg's cytoplasm (lower solvent concentration) via osmosis. If the other egg is placed in a saturated solution of sodium chloride, it will shrink. This occurs because the water inside the egg (higher solvent concentration compared to the saturated salt solution) will move out of the egg and into the salt solution through osmosis.
In simple words: An egg in pure water will swell as water moves in, while an egg in salty water will shrink as water moves out, both due to osmosis through its membrane.

🎯 Exam Tip: Focus on explaining the direction of water movement in each scenario based on relative solute concentrations and the definition of osmosis through a semi-permeable membrane.