Class 12 Chemistry Solutions IMPORTANT Questions CBSE #2024, Assignments of Chemistry

Download Class 12 Chemistry Solutions IMPORTANT Questions CBSE #2024 and more Assignments Chemistry in PDF only on Docsity! Chapter 2 – Solutions Page No 37: Question 2.1: Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Answer: Mass percentage of C6H6  Mass percentage of CCl4 Alternatively, Mass percentage of CCl4 = (100 − 15.28)% = 84.72% Question 2.2: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Answer: Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴Mass of carbon tetrachloride = (100 − 30)g = 70 g Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol−1 = 78 g mol−1 ∴Number of moles of  = 0.3846 mol Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 355 = 154 g mol−1 ∴Number of moles of CCl4  = 0.4545 mol Thus, the mole fraction of C6H6 is given as: = 0.458 Question 2.3: Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL. Answer: Molarity is given by: (a) Molar mass of Co (NO3)2.6H2O = 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g mol−1 ∴Moles of Co (NO3)2.6H2O = 0.103 mol Therefore, molarity  = 0.023 M (b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water. Moles of water  = 55.56 mol ∴Mole fraction of H2S, x = 0.0035 At STP, pressure (p) = 0.987 bar According to Henry’s law: p = KHx = 282 bar Question 2.7: Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K. Answer: It is given that: KH = 1.67 × 108 Pa  = 2.5 atm = 2.5 × 1.01325 × 105 Pa = 2.533125 × 105 Pa According to Henry’s law: = 0.00152 We can write,  [Since,  is negligible as compared to ] In 500 mL of soda water, the volume of water = 500 mL [Neglecting the amount of soda present] We can write: 500 mL of water = 500 g of water = 27.78 mol of water Now, Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g = 1.848 g Page No 47: Question 2.8: 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: It is given that: = 450 mm of Hg = 700 mm of Hg ptotal = 600 mm of Hg From Raoult’s law, we have: Therefore, total pressure,  Therefore,  = 1 − 0.4 = 0.6 Now, = 450 × 0.4 = 180 mm of Hg = 700 × 0.6 = 420 mm of Hg Now, in the vapour phase: Mole fraction of liquid A = 0.30 And, mole fraction of liquid B = 1 − 0.30 = 0.70 = 5.08 g (approx) Hence, 5.08 g of ascorbic acid is needed to be dissolved. Note: There is a slight variation in this answer and the one given in the NCERT textbook. Question 2.12: 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: It is given that: Volume of water, V = 450 mL = 0.45 L Temperature, T = (37 + 273)K = 310 K Number of moles of the polymer,  We know that: Osmotic pressure,  = 30.98 Pa = 31 Pa (approximately) Page No 59: Question 2.1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. Answer: Homogeneous mixtures of two or more than two components are known as solutions. There are three types of solutions. (i) Gaseous solution: The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution. (ii) Liquid solution: The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid. For example, a solution of ethanol in water is a liquid solution. (iii) Solid solution: The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution. Question 2.2: Give an example of solid solution in which the solute is a gas. Answer: In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas. Question 2.3: Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage. Answer: (i) Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture. i.e., Mole fraction of a component  Mole fraction is denoted by ‘x’. If in a binary solution, the number of moles of the solute and the solvent are nA and nBrespectively, then the mole fraction of the solute in the solution is given by, Similarly, the mole fraction of the solvent in the solution is given as: (ii) Molality Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as: Molality (m) (iii) Molarity Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of the solution. It is expressed as: Molarity (M) (iv) Mass percentage: The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution. It is expressed as: Mass % of a component  Question 2.4: Concentrated nitric acid used in laboratory work 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: Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution. Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1 Then, number of moles of HNO3  And, number of moles of NaHCO3  = 0.0053 mol HCl reacts with Na2CO3 and NaHCO3 according to the following equation. 1 mol of Na2CO3 reacts with 2 mol of HCl. Therefore, 0.0053 mol of Na2CO3 reacts with 2 × 0.0053 mol = 0.0106 mol. Similarly, 1 mol of NaHCO3 reacts with 1 mol of HCl. Therefore, 0.0053 mol of NaHCO3 reacts with 0.0053 mol of HCl. Total moles of HCl required = (0.0106 + 0.0053) mol = 0.0159 mol In 0.1 M of HCl, 0.1 mol of HCl is preset in 1000 mL of the solution. Therefore, 0.0159 mol of HCl is present in  = 159 mL of the solution Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na2CO3and NaHCO3, containing equimolar amounts of both. Question 2.7: 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: Total amount of solute present in the mixture is given by, = 75 + 160 = 235 g Total amount of solution = 300 + 400 = 700 g Therefore, mass percentage (w/w) of the solute in the resulting solution,  = 33.57% And, mass percentage (w/w) of the solvent in the resulting solution, = (100 − 33.57)% = 66.43% Question 2.8: An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) 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: Molar mass of ethylene glycol = 2 × 12 + 6 × 1 + 2 ×16 = 62 gmol−1 Number of moles of ethylene glycol  = 3.59 mol Therefore, molality of the solution  = 17.95 m Total mass of the solution = (222.6 + 200) g = 422.6 g Given, Density of the solution = 1.072 g mL−1 Volume of the solution  = 394.22 mL = 0.3942 × 10−3 L  Molarity of the solution  = 9.11 M Question 2.9: A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample. Answer: (i) 15 ppm (by mass) means 15 parts per million (106) of the solution. Therefore, percent by mass  = 1.5 × 10−3 % (ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5 = 119.5 g mol−1 Now, according to the question, 15 g of chloroform is present in 106 g of the solution. i.e., 15 g of chloroform is present in (106 − 15) ≈ 106 g of water. Molality of the solution = 1.26 × 10−4 m Question 2.10: What role does the molecular interaction play in a solution of alcohol and water? Answer: In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution. Question 2.11: Why do gases always tend to be less soluble in liquids as the temperature is raised? Answer: over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law. Vapour pressure of a two-component solution showing positive deviation from Raoult’s law Vapour pressure of a two-component solution showing negative deviation from Raoult’s law In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero. ΔsolH = 0 In the case of solutions showing positive deviations, absorption of heat takes place. ∴ΔsolH = Positive In the case of solutions showing negative deviations, evolution of heat takes place. ∴ΔsolH = Negative Question 2.15: 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: Here, Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar Vapour pressure of pure water at normal boiling point  Mass of solute, (w2) = 2 g Mass of solvent (water), (w1) = 98 g Molar mass of solvent (water), (M1) = 18 g mol−1 According to Raoult’s law, = 41.35 g mol−1 Hence, the molar mass of the solute is 41.35 g mol−1. Question 2.16: 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: Vapour pressure of heptane  Vapour pressure of octane  = 46.8 kPa We know that, Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1 = 100 g mol−1 Number of moles of heptane  = 0.26 mol Molar mass of octane (C8H18) = 8 × 12 + 18 × 1 = 114 g mol−1 Number of moles of octane = 0.31 mol Mole fraction of heptane, = 0.456 And, mole fraction of octane, x2 = 1 − 0.456 = 0.544 Now, partial pressure of heptane,  = 0.456 × 105.2 = 47.97 kPa Partial pressure of octane,  = 0.544 × 46.8 = 25.46 kPa Hence, vapour pressure of solution, ptotal = p1 + p2 = 47.97 + 25.46 = 73.43 kPa Question 2.17: The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it. Answer: 1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water). Molar mass of water = 18 g mol−1  Number of moles present in 1000 g of water  = 55.56 mol Therefore, mole fraction of the solute in the solution is . It is given that, Vapour pressure of water,  = 12.3 kPa After the addition of 18 g of water: Again, applying the relation: Dividing equation (i) by (ii), we have: Therefore, the molar mass of the solute is 23 g mol−1. (ii) Putting the value of ‘M’ in equation (i), we have: Hence, the vapour pressure of water at 298 K is 3.53 kPa. Question 2.20: 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: Here, ΔTf = (273.15 − 271) K = 2.15 K Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol−1 5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water. Now, number of moles of cane sugar = 0.0146 mol Therefore, molality of the solution,  = 0.1537 mol kg−1 Applying the relation, ΔTf = Kf × m = 13.99 K kg mol−1 Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol−1 5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.  Number of moles of glucose  = 0.0278 mol Therefore, molality of the solution,  = 0.2926 mol kg−1 Applying the relation, ΔTf = Kf × m = 13.99 K kg mol−1 × 0.2926 mol kg−1 = 4.09 K (approximately) Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K. Question 2.21: Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol−1. Calculate atomic masses of A and B. Answer: We know that, KCl < CH3OH < CH3CN < Cyclohexane Question 2.25: Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol. Answer: (i) Phenol (C6H5OH) has the polar group −OH and non-polar group −C6H5. Thus, phenol is partially soluble in water. (ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water. (iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water. Thus, formic acid is highly soluble in water. (iv) Ethylene glycol   has polar −OH group and can form H−bond. Thus, it is highly soluble in water. (v) Chloroform is insoluble in water. (vi) Pentanol (C5H11OH) has polar −OH group, but it also contains a very bulky non-polar ‐ −C5H11 group. Thus, pentanol is partially soluble in water. Question 2.26: If the density of some lake water is 1.25 g mL−1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake. Answer: Number of moles present in 92 g of Na+ ions = = 4 mol Therefore, molality of Na+ ions in the lake  = 4 m Question 2.27: If the solubility product of CuS is 6 × 10−16, calculate the maximum molarity of CuS in aqueous solution. Answer: Solubility product of CuS, Ksp = 6 × 10−16 Let s be the solubility of CuS in mol L−1. Now,  = s × s = s2 Then, we have, Ksp =  = 2.45 × 10−8 mol L−1 Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10−8 mol L−1. Question 2.28: Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN. Answer: 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN. Then, total mass of the solution = (6.5 + 450) g = 456.5 g Therefore, mass percentage ofC9H8O4  = 1.424% Question 2.29: Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10−3m aqueous solution required for the above dose. Answer: The molar mass of nalorphene   is given as: In 1.5 × 10−3m aqueous solution of nalorphene, 1 kg (1000 g) of water contains 1.5 × 10−3 mol Therefore, total mass of the solution  This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g. Therefore, mass of the solution containing 1.5 mg of nalorphene is: Hence, the mass of aqueous solution required is 3.22 g. Note: There is a slight variation in this answer and the one given in the NCERT textbook. Question 2.30: Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol. Answer: 0.15 M solution of benzoic acid in methanol means, 1000 mL of solution contains 0.15 mol of benzoic acid Therefore, 250 mL of solution contains =   mol of benzoic acid = 0.0375 mol of benzoic acid Molar mass of benzoic acid (C6H5COOH) = 7 × 12 + 6 × 1 + 2 × 16 = 122 g mol−1 Hence, required benzoic acid = 0.0375 mol × 122 g mol−1 = 4.575 g 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid. Answer: It is given that: We know that: Therefore, observed molar mass of  The calculated molar mass of  is: Therefore, van’t Hoff factor,  Let α be the degree of dissociation of  Now, the value of Ka is given as: Taking the volume of the solution as 500 mL, we have the concentration: Therefore, Question 2.34: Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water. Answer: Vapour pressure of water,  = 17.535 mm of Hg Mass of glucose, w2 = 25 g Mass of water, w1 = 450 g We know that, Molar mass of glucose (C6H12O6), M2 = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol−1 Molar mass of water, M1 = 18 g mol−1 Then, number of moles of glucose,  = 0.139 mol And, number of moles of water,  = 25 mol We know that, ⇒ 17.535 − p1 = 0.097 ⇒ p1 = 17.44 mm of Hg Hence, the vapour pressure of water is 17.44 mm of Hg. Question 2.35: Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg. Answer: Here, p = 760 mm Hg kH = 4.27 × 105 mm Hg According to Henry’s law, p = kHx It can be observed from the graph that the plot for the ptotal of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour. Question 2.38: Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. Answer: Molar mass of benzene  Molar mass of toluene  Now, no. of moles present in 80 g of benzene  And, no. of moles present in 100 g of toluene  ∴Mole fraction of benzene, xb  And, mole fraction of toluene,  It is given that vapour pressure of pure benzene,  And, vapour pressure of pure toluene,  Therefore, partial vapour pressure of benzene, And, partial vapour pressure of toluene, Hence, mole fraction of benzene in vapour phase is given by: Question 2.39: 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 Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water. Answer: Percentage of oxygen (O2) in air = 20 % Percentage of nitrogen (N2) in air = 79% Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) mm Hg = 7600 mm Hg Therefore, Partial pressure of oxygen,  = 1520 mm Hg Partial pressure of nitrogen,  = 6004 mmHg Now, according to Henry’s law: p = KH.x For oxygen: For nitrogen: Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10−5and 9.22 × 10−5respectively. Question 2.40: Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C. Answer: We know that, Here, R = 0.0821 L atm K-1mol-1 M = 1 × 40 + 2 × 35.5 = 111g mol-1