NCERT Class 12 Chemistry Solutions Chapter 1 PDF

NCERT Solutions for Class 12 Science Chapter 1- Solutions Intext Questions 1. Calculate the mass percentage of benzene ( C6H6 ) and carbon tetrachloride ( CCl 4 ) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Ans:Mass percentage is the ratio of...

NCERT Solutions for Class 12 Science Chapter 1- Solutions Intext Questions 1. Calculate the mass percentage of benzene ( C6H6 ) and carbon tetrachloride ( CCl 4 ) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Ans:Mass percentage is the ratio of mass of solute divided by the total mass of solution multiplied by 100. masssolute Mass%   100 masssolution Here, Mass percentage of both the compounds are given as; massC6H6 Mass percentage of C6H6 =  100 massC6H6  massCCl4 22 =  100 22  122 = 15.28 % Now, as the solution consists of only two components i.e., C6H6 and CCl4. Thus, Mass percentage of CCl4 = 100 – 15.28 = 84.72%  Mass percentages of C6H6 and CCl4 are 15.28% and 84.72% respectively. 2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Ans:Mole fraction is the ratio of moles of solute to the total moles of solution. n Mole fraction = solute n solution Here, We have given a solution containing 30% by mass of benzene in carbon tetrachloride; which clearly states that 30 g of benzene is present in 100 g of solution. Thus, carbon tetrachloride is 70 g in 100 g of solution. We know that, Molar mass of benzene = 78 g/mol Molar mass of carbon tetrachloride = 154 g/mol Class XII Chemistry www.vedantu.com 1 Thus, 30 Number of moles of benzene =  0.385mol 78 70 Number of moles of carbon tetrachloride =  0.455mol 154 Now, n C6 H 6 Mole fraction of benzene = n C6H6  n CCl4 0.385 = 0.385  0.455 0.385 =  0.458 0.84  Mole fraction of benzene is 0.458 in the solution. 3. Calculate the molarity of each of the following solution: (a) 30 g of Co  NO3  2.6H2O in 4.3 L of solution. Ans:Molarity is the number of moles of solute per liter of solution. n Molarity = solute M Vsolution Here, Molarity of 30 g of Co  NO3 2.6H 2O in 4.3 L of solution is given as; Molar mass of Co  NO3 2.6H 2O = 310.7 g/mol. 30 Number of moles =  0.0966 310.7 0.0966 Thus, molarity =  0.022M 4.3 (b) 30 mL of 0.5 M H2SO4 diluted to 500 ml. Ans: Molarity of 30 mL of 0.5 M H 2SO4 diluted to 500 ml. 1000 ml of 0.5 M H 2SO4 contains 0.5 moles of H 2SO4. Thus, 30 ml of 0.5 M H 2SO4 contains; 0.5 =  30  0.015moles H 2SO4. 1000 Also, volume of solution = 500 ml = 0.5 L. Now, Class XII Chemistry www.vedantu.com 2  0.015  Molarity =  0.03M 0.5 4. Calculate the mass of urea ( NH2CONH 2 ) required in making 2.5 kg of 0.25 molal aqueous solution. Ans: Molality is given as ratio of moles of solute per gm of mass of solvent. n Molality = solute m msolvent Here, 0.25 molal solution of urea states that 0.25 moles of urea is present in 1000 gm of solvent. Molar mass of solute i.e., urea = 60 g/mol Thus, Mass of urea in the solution = 0.25  60  15g Total mass of solution = 1000 gm solvent + 15 gm of solute = 1015 gm = 1.015 Kg This states that, 1.015 Kg solution has 15 gm urea. Thus, 2.5 Kg solution will have; 15  2.5  37g urea. 1.015  Mass of urea required will be 37 g. 5. Calculate (a) Molality Ans: Given that, 20% of aqueous KI solution has 20 gm of KI in 100 gm of solution. Mass of water (solvent) = 100 – 20 = 80 gm. Volume of solution is given as, mass 100 Volume =   83.194ml density 1.202 Molar mass of KI = 166.0028 g/mol Thus, 20 Number of moles of KI =  0.120mol 166.0028 Molar mass of water = 18 g/mol Thus, Class XII Chemistry www.vedantu.com 3  80 Number of moles of water =  4.44mol 18 Molality of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g / ml ; 0.120 Molality =  1.50m 0.08 (b) Molarity and Ans: Molarity of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g / ml. 0.120 Molarity =  1.44M 0.083194 (c) Mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g/ml. Ans: Mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g / ml. 0.120 Mole fraction =  0.0263 0.120  4.44 6. H 2S , a toxic gas with a rotten egg like smell, is used for the qualitative analysis. If the solubility of H 2S in water at STP is 0.195 m, calculate Henry’s law constant. Ans: Molality (solubility) of H2S in water = 0.195 m which states that, 0.195 moles of H2S in 1Kg of water (1000 g). Now, Molar mass of water = 18 g/mol. 1000 Number of moles of water =  55.55mol 18 Mole fraction of H2S ; 0.195 X H2S   0.00349 0.195  55.55 Pressure at STP is given as 0.987 bar. Thus, applying Henry’s law we get; PH2S  K H  X H2S PH2S 0.987 KH    282.80bar X H2S 0.00349  Henry’s constant will be 282.80 bar. 7. Henry’s law constant for CO2 in water is 1.67  108 Pa at 298 K. Calculate Class XII Chemistry www.vedantu.com 4 the quantity of CO2 in 500 ml of soda water when packed under 2.5 atm CO2 pressure at 298 K. Ans: Given that, Henry’s law constant, K H  1.67  108 Pa Pressure, PCO2  2.5atm  2.5  101325  253312.5Pa Now, by Henry’s law; PCO2  K H  X CO2 253312.5 XCO2   1.5168  103 1.67  108 Now, 500 ml soda water is equivalent to 500 ml water which indirectly signifies 500 gm of the same. Thus, We know that, molar mass of water = 18 g/mol 500 Number of moles of water =  27.78mol 18 Number of moles of CO2 is given as, n CO2 X CO2  n CO2  n water  n CO2  X CO2 n CO2  n water  n CO2  42.20  103 mol Mass of CO2 is given as, Molar mass of CO2 = 44 g/mol. Thus, Mass = n CO2  44  1.857gm 8. The vapor 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 vapor pressure is 600 mm Hg. Also find the composition of the vapor phase. Ans: Vapor pressures of the given pure liquids are; PA 0  450mm Hg, PB0  700mm Hg and PTotal  600mm Hg By Raoult’s law; PTotal  PA  PB PTotal  X A PA 0  X BPB0  X A PA 0  1  X A  PB0 PTotal  PB0   PA 0  PB0  X A 600 = 700 + (450-700) XA Class XII Chemistry www.vedantu.com 5  XA  0.4 Thus, XB  1  0.4  0.6 PA  X A PA 0  0.4  450  180mm Hg PB  X BPB0  0.6  700  420mm Hg Thus, 180 Mole fraction of A in vapor phase =  0.3 180  420 Mole fraction of B in vapor phase = 1 – 0.3 = 0.7 9. Vapor pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea ( NH2CONH 2 ) is dissolved in 850 g of water. Calculate the vapor pressure of water for this solution and its relative lowering. Ans: Given that, Vapor pressure of pure water, P0 = 23.8 mm Hg Mass of urea, W2 = 50 g Molar mass of urea, M2 = 60 g/mol Mass of water, W1 = 850 g Molar mass of water, M1 = 18 g/mol By Raoult’s law; P  P0  Ps  X 2  P 0 W2 P 0  Ps n2 M2  X   n1  n 2 W1 2 P0 W  2 M1 M2 X2  0.017 Now, We have given that P0 = 23.8 mm Hg; 23.8  Ps  0.017 23.8 Thus, V. P. of solution, Ps = 23.4 mmHg. 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. Ans: Given that, Boiling point of water = 100  C Boiling point of water at 750 mm Hg = 99.63  C Elevation in boiling point, Tb  100  99.63  0.37 Class XII Chemistry www.vedantu.com 6 Ebullioscopic constant, K b  0.52 Mass of water, W1 = 500 g Molar mass of water, M1 = 18 g/mol Molar mass of sucrose, M2 = 342 g/mol Mass of sucrose, W2  ? The elevation in boiling point is given as; Tb  K b  m n W 1000 Tb  K b  2  K b  2  W1 M 2 W1 T  M 2  W1 W2  b  121.67g W2  1000  121.67 g sucrose to be added in water such that it boils at 100  C. 11. Calculate the mass of ascorbic acid (Vitamin C, C6H8O6 ) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5  C. K f  3.9Kg / mol. Ans: Given that, Lowering in freezing point, Tf = 1.5  C Cryoscopic constant, Kf  3.9Kg / mol Mass of acetic acid, W1 = 75 g Molar mass of acetic acid, M1 = 60 g/mol Molar mass of ascorbic acid, M2 = 176 g/mol Mass of ascorbic acid, W2  ? The lowering in freezing point is given as; Tf  K f  m n W 1000 Tf  K f  2  K f  2  W1 M 2 W1 T  M 2  W1 W2  f  5.077g K f  1000  Mass of ascorbic acid is 5.077 g. 12. Calculate the osmotic pressure in pascals exerted by solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37  C. Ans: Given that, Mass of polymer, W = 1 g Molar mass of polymer, M = 185000 g/mol Volume of water, V = 450 ml = 0.45 L Class XII Chemistry www.vedantu.com 7 Temperature, T = 37  C = 310 K Gas constant, R = 8.314kPa.L.K 1mol1  8.314  103 Pa.L.K 1mol1 Now, using formula for osmotic pressure; n W 1   CRT  RT    RT V M V   30.96Pa  Osmotic pressure is 30.96 Pa. NCERT Exercise 1. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. Ans: A homogeneous mixture of two or more chemically non-reacting substances is called a solution. There are nine types of solution under the main 3 heads i.e., gaseous solution, liquid solution and solid solution. Types of Solution with examples; Gas in gas: Air i.e., mixture of O2 and N 2 , etc. Liquid in gas: Water vapor. Solid in gas: Smoke, Camphor vapors in N 2 gas, etc. Gas in liquid: Aerated water, O2 dissolved in water, etc. Liquid in liquid: Vinegar solution, etc. Solid in liquid: Glucose dissolved in water, saline water, etc. Gas in solid: Solution of hydrogen in platinum, etc. Liquid in solid: Amalgams eg. Mg-Hg. Solid in solid: Ornaments (Cu/Ag with Au). 2. Give an example of a solid solution in which the solute is a gas. Ans: It is said that gas is solute in the solid solution i.e., gas – solid solution. The examples are solid carbon dioxide in fire extinguishers, solution of hydrogen in palladium, dissolved gases in underground minerals and many more. 3. Define the following terms: (i) Mole fraction Ans: Mole fraction is the ratio of moles of solute to the total moles of solution. n Mole fraction = solute n solution (ii) Molality Ans: Molality is given as ratio of moles of solute per gm of mass of solvent. Class XII Chemistry www.vedantu.com 8  n solute Molality = m msolvent It is better way of expressing concentration of the solute because this does not change with the change in temperature as that of molarity (dependent on volume of solution). (iii) Molarity Ans: Molarity is the number of moles of solute per liter of solution. n Molarity = solute M Vsolution (iv) Mass percentage Ans: Mass Percentage: Mass percentage is the ratio of mass of solute divided by the total mass of solution multiplied by 100. masssolute Mass%   100 masssolution 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? Ans: Given that, We have 68% nitric acid by mass in aqueous solution which means 68 g nitric acid is present in 100 g of solution. Molar mass of nitric acid = 63 g/mol 68 Number of moles of nitric acid =  1.079mol 63 Also, Density of the solution is given as 1.504 g/ml. Thus, mass 100 Volume =   66.489ml density 1.504 Now, Molarity is given as; Thus, 1.079  1000 Molarity =  16.22M 66.489 5. A solution of glucose in water is labelled as 10% w/w, that would be the molality and mole fraction of each component in the solution? If the density Class XII Chemistry www.vedantu.com 9  of solution is 1.2 g/ml then what shall be the molarity of the solution? Ans: We have given that, 10% w/w solution of glucose in water i.e., 10 g of glucose in 90 g of water. Now, as we know; Molar mass of glucose = 180 g/mol Molar mass of water = 18 g/mol Thus, 10 Number of moles of glucose in the solution =  0.055mol 180 90 Number of moles of water in the solution =  5mol 18 Taking into consideration the above values; Molality is given as – 0.055  1000 Molality =  0.617m 90 Mole fraction of each component can be given as – 0.055 Mole fraction of glucose =  0.0108 0.055  5 Mole fraction of water = 1 – 0.0108 = 0.9892 Again, We have given, density of solution is 1.2 g/ml. Thus, mass 100 Volume =   83.33ml density 1.2 Thus, Molarity can be given as; 0.055  1000 Molarity =  0.66M 83.33 6. How many ml of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both? Ans: Let us consider that we have x g of Na 2CO3 in 1g mixture. Thus, we will have (1 - x) g of NaHCO3 in the same. We know that, Molar mass of Na 2CO3 = 106 g/mol Molar mass of NaHCO3 = 84 g/mol Thus, x Number of moles of Na 2CO3 in mixture = 106 Class XII Chemistry www.vedantu.com 10 Number of moles of NaHCO3 in mixture = 1  x  84 Now, as we have given that they are equimolar; thus, x  1  x  106 84 84x  106  106x  x  0.557 Hence, 0.557 Number of moles of Na 2CO3 in mixture =  0.00526 106 Number of moles of NaHCO3 in mixture = 1  0.577   0.00503 84 To calculate how many ml of 0.1 M HCl is required to react completely with 1 g mixture of Na 2CO3 and NaHCO3 we need to analyze reactions for both; Na 2CO3  2HCl 2NaCl  H2O  CO2 Here, each mole of Na 2CO3 requires 2 moles of HCl. Thus, 0.00526 moles of Na 2CO3 requires = 0.00526  2  0.01052 moles of HCl. NaHCO3  HCl NaCl  H2O  CO2 Here, each mole of NaHCO3 requires a mole of HCl. Thus, 0.00503 moles of NaHCO3 will require 0.00503 HCl. Total moles of HCl required = 0.01052 + 0.00503 = 0.01555 moles Now, 1.1 moles of 0.1 M HCl are present in 1000 ml of solution. Thus, 0.01555  1000 0.01555 moles will be present in =  155.5ml of solution. 0.1  Volume required to react completely with the mixture will be 155.5 ml. 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. Ans: Mass percentage is defined as the ratio of mass of solute by the mass of solution multiplied by 100. masssolute Mass percentage =  100 masssolution Here, We have given a mixture of 300 g of 25% solution and 400 g of 40% solution by mass. Thus, 25  300 300 g of 25% solution will contain =  75 g of solute 100 Class XII Chemistry www.vedantu.com 11  40  400 400 g of 40% solution will contain =  160 g of solute 100 Now, the resulting solution will have contents as; Total mass of solute = 75 + 160 = 235 g Total mass of solution = 300 + 400 = 700 g Thus, Mass percentage is given as, 235 Mass percentage of solute in the solution =  100  33.57 % 700 Similarly, Mass percentage of water in the solution = 100 – 33.57 = 66.42% 8. An antifreeze solution is prepared from 222.6 g of ethylene glycol ( C2H6O2 ) and 200g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g/ml, then what shall be the molarity of the solution? Ans: Given that, Mass of solute ( C2H6O2 ) = 222.6 g Molar mass of C2H6O2 = 62 g/mol 222.6 Number of moles of C2H6O2 =  3.59mol 62 Mass of solvent (water) = 200 g Total mass of solution = 422.6 g Density of solution = 1.072 g/ml mass 422.6 Volume =   394.21ml density 1.072 Thus, Molality is given as, 3.59 Molality =  17.95m 200 And, molarity is given as, 3.59  1000 Molarity =  9.106M 394.21 9. 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). (i) Express this in percent by mass. Ans: We know that ppm is the units which expresses the quantity in parts per million i.e., parts per 106 of solution. Class XII Chemistry www.vedantu.com 12 Here, 15 ppm means 15 parts per 106 of solution. Thus, mass percentage is given as; 15 Mass percentage = 6  100  15  104 10 (ii) determine the molality of chloroform in water samples. Ans: 15 ppm states that we have 15 g chloroform in 106 g of solution i.e., mass of solvent = 106 g Molar mass of chloroform = 119.5 g/mol 15 Number of moles of chloroform =  0.125mol 119.5 Thus, Molality is given as, 0.125  1000 Molality = 6  125  106 m 10 10. What role does molecular interaction play in a solution of alcohol and water? Ans: Alcohol and water both possess a strong tendency to form intermolecular hydrogen bonding. When we mix the two liquids, a solution is formed as a result of formation of H-bonds between alcohol and H2O molecules. These interactions are weaker and less extensive than those in pure H2O. Hence, they show a positive deviation from ideal behavior. As a result, the solution of alcohol and water will have higher vapor pressure and lower boiling point than that of pure water and pure alcohol. 11. Why do gases always tend to be less soluble in liquids as the temperature is raised? Ans: When gases dissolve in water, the process is accompanied by release of excess heat energy, i.e., exothermic. According to Le Chatelier’s principle, when the temperature of the process is increased further the equilibrium shifts in backward direction. Hence, gases become less soluble in liquids. 12. State Henry’s law and mention some important applications. Ans: Henry’s law: It states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas. The effect of pressure on the solubility of a gas in a liquid is governed by this law. Mathematically; P  KH x Class XII Chemistry www.vedantu.com 13  where, P is the partial pressure of the gas x is the mole fraction of the gas in the solution K H is Henry’s Law constant. Applications of Henry’s law: a. In the sea diving. b. In the production of carbonated beverages. 13. The partial pressure of ethane over a solution containing 6.56  103 g of ethane is 1 bar. If the solution contains 5  102 g of ethane, then what shall be the partial pressure of the gas? Ans: By Henry’s law, we know that the solubility of gas in liquid is directly proportional to the pressure of the gas. The proportionality sign is then replaced by Henry’s constant. This is stated as; m  KH  P Also, mole fraction is directly proportional to the mass of the ethane. Thus, For case 1, 6.56  103  K H  1 For case 2, 5  102  K H  x where, x is the partial pressure of gas when the solution contains 5  102 g of ethane. Now, equating both the above equations; 3 5  102 6.56  10  x Thus, The partial pressure of gas, x  7.62 bar 14. What is meant by positive and negative deviations from Raoult’s law and how is the sign of  mix H related to positive and negative deviations from Raoult’s law? Ans: Positive deviation from Raoult’s law: Solutions exhibit positive deviation from Raoult’s law when they have vapor pressure more than expected from the law. Here,  mix H is positive as the energy is consumed for breaking the strong interaction and form weaker interactions. In a similar way,  mix V is positive as the expansion of volume takes place. - Negative deviation from Raoult’s law: Class XII Chemistry www.vedantu.com 14 Solutions exhibit negative deviation from Raoult’s law when they have vapor pressure less than expected from the law. Here,  mix H is negative as the energy is released due to replacement of weaker interactions by stronger ones. 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? Ans: Given that, Vapor pressure of pure water at boiling point, P0 = 1.103 bar Vapor pressure of solution, Ps = 1.004 bar An aqueous solution has 2% non-volatile solute. Thus, this states that the solution is 100 g from which 2 g  W2  is the non-volatile solute and the solvent is 98 g  W1 . Molar mass of water  M1  = 18 g/mol. Now, by Raoult’s law for dilute solutions; P0  Ps n2 n2   P0 n1  n 2 n1 W2 P  Ps 0 M2 0  P W1 M1 Thus, The molar mass of solute, M2  41.34g / mol 16. Heptane and octane form an ideal solution. At 373 K, the vapor pressure of the two liquid components is 105.2 kPa and 46.8 kPa respectively. What will be the vapor pressure of a mixture of 26.0 g of heptane and 35.0 g of octane? Ans: We have given the vapor pressures of the two liquid components i.e. V. P. of heptane = 105.2 kPa V. P. of octane = 46.8 kPa Mass of heptane = 26 g Mass of octane = 35 g Molar mass of heptane = 100 g/mol Molar mass of octane = 114 g/mol Thus, 26 Number of moles of heptane present =  0.26mol 100 35 Number of moles of octane present =  0.307mol 114 Class XII Chemistry www.vedantu.com 15  0.26 Mole fraction of heptane =  0.458 0.26  0.307 Mole fraction of octane = 1 – 0.458 = 0.541 Now, By Raoult’s law; Vapor pressure of heptane = x H  P 0 = 0.458 105.2  48.1816kPa Vapor pressure of octane = x O  P 0 = 0.541 46.8  25.3188kPa Thus, Vapor pressure of mixture = 48.1816 + 25.3188 = 73.5004kPa 17. The vapor pressure of water is 12.3 kPa at 300 K. Calculate vapor pressure of 1 molal solution of a non-volatile solute in it. Ans: We have given 1 molal solution i.e., 1 mole of non-volatile solute in 1000 g water. Vapor pressure of water, P0 = 12.3 kPa Now, Molar mass of water = 18 g/mol 1000 Number of moles of water in the solution =  55.56mol 18 1 Mole fraction of solute =  0.0176 1  55.56 Thus, By Raoult’s law, P0  Ps x P0 12.3  Ps  0.0176 12.3 Thus, V. P. of solution, Ps  12.082kPa 18. Calculate the mass of a non-volatile solute (molar mass 40 g/mol) which should be dissolved in 114 g octane to reduce its vapor pressure to 80%. Ans: Given that, Reduced vapor pressure = 80% of vapor pressure of pure components. 80 0  Ps  P  0.8P0 100 Now, let us consider the mass of a non-volatile solute as W g. Class XII Chemistry www.vedantu.com 16 Molar mass of the same solute = 40 g/mol W So, number of moles of same solute = mol 40 Mass of octane = 144 g Molar mass of octane = 144 g/mol So, number of moles of octane = 1 mol Now, W Mole fraction of non-volatile solute = 40 W 1 40 By Raoult’s law, P0  Ps x P0 W P 0  0.8P 0 40 x 1 0 P W 40 Thus, Mass of solute, W = 10 g 19. A solution containing 30g of non-volatile solute exactly in 90 g of water has a vapor pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new of vapor pressure becomes 2.9 kPa at 298 K. Calculate (i) molar mass of the solute Ans: Given that, Mass of non-volatile solute = 30 g Let molar mass of the same be M g/mol 30 Thus, number of moles of solute = mol M Case 1 Mass of water = 90 g Molar mass of water = 18 g/mol 90 So, number of moles of water involved =  5mol 18 Vapor pressure of solution, Ps = 2.8 kPa 30 Mole fraction of solute = M 30  5 M By Raoult’s law, Class XII Chemistry www.vedantu.com 17 P0  Ps x P0 30 P 0  2.8 M  5 0 P 30 M P0 6  1 2.8 M Case 2 Mass of water = 90 + 18 = 108 g Molar mass of water = 18 g/mol 108 So, number of moles of water involved =  6mol 18 Vapor pressure of solution, Ps = 2.9 kPa 30 Mole fraction of solute = M 30  6 M By Raoult’s law, P0  Ps x P0 30 P 0  2.8 M  6 0 P 30 M P0 5  1 2.9 M Now, Considering both the above equations, we get; 2.9 1  M 6  2.8 1  5 M  Molar mass of solute; M = 23 g/mol (ii) vapor pressure of water at 298 K. Ans: Now, putting this value in equation (1.1) P0 6 6 1 1 2.8 M 23  V. P. of water, P0  3.53kPa Class XII Chemistry www.vedantu.com 18 20. A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15K. Ans: Given that, Freezing point of pure water = 273.15 K Case 1 – A solution of 5% of cane sugar is in water i.e., 5 g of cane sugar is present in 100 g of water. Molar mass of cane sugar = 342 g/mol Molality is given as, 5  1000 Molality, m =  0.146m 342  100 Freezing point of solution = 271 K Tf  273.15  271  2.15K Now, Lowering in freezing point is given as, Tf  Kf m 2.15 Kf   14.726 0.146 Case 2 – A solution of 5% glucose is in water i.e. 5 g of glucose in 100 g of water. Molar mass of glucose = 180 g/mol Molality is given as, 5  1000 Molality, m =  0.277m 180  100 Kf  14.726 Lowering in freezing point is given as, Tf  Kf m = 14.726  0.277  4.079 Thus, Tf  273.15  T  4.079 Thus, the freezing point; T = 269.07 K 21. Two elements A and B form compounds having formula AB 2 and AB4. When dissolved in 20g of benzene ( C6H6 ). 1 g of AB 2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K.Kg / mol. Calculate atomic masses of A and B. Class XII Chemistry www.vedantu.com 19 Ans: We know that the lowering in freezing point is stated as, Tf  Kf m which also can be evaluated as; W  1000 Tf  K f  2 M 2  W1 Mass of solvent (benzene), W1 = 20 g Molal depression constant, K f = 5.1 K.Kg / mol Now, Case 1 – Solute is AB2 ; Mass of solute, W2 = 1 g Tf = 2.3 K Thus, Molar mass for the solute i.e., AB2 is given as, K  W2  1000 M2  f Tf  W1 5.1 1 1000 M2   110.869g / mol 2.3  20 Case 2 – Solute is AB4 ; Mass of solute, W2 = 1 g Tf = 1.3 K Thus, Molar mass for the solute i.e., AB4 is given as, K  W2  1000 M2  f Tf  W1 5.1  1  1000 M2   196.153g / mol 1.3  20 Now, Molar mass for AB2 is given as; A + 2 (B) = 110.869 g/mol Molar mass for AB4 is given as; A + 4 (B) = 196.153 g/mol Therefore, Solving above 2 equations we get, the atomic masses as; A = 25.59 u B = 42.64 u Class XII Chemistry www.vedantu.com 20 22. At 300 K, 36g of glucose present in a liter of its solution has an osmotic pressure of 4.08 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration? Ans: The formulation for the osmotic pressure is given as;   CRT Case 1 – T = 300 K Mass of glucose = 36 g Molar mass of glucose = 180 g/mol Osmotic pressure = 4.08 bar According to the formula; W   RT M 36 4.08   R  300 180 R = 0.068 units Case 2 – Osmotic pressure = 1.52 bar T = 300 K Thus, by given formula;   CRT Thus, the concentration will be, 1.52 C=  0.0745M 0.068  300 23. Suggest the most important type of intermolecular attractive interaction in the following pairs: (i) n-hexane and n-octane Ans: Both are nonpolar and hence, the intermolecular interactions will be London dispersion forces. (ii) I 2 and CCl 4. Ans: Both are nonpolar and hence, the intermolecular interactions will be London dispersion forces. (iii) NaClO4 and water Ans: The intermolecular interactions will be ion-dipole interactions as NaClO4 is an ionic compound and water is a polar molecule. (iv) methanol and acetone Class XII Chemistry www.vedantu.com 21 Ans: Both are polar molecules and hence, intermolecular interactions will be dipole- dipole interactions. (v) acetonitrile ( CH 3CN ) and acetone ( C3H6O ) Ans: Both are polar molecules and hence, intermolecular interactions will be dipole- dipole interactions. 24. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH , CH 3CN. Ans: The arrangement according to the increasing order of solubility in n-octane is; KCl < CH3OH < CH3CN < Cyclohexane. This is because; a. KCl is an ionic compound and will not dissolve in n-octane. b. CH3OH is a polar molecule and hence, will dissolve in n-octane. c. CH3CN is also a polar molecule but less than that of CH3OH. Thus, it will dissolve in n-octane to a great extent. d. Cyclohexane is also a polar and simpler molecule which will cause its dissolving in all proportions. 25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water. (i) Phenol Ans: Insoluble in water: (ii) Toluene Ans: Toluene is not soluble in one another due to their non-polar nature. (iii) Formic acid Ans: Partially soluble in water: (iv) Ethylene glycol Ans: Partially soluble (v) Chloroform Ans: Highly insoluble in water: (vi) Pentanol Ans: Partially soluble in water due to the presence of the alcohol functional group. Class XII Chemistry www.vedantu.com 22 26. If the density of some lake water is 1.25 g/ml and contains 92g of Na ions per kg of water, calculate the molality of Na ions in the lake. Ans: Given that, Mass of Na  ions = 92 g Molar mass of Na  ions = 23 g/mol Mass of water = 1000 g  Molality is given as; 92  1000 Molality, m =  4m 23  1000 27. If the solubility product of CuS is 6  1016 , calculate the maximum molarity of CuS aqueous solution. Ans: The dissociation reaction is given as; CuS Cu 2  S2 The solubility product is given as; K sp  6  1016 Let the solubility of each ion be denoted as X mol/L. Thus, K sp  Cu 2  S2   X  X X 2  6  1016 X  6  1016  Maximum molarity of CuS, X  2.44  108 mol / L 28. Calculate the mass percentage of aspirin ( C9H8O4 ) in acetonitrile ( CH 3CN ) when 6.5 g of C9H8O4 is dissolved in 450 g of CH 3CN. Ans: Given that, Mass of aspirin = 6.5 g Mass of acetonitrile = 450 g Total mass of solution = 6.5 + 450 = 456.5 g Mass percentage is given as, masssolute Mass percentage =  100 masssolution 6.5 =  100  1.423% 456.5 29. Nalorphene ( C19H21NO3 ), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given Class XII Chemistry www.vedantu.com 23  is 1.5 mg. Calculate the mass of 1.5  103 m aqueous solution required for the above dose. Ans: We have given, Molality of solution = 1.5  103 m; which states that the 1.5  103 moles are present in 1 kg of solvent. Molar mass of nalorphene = 311 g/mol Thus, Mass of solute = 1.5  103  311  0.4665g Total mass of the solution = 0.4665 + 1000 = 1000.4665 g From this we can say that 0.4665 g narlophene requires total solution of 1000.4665 g. Likewise; 1.5 mg narlophene will have; 1.5  103  1000.4665 Mass of solution =  3.216g 0.4665 30. Calculate the amount of benzoic acid ( C6H5COOH ) required for preparing 250 mL of 0.15M solution in methanol. Ans: Given that, We have a 0.15M solution which states that the solution of 1000 ml has 0.15 moles of solute. 250  0.15 Thus, to prepare 250 ml of solution we will require =  0.0375mol 1000 Molar mass of benzoic acid = 122 g/mol Therefore, mass of benzoic acid is given as, Mass = 0.0375  122  4.575g 31. 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 below. Explain briefly. Acetic acid < trichloroacetic acid < trifluoroacetic acid. Ans: The structures for the given acids are; We know that, H is least electronegative whereas, F is most electronegative. Thus, F can withdraw electrons towards itself more than Cl and H. This signifies that, Class XII Chemistry www.vedantu.com 24 trifluoroacetic acid can easily lose H  ions i.e., ionizes to the greatest extent. The more ions produced, the greater the depression of the freezing point. Therefore, the depression in the freezing point increases in the order: Acetic acid < trichloroacetic acid < trifluoroacetic acid. 32. Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. 3 K a  1.4  10 ,K f  1.86K.Kg / mol. Ans: Given that, Mass of CH3CH2CHClCOOH = 10 g Molar mass of CH3CH2CHClCOOH = 122.5 g/mol 10 Number of moles of the same, n =  0.0816moles 122.5 Mass of water (solvent), W1 = 250 g Molality is given by; n 0.0816  1000 Molality =   3.265  101 m W1 250 Also, we know that, K a  1.4  103 ,K f  1.86K.Kg / mol. Now, The dissociation reaction is given as,  CH 3CH 2CHClCOOH  CH 3CH 2CHClCOO   H at in itial conc. 0 C mol L-1 0  CH 3CH 2CHClCOOH  CH 3CH 2CHClCOO   H at equilibrium C(1 ) C C where,  is the degree of dissociation. Now, Dissociation constant is given as; C 2 2 Ka  C 1    As, 