Class 12 Chemistry
Chapter 2 - Solutions
Very Important Questions (Magical Series)
Q.1. Define Molarity (M).
Q.2. Define Molality (m).
Q.3. State the main advantage of molality over molarity as the unit of concentration.
Q.4. Define Mole fraction.
Q.5. Explain the Henry's law about dissolution of a gas in a liquid.
Q.6. State important applications of Henry's law.
Q.7. Aquatic animals are more comfortable in cold water than in warm water.
Q.8. Calculate the molarity of 9.8% (w/w) solution of H2SO4 if the density of the solution is 1.02 g mL–1. (Molar mass of H2SO4 = 98 g mol–1). [Ans: 1.02 M]
Q.9. Differentiate between molarity and molality of a solution. How can we change molality value of a solution into molarity value?
Q.10. A solution of glucose (molar mass = 180 g mol–1) in water is labelled as 10% (by mass). Calculate molality and molarity of the solution. (Density of solution = 1.2 g mL–1).
[Ans: 0.62 m, 0.66 M]
Q.11. Gas (A) is more soluble in water than gas (B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry's constant) and why?
Q.12. What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of 25°C, a total pressure of l atmosphere and mole fraction of nitrogen in air of 0.78. [KH for nitrogen= 8.42 × 10–7 M/mm Hg].
[Ans: 4.99 × 10–4]
Q.13. The partial pressure of ethane over a saturated solution containing 6.56 × 10–2 g of ethane is 1 bar. If the solution contains 5.0 × 10–2 g of ethane, then what will be the partial pressure of the gas? [Ans: 0.762 bar]
Q.14. If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water? Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N2 at 293 K is 76.48 k bar. [Ans: 0.716 millimoles]
Q.15. Define Raoult's law. How is it formulated for solutions of non-volatile solutes?
Q.16. Name the solution which follows Raoult's law at all concentrations and temperatures.
Q.17. Define Ideal solution.
Q.18. In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?
Q.19. What are 'azeotropes'?
Q.20. Write two differences between ideal solutions and non-ideal solutions.
Q.21. What is the similarity between Raoult's law and Henry's law?
Q.22. 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.
[Ans: Liquid mixture: A = 0.40, B = 0.60. Vapour phase: A: 0.30, B: 0.70]
Q.23. What is meant by positive deviations from Raoult's law? Give an example. What is the sign of ∆mixH for positive deviation?
Q.24. What type of azeotrope is formed by positive deviation from Raoult's law? Give an example.
Q.25. What is meant by negative deviation from Raoult's law? Give an example. What is the sign of ∆mixH for negative deviation?
Q.26. What type of azeotrope is formed by negative deviation from Raoult's law? Give an example.
Q.27. What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
Q.28. Explain why a solution of chloroform and acetone shows negative deviation from Raoult's law.
Q.29. What type of intermolecular attraction exists in each of the following pairs of compounds:
(i) n-hexane and n-octane
(ii) methanol and acetone
Q.30. Define Colligative properties.
Q.31. What are isotonic solutions?
Q.32. Define Molal elevation constant (Boiling point elevation constant), Kb.
Q.33. How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it?
Q.34. Define Osmotic pressure.
Q.35. Calculate the freezing point of a solution containing 60 g of glucose (molar mass = 180 g Mol–1) in 250 g of Water. (Kf of water = 1.86 K kg mol–1) [Ans: 270.52 K]
Q.36. Give reasons for the following:
(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
(b) Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.
Q.37. (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution?
Q.38. Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?
Q.39. Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
Q.40. Calculate the mass of compound (molar mass = 256 g mol–1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K. (Kf = 5.12 K kg mol–1). [Ans: 1.8 g]
Q.41. 18 g of glucose, C6H12O6 (Molar mass = 180 g mol–1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water = 0.52 K kg mol–1, boiling point of pure water = 373.15 K) [Ans: 373.202 K]
Q.42. An aqueous solution of sodium chloride freezes below 273 K. Explain the lowering in freezing point of water with the help of a suitable diagram.
Q.43. Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
Q.44. List any four factors on which the colligative properties of a solution depend.
Q.45. Outer hard shells of two eggs are removed. One of the eggs is placed in pure water and the other is placed in saturated solution of sodium chloride. What will be observed and why?
Q.46. Find the boiling point of a solution containing 0.520 g of glucose (C6H12O6) dissolved in 80.2 g of water. [Given: Kb for water = 0.52 K/m] [Ans: 373.019 K]
Q.47. A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K. (Given: Molar mass of sucrose = 342 g mol–1, molar mass of glucose = 180 g mol–1).
[Ans: 265.55 K]
Q.48. 30 g of urea (M = 60 g mol–1] is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg. [Ans: 23.5 mm Hg]
Q.49. Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water. (Kf for water = 1.86 K kg mol–1). [Ans: 270.52 K]
Q.50. A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass= 180 g mol−1) in water. Calculate the mass of glucose present in one litre of its solution. [Ans: 45 g]
Q.51. Calculate the boiling point elevation for a solution prepared by adding 10 g of CaCl2 to 200 g of water. (Kb for water = 0.52 K kg mol–1, molar mass of CaCl2 = 111 g mol–1)
[Ans: 0.2306 K]
Q.52. Some ethylene glycol, HOCH2CH2OH, is added to your car's cooling system along with 5 kg of water. If the freezing point of water-glycol solution is –15.0 °C, what is the boiling point of the solution? (Kb = 0.52 K kg mol-1 and Kf = 1.86 K kg mol-1 for water)
[Ans: 453.51 K]
Q.53. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol-1. Find the molar mass of the solute. [Ans: 256 g mol–1]
Q.54. A 5% solution (by mass) of cane-sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K. [Molecular masses: Glucose C6H12O6: 180 amu; Cane-sugar C12H22O11: 342 amu] [Ans: 269.07 K]
Q.55. A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42 °C while pure water boils at 100°C. What mass of glycerol was dissolved to make the solution? (Kb for water = 0.512 K kg mol–1). [Ans: 3 7.73 g]
Q.56. 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at –0.34 °C. What is the molar mass of this material? (Kf or water = 1.86 K kg mol-1). [Ans: 182.35 g mol–1]
Q.57. 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 added to this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) the molecular mass of solute and
(ii) vapour pressure of water at 298 K.
[Ans: (i) 34 g mol–1, (ii) 3.4 kPa]
Q.58. Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.00 g of water. (Kb for water = 0.512 K kg mol–1. (Molar mass of NaCl = 58.44 g).
[Ans: 374.20 K]
Q.59. A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is a non-electrolyte, determine its molar mass.
[Ans: 14193.29 g mol–1]
Q.60. What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g of chloroform to form a solution that has a boiling point of 68.04 °C. The boiling point of pure chloroform is 61.7 °C and the boiling point elevation constant, Kb for chloroform is 3.63 °C/m. [Ans: 148.15 g mol–1]
Q.61. What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50 °C? The freezing point depression constant (Kf) for water is 1.86 °C/m. Assume van't Hoff factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g mol–1).
[Ans: 8.199 g]
Q.62. Calculate the boiling point of one molar aqueous solution (density 1.06 g mol–1) of KBr.
[Given: Kb for H2O = 0.52 K kg mol–1, atomic mass: K = 39, Br = 80] [Ans: 373.553 K]
Q.63. A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31 oc. Determine the molar mass of this compound. (B.p. of pure benzene = 80.10 °C and Kb for benzene = 2.53 °C kg mol-1)
[Ans: 152 g mol–l]
Q.64. What mass of ethylene glycol (molar mass = 62.0 g mort) must be added to 5.50 kg of water to lower the freezing point of water from 0 °C to –10.0 °C? (Kf for water 1.86 K kg mol–1).
[Ans: 1833.33 g]
Q.65. 100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein? (R = 0.0821 L atm mol-1 K–1 and 760 mm Hg = 1 atm.).
[Ans: 13980.45 g mol–1]
Q.66. Calculate the amount of sodium chloride which must be added to one kilogram of water so that the freezing point of water is depressed by 3 K. [Given: Kf = 1.86 K kg mol–1, atomic mass: Na = 23.0, Cl = 35.5]
[Ans: 47.18 g]
Q.67. x g of a non-electrolytic compound (molar mass = 200) is dissolved in 1.0 L of 0.05 M NaCl aqueous solution. The osmotic pressure of this solution is found to be 4.92 atm at 27 °C. Calculate the value of x. Assume complete dissociation of NaCl and ideal behaviour of the solution. (R = 0.082 L atm mort K–1).
[Ans: 20 g]
Q.68. Calculate the freezing point of a solution containing 18 g glucose, C6H12O6 and 68.4 g sucrose, C12H22O11 in 200 g of water. The freezing point of pure water is 273 K and Kf for water is 1.86 K mol–1. [Ans: 270.21 K]
Q.69. Define the van't Hoff factor.
Q.70. Define the Abnormal molar mass.
Q.71. What types of values can it have if in forming the solution the solute molecules undergo
(i) dissociation
(ii) association?
Q.72. Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber's salt, Na2SO4.10H2O in 0.100 kg of water. (Kf or water = 1.86 K kg mol–1, atomic masses: Na = 23, S = 32, O = 16, H = 1).
[Ans: 271.96 K]
Q.73. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.18 °C. Determine the van't Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol–1). [Ans: 0.35]
Q.74. Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol–1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization. (Kf for water = 1.86 K kg mol–1)
[Ans: 270.77 K]
Q.75. When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx). (Kf the CS2 = 3.83 K kg mol–1, atomic mass of sulphur = 32 g mol–1). [Ans: S8]
Q.76. Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g mol–1) (Kf for water = 1.86 K kg mol–1)
[Ans: 271.408 K]
Q.77. A 0.561 m solution of an unknown electrolyte depresses the freezing point of water by 2.93 °C. What is van't Hoff factor for this electrolyte? The freezing point depression constant (Kf) for water is 1.86 °C kg mol–1. [Ans: 2.807]
Q.78. Phenol associates in benzene to a certain extent to form a dimer. A solution containing 20 g of phenol in 1.0 kg of benzene has its freezing point lowered by 0.69 K. Calculate the fraction of phenol that has dimerised [Given Kf for benzene= 5.1 K mol–1]
[Ans: 73 %]
Q.79. An aqueous solution containing 12.48 g of barium chloride in 1.0 kg of water boils at 373.0832 K. Calculate the degree of dissociation of barium chloride. [Given Kb for H2O = 0.52 K mol–1; Molar mass of BaCl2 = 208.34 g mol–1].
[Ans: 83.5 %]
Q.80. A decimolar solution of potassium ferrocyanide K4[Fe(CN)6] is 50% dissociated at 300 K. Calculate the value of van't Hoff factor for potassium ferrocyanide. Also calculate the osmotic pressure.
[Ans: 3, 7.389 atm]
Q.81. Calculate the boiling point of solution when 4 g of MgSO4 (M = 120 g mol–1) was dissolved in 100 g of water, assuming MgSO4 undergoes complete ionization. (Kb for water = 0.52 K kg mol–1).
[Ans: 100.34 °C]
Q.82. 3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated). (Given: Molar mass of benzoic acid = 122 gm, Kf for benzene = 4.9 K kg mol–1). [Ans: 0.506, associated]
Q.83. Calculate the mass of NaCl (molar = 58.5 g mol–1) to be dissolved in 37.2 g of water to lower the freezing point by 2 °C, assuming that NaCl undergoes complete dissociation. (Kf for water = 1.86 K kg m or 1). [Ans: 1.17 g]
Q.84. Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in 2 L of water at 25°C, assuming that it is completely dissociated. (R = 0.0821 L atm K–1 mol–1, molar mass of K2SO4 = 174 g mol–1).
[Ans: 0.1758 ×10–2 atm]
Q.85. Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K (the Kf for water = 1.86 K kg mol–1). [Ans: 40.05 g]
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