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Page 1 1. (b) According to Le-chateliers principle, for an exothermic reaction (??< 0) increase in temperature decreases the solubility. 2. (b) = + = 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg The observed is 100 mm Hg which is less than 104 mm Hg. Hence the solution shows negative deviation. 3. (d) According to Henry’s law, m = k × p given K H = 1.4 × 10 –3 = 0.5 or No. of moles; 4. (c) These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B. 5. (c) There is no change in vapour pressure. 6. (b) = This is condition for ideal solution. 7. (d) C 2 H 5 I and C 2 H 5 OH form non-ideal solution. Page 2 1. (b) According to Le-chateliers principle, for an exothermic reaction (??< 0) increase in temperature decreases the solubility. 2. (b) = + = 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg The observed is 100 mm Hg which is less than 104 mm Hg. Hence the solution shows negative deviation. 3. (d) According to Henry’s law, m = k × p given K H = 1.4 × 10 –3 = 0.5 or No. of moles; 4. (c) These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B. 5. (c) There is no change in vapour pressure. 6. (b) = This is condition for ideal solution. 7. (d) C 2 H 5 I and C 2 H 5 OH form non-ideal solution. 8. (b) For this solution intermolecular interactions between n-heptane and ethanol are weaker than n-heptane-n-heptane & ethanol- ethanol interactions hence the solution of n-heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law. 9. (a) Liquid solvent and solid solvent are in equilibrium. 10. (b) The molality involves weights of the solute and the solvent. Since weight does not change with the temperature, therefore molality does not depend upon the temperature. 11. (d) Molarity (M) = wt. = g wt. of 70% acid = = 45 g 12. (c) Vant Hoff factor i = 4 in case of (NH 4 ) 3 PO 4 , (NH 4 ) 3 PO 4 13. (d) ?T b = iK b m Given, (?T b ) X > (?T b ) Y ? i X K b m > i Y K b m (K b is same for same solvent) i X > i Y So, x is undergoing dissociation in water. 14. (a) Depression in F.P. No. of particles. provides five ions on ionisation while KCl provides two ions Page 3 1. (b) According to Le-chateliers principle, for an exothermic reaction (??< 0) increase in temperature decreases the solubility. 2. (b) = + = 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg The observed is 100 mm Hg which is less than 104 mm Hg. Hence the solution shows negative deviation. 3. (d) According to Henry’s law, m = k × p given K H = 1.4 × 10 –3 = 0.5 or No. of moles; 4. (c) These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B. 5. (c) There is no change in vapour pressure. 6. (b) = This is condition for ideal solution. 7. (d) C 2 H 5 I and C 2 H 5 OH form non-ideal solution. 8. (b) For this solution intermolecular interactions between n-heptane and ethanol are weaker than n-heptane-n-heptane & ethanol- ethanol interactions hence the solution of n-heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law. 9. (a) Liquid solvent and solid solvent are in equilibrium. 10. (b) The molality involves weights of the solute and the solvent. Since weight does not change with the temperature, therefore molality does not depend upon the temperature. 11. (d) Molarity (M) = wt. = g wt. of 70% acid = = 45 g 12. (c) Vant Hoff factor i = 4 in case of (NH 4 ) 3 PO 4 , (NH 4 ) 3 PO 4 13. (d) ?T b = iK b m Given, (?T b ) X > (?T b ) Y ? i X K b m > i Y K b m (K b is same for same solvent) i X > i Y So, x is undergoing dissociation in water. 14. (a) Depression in F.P. No. of particles. provides five ions on ionisation while KCl provides two ions C 6 H 12 O 6 and C 12 H 22 O 11 are not ionised so they have single particle. Hence, Al 2 (SO 4 ) 3 have maximum value of depression in F.P or lowest F.P 15. (b) Sodium sulphate dissociates as hence van’t hoff factor i = 3 Now = 3 × 1.86 × 0.01 = 0.0558 K 16. (a) and both dissociates to give 5 ions or i = 5 K 4 [Fe(CN) 6 ] 4K + + [Fe(CN) 6 ] 4– and 17. (b) 1 molal solution means 1 mole of solute dissolved in 1000 gm solvent. ? n solute = 1 w solvent = 1000g ? n solvent = = 55.56 X solute = = 0.0177 18. (b) Concentration of = 0.955 M = 2 × 0.955 = 1.91 M = 0.955 M 19. (d) Azeotrope of HCl + H 2 O contains 20.2% HCl. 20. (a) Mole fraction of any component A Page 4 1. (b) According to Le-chateliers principle, for an exothermic reaction (??< 0) increase in temperature decreases the solubility. 2. (b) = + = 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg The observed is 100 mm Hg which is less than 104 mm Hg. Hence the solution shows negative deviation. 3. (d) According to Henry’s law, m = k × p given K H = 1.4 × 10 –3 = 0.5 or No. of moles; 4. (c) These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B. 5. (c) There is no change in vapour pressure. 6. (b) = This is condition for ideal solution. 7. (d) C 2 H 5 I and C 2 H 5 OH form non-ideal solution. 8. (b) For this solution intermolecular interactions between n-heptane and ethanol are weaker than n-heptane-n-heptane & ethanol- ethanol interactions hence the solution of n-heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law. 9. (a) Liquid solvent and solid solvent are in equilibrium. 10. (b) The molality involves weights of the solute and the solvent. Since weight does not change with the temperature, therefore molality does not depend upon the temperature. 11. (d) Molarity (M) = wt. = g wt. of 70% acid = = 45 g 12. (c) Vant Hoff factor i = 4 in case of (NH 4 ) 3 PO 4 , (NH 4 ) 3 PO 4 13. (d) ?T b = iK b m Given, (?T b ) X > (?T b ) Y ? i X K b m > i Y K b m (K b is same for same solvent) i X > i Y So, x is undergoing dissociation in water. 14. (a) Depression in F.P. No. of particles. provides five ions on ionisation while KCl provides two ions C 6 H 12 O 6 and C 12 H 22 O 11 are not ionised so they have single particle. Hence, Al 2 (SO 4 ) 3 have maximum value of depression in F.P or lowest F.P 15. (b) Sodium sulphate dissociates as hence van’t hoff factor i = 3 Now = 3 × 1.86 × 0.01 = 0.0558 K 16. (a) and both dissociates to give 5 ions or i = 5 K 4 [Fe(CN) 6 ] 4K + + [Fe(CN) 6 ] 4– and 17. (b) 1 molal solution means 1 mole of solute dissolved in 1000 gm solvent. ? n solute = 1 w solvent = 1000g ? n solvent = = 55.56 X solute = = 0.0177 18. (b) Concentration of = 0.955 M = 2 × 0.955 = 1.91 M = 0.955 M 19. (d) Azeotrope of HCl + H 2 O contains 20.2% HCl. 20. (a) Mole fraction of any component A As total no. of moles > no. of moles of A thus x can never be equal to one or zero. 21. (50) = 0.3 °C 0.3 °C = 0.3 = = 50 g The amount used should be more than 50 g. 22. (0.83) Let a be the give of dissociation then Van’t Hoff's factor i = Again Van’t Hoff’s factor Equating to both values of i, 23. (2) ?T f = 0 – (– 0.00732°) = 0.00732 ?T f = i × K f × m 24. (–0.654) As ?T f = K f . m ?T b = K b . m Hence, we have Page 5 1. (b) According to Le-chateliers principle, for an exothermic reaction (??< 0) increase in temperature decreases the solubility. 2. (b) = + = 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg The observed is 100 mm Hg which is less than 104 mm Hg. Hence the solution shows negative deviation. 3. (d) According to Henry’s law, m = k × p given K H = 1.4 × 10 –3 = 0.5 or No. of moles; 4. (c) These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B. 5. (c) There is no change in vapour pressure. 6. (b) = This is condition for ideal solution. 7. (d) C 2 H 5 I and C 2 H 5 OH form non-ideal solution. 8. (b) For this solution intermolecular interactions between n-heptane and ethanol are weaker than n-heptane-n-heptane & ethanol- ethanol interactions hence the solution of n-heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law. 9. (a) Liquid solvent and solid solvent are in equilibrium. 10. (b) The molality involves weights of the solute and the solvent. Since weight does not change with the temperature, therefore molality does not depend upon the temperature. 11. (d) Molarity (M) = wt. = g wt. of 70% acid = = 45 g 12. (c) Vant Hoff factor i = 4 in case of (NH 4 ) 3 PO 4 , (NH 4 ) 3 PO 4 13. (d) ?T b = iK b m Given, (?T b ) X > (?T b ) Y ? i X K b m > i Y K b m (K b is same for same solvent) i X > i Y So, x is undergoing dissociation in water. 14. (a) Depression in F.P. No. of particles. provides five ions on ionisation while KCl provides two ions C 6 H 12 O 6 and C 12 H 22 O 11 are not ionised so they have single particle. Hence, Al 2 (SO 4 ) 3 have maximum value of depression in F.P or lowest F.P 15. (b) Sodium sulphate dissociates as hence van’t hoff factor i = 3 Now = 3 × 1.86 × 0.01 = 0.0558 K 16. (a) and both dissociates to give 5 ions or i = 5 K 4 [Fe(CN) 6 ] 4K + + [Fe(CN) 6 ] 4– and 17. (b) 1 molal solution means 1 mole of solute dissolved in 1000 gm solvent. ? n solute = 1 w solvent = 1000g ? n solvent = = 55.56 X solute = = 0.0177 18. (b) Concentration of = 0.955 M = 2 × 0.955 = 1.91 M = 0.955 M 19. (d) Azeotrope of HCl + H 2 O contains 20.2% HCl. 20. (a) Mole fraction of any component A As total no. of moles > no. of moles of A thus x can never be equal to one or zero. 21. (50) = 0.3 °C 0.3 °C = 0.3 = = 50 g The amount used should be more than 50 g. 22. (0.83) Let a be the give of dissociation then Van’t Hoff's factor i = Again Van’t Hoff’s factor Equating to both values of i, 23. (2) ?T f = 0 – (– 0.00732°) = 0.00732 ?T f = i × K f × m 24. (–0.654) As ?T f = K f . m ?T b = K b . m Hence, we have or = 0.18 × = 0.654°C As the Freezing Point of pure water is 0 °C, ?T f = 0 –T f 0.654 = 0 – T f ? T f = – 0.654 Thus the freezing point of solution will be – 0.654 °C. 25. (0.052) ; ?Read More
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