Chemistry_Class-12_Ch-1_Solution_Topic-3_Colligative Properties 1 / 30 1. Equimolal solutions in the same solvent have A) Same boiling point but different freezing point B) Same freezing point but different boiling point C) Same boiling and same freezing points D) Different boiling and different freezing points 2 / 30 2. Which of the following is a colligative property A) Osmotic pressure B) Boiling point C) Vapour pressure D) ) Freezing point 3 / 30 3. The colligative properties of a solution depend on A) Nature of solute particles present in it B) Nature of solvent used C) Number of solute particles present in it D) Number of moles of solvent only 4 / 30 4. Which of the following aqueous solutions has the highest boiling point A) 0.1 M KNO₃ B) 0.1 M Na₃PO₄ C) 0.1 M BaCl₂ D) 0.1 M K₂SO₄ 5 / 30 5. Which of the following statements is false A) Two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point B) The osmotic pressure of a solution is given by the equation 𝝅 = CRT (where. C is the molarity of the solution) C) Decreasing order of osmotic pressure for 0.01 M aqueous solutions of barium chloride, potassium chloride. acetic acid and sucrose BaCl2 > KCI > CH3COOH > sucrose is D) According to Raoult's law, the vapour pressure exerted by a non volatile component of a solution is directly proportional to its mole fraction in the solution 6 / 30 6. The magnitude of colligative properties in all colloidal dispersions is___________ than solution A) Lower B) Higher C) Both D) None 7 / 30 7. Which has highest freezing point A) 1 m K₄[Fe(CN)₆] solution B) 1 m C₆H₁₂O₆ solution C) 1 m KCl solution D) 1 m rock salt solution 8 / 30 8. Colligative properties are used for the determination of A) ) Molar Mass B) Equivalent weight C) Arrangement of molecules D) Melting point and boiling point 9 / 30 9. Which colligative property is more useful to determine the molecular weight of the substances like proteins and polymers A) Lowering of vapour pressure B) Elevation in boiling point C) Depression of freezing point D) Osmotic pressure 10 / 30 10.Vapour pressure of CCl4 at 25°C is 143 mm of Hg. 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in 100 mL CCl4. Find the vapour pressure of the solution (Density of CCl4 = 1.58 g/cm³) A) 141.93 mm B) 199.34 mm C) 94.39 mm D) 143.99 mm 11 / 30 11. Vapour pressure of chloroform (CHCI3) and dichloromethane (CH₂Cl₂) at 25°C are 200mm Hg and 41.5mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH₂Cl₂ at the same temperature will be (Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85u) A) 90.92 mm Hg B) 347.9 mm Hg C) 615.0 mm Hg D) 285.5 mm Hg 12 / 30 12. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be respectively A) 200 and 300 B) 300 and 400 C) 400 and 600 D) 500 and 600 13 / 30 13. The vapour pressure of water at 20°C is 17.5 mmHg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20°C. the vapour pressure of the resulting solution will be A) 15.750 mm Hg B) 16.500 mm Hg C) 17.325 mm Hg D) 17.675 mm Hg 14 / 30 14. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6. its vapour pressure (in mm) at the same temperature will be A) 350 B) 300 C) 700 D) 360 15 / 30 15. . At 300 K, when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be A) 0.005 B) 0.010 C) 0.100 D) 0.900 16 / 30 16. Benzene and toluene form nearly ideal solutions. At 20°C. the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20°C for a solution containing 78 g of benzene and 46 g of toluene in torr is A) 50 B) 25 C) 37.5 D) 53.5 17 / 30 17. According to Raoult's law the relative lowering of vapour pressure of a solution of non-volatile substance is equal to [IIT JEE 1983; CBSE PMT 1995; BHU 2001;( A) Mole fraction of the solvent B) Mole fraction of the solute C) Weight percentage of a solute D) Weight percentage of a solvent 18 / 30 18. When a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in A) An increase in the boiling point of the solution B) A decrease in the boiling point of solvent C) The solution having a higher freezing point than the solvent D) The solution having a lower osmotic pressure than the solvent 19 / 30 19. "Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute". Above statement is A) Henry law B) Dulong and Petit law C) Raoult's law D) Le-Chatelier's principle 20 / 30 20. The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be A) 140 torr B) 20 torr C) 68 torr D) 72 torr 21 / 30 21. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte solid weighing 2.175g is added to 39.08g of benzene. The vapour pressure of the solution is 600mm of Hg. What is the molecular weight of solid substance A) 49.50 B) 59.6 C) 69.5 D) 79.8 22 / 30 22. 60 g of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is Po, the vapour pressure of solution is A) 0.10 P° B) 1.10 P° C) 0.90 P° D) 0.99 P° 23 / 30 23. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass 58) at 298K. The vapour pressure of the solution was found to be 192.5 mmHg. The molecular weight of the solute is (vapour pressure of acetone = 195 mmHg) A) 25.24 B) 35.24 C) 50 D) 55.24 24 / 30 24. The vapour pressure of a solvent decreased by 10mm of mercury, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if decrease in the vapour pressure is to be 20mm of mercury A) 0.8 B) 0.6 C) 0.4 D) 0.2 25 / 30 25. The vapour pressure of a solvent A is 0.80 atm When a non-volatile substance B is added to this solvent its vapour pressure drops to 0.6 atm. What is mole fraction of B in solution A) 250 mm, 550 mm B) 350 mm, 450 mm C) 350 mm, 700 mm D) 550 mm, 250 mm 26 / 30 26. The mass of a non-volatile solute of molar mass 40 gmol-1 that should be dissolved in 114g of octane to lower its vapour pressure by 20% is A) 10 g B) 11.4g C) 9.8g D) 12.8g 27 / 30 27. Which of the following can be measured by the Ostwald- Walker dynamic method A) Vapour pressure of the solvent B) Relative lowering of vapour pressure C) Lowering of vapour pressure D) All of these 28 / 30 28. Vapour pressure of pure 'A' is 70 mm of Hg at 25°C. it forms an ideal solution with 'B' in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at 25°C, the vapour pressure of pure 'B' at 25°C is A) 56 mm B) ) 70 mm C) 140 mm D) 28 mm 29 / 30 29. At 300 K two pure liquids A and B have vapour pressure respectively 150 mm Hg and 100 mm Hg. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is A) 0.6 B) 0.5 C) 0.8 D) 0.4 30 / 30 30. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0g of heptane and 35g of octane will be (molar mass of heptane = 100 g/mol and of octane = 114 g/mol ) A) 144.5 kPa B) 72.0 kPa C) 36.1 kPa D) 96.2 kPa Your score isThe average score is 23% 0% Exit
