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Solutions Class 12 MCQ Questions with Answers
Class 12 Chemistry MCQ with answers are given here to chapter Solution. These MCQs are based on the latest CBSE board syllabus and relate to the latest Class 12 Chemistry syllabus. By Solving these Class 12 MCQs, you will be able to analyze all of the concepts quickly in the chapter and get ready for the Class 12 Annual exam.
Learn Class 12 Chemistry Chapter 2 MCQ with answers pdf free download according to the latest CBSE and NCERT syllabus. Students should prepare for the examination by solving Solutions Class 12 Chemistry MCQ with answers given below.
Question 1. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3? The concentrated acid is 70% HNO3.
(a) 70.0 g conc. HNO3
(b) 54.0 g conc. HNO3
(c) 45.0 g conc. HNO3
(d) 90.0 g conc. HNO3
Answer
C
Question 2. Which of the following is dependent on temperature?
(a) Molarity
(b) Mole fraction
(c) Weight percentage
(d) Molality
Answer
A
Question 3. What is the mole fraction of the solute in a 1.00 m aqueous solution?
(a) 1.770
(b) 0.0354
(c) 0.0177
(d) 0.177
Answer
C
Question 4. Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL–1. Volume of acid required to make one litre of 0.1 M H2SO4 solution is
(a) 16.65 mL
(b) 22.20 mL
(c) 5.55 mL
(d) 11.10 mL
Answer
C
Question 5. The mole fraction of the solute in one molal aqueous solution is
(a) 0.009
(b) 0.018
(c) 0.027
(d) 0.036
Answer
B
Question 6. Which of the following compounds can be used as antifreeze in automobile radiators?
(a) Methyl alcohol
(b) Glycol
(c) Nitrophenol
(d) Ethyl alcohol
Answer
B
Question 7. What is the molarity of H2SO4 solution, that has a density 1.84 g/cc at 35°C and contains 98% by weight?
(a) 18.4 M
(b) 18 M
(c) 4.18 M
(d) 8.14 M
Answer
A
Question 8. 2.5 litre of 1 M NaOH solution is mixed with another 3 litre of 0.5 M NaOH solution. Then find out molarity of resultant solution.
(a) 0.80 M
(b) 1.0 M
(c) 0.73 M
(d) 0.50 M
Answer
C
Question 9. How many g of dibasic acid (mol. weight 200) should be present in 100 mL of the aqueous solution to give strength of 0.1 N?
(a) 10 g
(b) 2 g
(c) 1 g
(d) 20 g
Answer
C
Question 10. How many grams of CH3OH should be added to water to prepare 150 mL solution of 2 M CH3OH?
(a) 9.6 × 103
(b) 2.4 × 103
(c) 9.6
(d) 2.4
Answer
C
Question 11. In water saturated air, the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is
(a) 1.18 atm
(b) 1.76 atm
(c) 1.176 atm
(d) 0.98 atm.
Answer
C
Question 12. The concentration unit, independent of temperature, would be
(a) normality
(b) weight volume percent
(c) molality
(d) molarity.
Answer
C
Question 13. A solution has a 1 : 4 mole ratio of pentane to hexane.
The vapour pressures of the pure hydrocarbons at 20 °C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be
(a) 0.200
(b) 0.549
(c) 0.786
(d) 0.478
Answer
D
Question 14. pA and pB are the vapour pressures of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be
(a) pA + xA(pB – pA)
(b) pA + xA ( pA – pB)
(c) pB + xA(pB – pA)
(d) pB + xA ( pA – pB)
Answer
D
Question 15. Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25°C are 200 mm Hg and 41.5 mm 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 = 85 u)
(a) 173.9 mm Hg
(b) 615.0 mm Hg
(c) 347.9 mm Hg
(d) 285.5 mm Hg
Answer
None
Question 16. The mixture which shows positive deviation from Raoult’s law is
(a) ethanol + acetone
(b) benzene + toluene
(c) acetone + chloroform
(d) chloroethane + bromoethane.
Answer
A
Question 17. For an ideal solution, the correct option is
(a) DmixG = 0 at constant T and P
(b) DmixS = 0 at constant T and P
(c) Dmix V ≠ 0 at constant T and P
(d) Dmix H = 0 at constant T and P.
Answer
D
Question 18. 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 mol of Q would be
(a) 72 torr
(b) 140 torr
(c) 68 torr
(d) 20 torr
Answer
A
Question 19. Which one of the following is incorrect for ideal solution?
(a) DHmix = 0
(b) DUmix = 0
(c) DP = Pobs – Pcalculated by Raoult’s law = 0
(d) DGmix = 0
Answer
D
Question 20. The mixture that forms maximum boiling azeotrope is
(a) heptane + octane
(b) water + nitric acid
(c) ethanol + water
(d) acetone + carbon disulphide.
Answer
B
Question 21. Which of the following statements is correct regarding a solution of two components A and B exhibiting positive deviation from ideal behaviour?
(a) Intermolecular attractive forces between A-A and B-B are stronger than those between A-B.
(b) Δmix H = 0 at constant T and P.
(c) Δmix V = 0 at constant T and P.
(d) Intermolecular attractive forces between A-A and B-B are equal to those between A-B.
Answer
A
Question 22. The osmotic pressure of 0.2 molar solution ofw·ea at 27°C (R = 0.082L atrn mol-1 K-1) is
(a) 4.92 atm
(b) 1 atrn
(c) 0.2 atm
(d) 27 atrn
Answer
A
Question 23.Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25°C.
(Given, vapour pressure data at 25°C, benzene = 12.8 kPa, toluene = 3.85 kPa)
(a) The vapour will contain equal amounts of benzene and toluene.
(b) Not enough information is given to make a prediction.
(c) The vapour will contain a higher percentage of benzene.
(d) The vapour will contain a higher percentage of toluene.
Answer
C
Question 24. Calculate the molal depression constant of a solvent which has freezing point 16.6°C and latent heat of fusion 180.75 Jg-1 ·
(a) 2.68
(b) 3.86
(c) 4.68
(d) 2.86
Answer
B
Question 25. Dry air is passed through a solution containing 10 g of a solute in 90 g of water and then through pure water. The loss in weight of solution is 2.5 g and that of pure solvent is 0.05 g. Calculate the molecular weight of the solute.
(a) 50
(b) 180
(c) 102
(d) 25
(e) 51
Answer
C
Question 26. By dissolving 5 g substance in 50 g of water, the decrease in freezing point is 1.2°C. The gram molal depression is 1.85°C. The molecular weight of substance is
(a) 105.4
(b) 118.2
(c) 137.2
(d) 154.2
Answer
D
Question 27. In an osmotic pressure measurement experiment, a 5% solution of compound’ X’ is found to be isotonic with a 2% acetic acid solution. The gram molecular mass of X’ is
(a) 24
(b) 60
(c) 150
(d) 300
Answer
C
Question 28. van’t Hoff factor of Ca(NO3 )2 is
(a) one
(b) two
(c) three
(d) four
Answer
C
Question 29. The volume of water to be added to 100cm3 of 0.5 N H2SO4 to get decinormal concentration is
(a) 400cm3
(b) 450cm3
(c) 500cm3
(d) 100cm3
Answer
A
Question 30. The vapour pressure of water at 23 °C is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm) of the resultant solution ?
(a) 19.0
(b) 19.602
(c) 19.402
(d) 19.202
Answer
B
Question 31. Abnormal colligative properties are observed only when the dissolved non-volatile solute in a given dilute solution
(a) is a non-electrolyte
(b) offers an intense colour
(c) associates or dissociates
(d) offers no colour
Answer
C
Question 32. van ‘t Hoff factor more than unity indicates that the solute in solution has
(a) dissociated
(b) associated
(c) Both (a) and (b)
(d) cannot say anything
Answer
A
Question 33. Osmotic pressure observed when benzoic acid is dissolved in benzene is less than that expected from theoretical considerations. This is because
(a) benzoic acid is an organic solute
(b) benzoic acid bas higher molar mass than benzene
(c) benzoic acid gets associated in benzene
(d) benzoic acid gets dissociated in benzene
Answer
C
Question 34. When 20 g ofnaphthoic acid (C11H8O2 )is dissolved in 50 g of benzene ( Kf = 1.72 K kg mol-1 ), a freezing point depression of 2 ‘lis observed. The van’t Hoff factor (i) is
(a) 0.5
(b) 1
(c) 2
(d) 3
Answer
A
Whoever needs to take the CBSE Class 12 Board Exam should look at this MCQ. To the Students who will show up in CBSE Class 12 Chemistry Board Exams, It is suggested to practice more and more questions. Aside from the sample paper you more likely had solved. These Solutions Class 12 MCQ are ready by the subject specialists themselves.
Question 35. Phenol dimerises in benzene having van’t Hoff factor 0.54. What is the degree of association ?
(a) 1.92
(b) 0.98
(c) 1.08
(d) 0.92
Answer
D
Question 36. Distribution law was given by
(a) Henry
(b) van’t Hoff
(c) Nemst’s
(d) Ostwald
Answer
C
Question 37. Observe the following abbreviations πobs = observed colligative property, πcal = theoretical colligative property assuming normal behaviour of solute. van ‘t Hoff factor (i) is given by
(a) i = πobs X πcal
(b) i = πobs + πcal
(c) i = πobs – πcal
(d) i = πobs / πcal
Answer
D
Question 38. The elevation in boiling point of a solution of 13.44 g of CuCl2 in 1 kg of water using the following infom1ation will be (molecular weight of CuCl2 = 134. 4 and kb = 0.52Km-1)
(a) 0.16
(b) 0.05
(c) 0.1
(d) 0.2
Answer
A
Question 39. The van’t Hoff factor of BaCl2 at 0.01 M concentration is 1.98. The percentage of dissociation of BaCl2 at this concentration is
(a) 4
(b) 69
(c) 89
(d) 98
(e) 100
Answer
A
Question 40. One gram of silver gets distributed between 10 cm3 of molten zinc and 100 cm3 of molten lead at 800°C. The percentage of silver still left in the lead layer in approximately. Distributed coefficient = 300
(a) 2
(b) 5
(c) 3
(d) 1
Answer
C
Question 41. Ifa is the degree of dissociation of Na2SO4 , the van ‘t Hoff factor (i) used for calculating the molecular mass is
(a) 1 – 2α
(b) 1 + 2α
(c) 1 – α
(d) 1 + α
Answer
B
Question 42. Which of the following is incorrect?
(a) Relative lowering of vapour pressure is independent of the nature of the solute and the solvent.
(b) The relative lowering of vapour pressure is a colligative property.
(c) Vapour pressure of a solution is lower than the vapour pressure of the sol vent.
(d) The relative lowering of vapour pressure is directly proportional to the original pressure.
Answer
D
Question 43. Solution A contains 7 g/L of MgCl2 and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of
(a) solution A is greater than B
(b) both have same osmotic pressure
(c) solution B is greater than A
(d) cannot be determine
Answer
C
Question 44. X is dissolved in water. Maximum boiling point is observed when X is … (0.1 M each)
(a) CaSO4
(b) BaCl2
(c) NaCl
(d) urea
Answer
B
Question 45. Which one of the following concentration units is independent of temperature ?
(a) Normality
(b) Molarity
(c) Molality
(d) ppm
Answer
C
Question 46. The density (in g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol-1 ) by mass will be
(a) 1.64
(b) 1.88
(c) 1.22
(d) 1.45
Answer
C
Question 47. At certain temperature a 5.12% solution of cane sugar is isotonic with a 0.9% solution of an unknown solute. The molar mass of solute is
(a) 60
(d) 90
(b) 46.17
(e) 92.34
(c) 120
Answer
A
Question 48. Osmotic pressure of 0.4% urea solution is 1.64 atrn and that of 3.42% cane sugar is 2.46 atm. When the above two solutions are mixed, the osmotic pressure of the resulting solution is
(a) 0.82 atrn
(b) 2.46 atm
(c) 1.64 atm
(d) 4.10 atm
Answer
D
Question 49. Which bas minimum osmotic pressure ?
(a) 200 mL of 2 M NaCl solution
(b) 200 mL of 1 M glucose solution
(c) 200 mL of 2 M urea solution
(d) All have same osmotic pressure
Answer
B
Question 50. The vapour pressure will be lowest for
(a) 0. 1 M sugar solution
(b) 0.1 M KCl solution
(c) 0. 1 M Cu(NO3 )2 solution
(d) 0.1 M AgNO3 solution
Answer
C
Question 51. A solution of acetone in ethanol
(a) obeys Raoult’s law
(b) shows a negative deviation from Raoult’s law
(c) shows a positive deviation from Raoult’s law
(d) behaves like a near ideal solution.
Answer
C
Question 52. A solution containing components A and B follows Raoult’s law
(a) A – B attraction force is greater than A – A and B – B
(b) A – B attraction force is less than A – A and B – B
(c) A – B attraction force remains same as A – A and B – B
(d) volume of solution is different from sum of volume of solute and solvent.
Answer
C
Question 53. Which condition is not satisfied by an ideal solution?
(a) DmixV = 0
(b) DmixS = 0
(c) Obeyance to Raoult’s Law
(d) DmixH = 0
Answer
B
Question 54. The freezing point depression constant (Kf) of benzene is 5.12 K kg mol–1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places)
(a) 0.20 K
(b) 0.80 K
(c) 0.40 K
(d) 0.60 K
Answer
C
Question 55. All form ideal solution except
(a) C6H6 and C6H5CH3
(b) C2H6 and C2H5I
(c) C6H5Cl and C6H5Br
(d) C2H5I and C2H5OH
Answer
D
Question 56. An ideal solution is formed when its components
(a) have no volume change on mixing
(b) have no enthalpy change on mixing
(c) have both the above characteristics
(d) have high solubility.
Answer
C
Question 57. At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be
(a) 102 °C
(b) 103 °C
(c) 101 °C
(d) 100 °C
Answer
C
Question 58. 200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10–3 bar. The molar mass of protein will be (R = 0.083 L bar mol–1 K–1)
(a) 51022 g mol–1
(b) 122044 g mol–1
(c) 31011 g mol–1
(d) 61038 g mol–1
Answer
D
Question 59. If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be
(a) halved
(b) tripled
(c) unchanged
(d) doubled.
Answer
C
Question 60. 1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by
(a) 0.2 K
(b) 0.4 K
(c) 0.3 K
(d) 0.5 K
Answer
B
Question 61. A solution of sucrose (molar mass = 342 g mol–1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol–1)
(a) – 0.372 °C
(b) – 0.520 °C
(c) + 0.372 °C
(d) – 0.570 °C
Answer
A
Question 62. During osmosis, flow of water through a semipermeable membrane is
(a) from solution having lower concentration only
(b) from solution having higher concentration only
(c) from both sides of semipermeable membrane with equal flow rates
(d) from both sides of semipermeable membrane with unequal flow rates.
Answer
A
Question 63. A solution of urea (mol. mass 56 g mol–1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol–1 respectively, the above solution will freeze at
(a) 0.654°C
(b) – 0.654°C
(c) 6.54°C
(d) – 6.54°C
Answer
B
Question 64. Pure water can be obtained from sea water by
(a) centrifugation
(b) plasmolysis
(c) reverse osmosis
(d) sedimentation.
Answer
C
Question 65. A solution containing 10 g per dm3 of urea (molecular mass = 60 g mol–1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is
(a) 200 g mol–1
(b) 250 g mol–1
(c) 300 g mol–1
(d) 350 g mol–1
Answer
C
Question 66. From the colligative properties of solution, which one is the best method for the determination of molecular weight of proteins and polymers?
(a) Osmotic pressure
(b) Lowering in vapour pressure
(c) Lowering in freezing point
(d) Elevation in boiling point
Answer
A
Question 67. If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by 0.216°C, than that of the pure solvent. The molecular weight of the substance (Molal elevation constant for the solvent is 2.16°C) is
(a) 10.1
(b) 100
(c) 1.01
(d) 1000
Answer
B
Question 68. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte solid, weighing 2.175 g is added to 39.08 of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of solid substance?
(a) 69.5
(b) 59.6
(c) 49.50
(d) 79.8
Answer
A
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