CBSE Class 12 Chemistry Solutions Question Bank Set 02

Question. Which of the following is dependent on temperature?
(a) Molarity
(b) Mole fraction
(c) Weight percentage
(d) Molality
Answer. A

Question. 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. 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. 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. 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. 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. 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. 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. 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. The concentration unit, independent of temperature, would be
(a) normality
(b) weight volume percent
(c) molality
(d) molarity. 
Answer. C

Question. 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. 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. 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) p_A + xA(p_B – p_A)
(b) p_A + xA ( pA – p_B)
(c) p_B + xA(p_B – p_A)
(d) p_B + xA ( p_A – p_B)
Answer. D

Question. 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. 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. 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. 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. 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. 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. 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
Answer. A

Question. Which one of the following is incorrect for ideal solution?
(a) ΔH_mix = 0
(b) ΔU_mix = 0
(c) ΔP = P_obs – Pcalculated by Raoult’s law = 0
(d) ΔG_mix = 0 
Answer. D

Question. 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. Which condition is not satisfied by an ideal solution?
(a) Δ_mixV = 0
(b) Δ_mixS = 0
(c) Obeyance to Raoult’s Law
(d) Δ_mixH = 0 
Answer. B

Question. 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. 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. All form ideal solution except
(a) C₆H₆ and C₆H₅CH₃
(b) C₂H₆ and C₂H5I
(c) C₆H₅Cl and C₆H₅Br
(d) C₂H₅I and C₂H₅OH
Answer. D

Question. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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 0.5 12 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. Pure water can be obtained from sea water by
(a) centrifugation
(b) plasmolysis
(c) reverse osmosis
(d) sedimentation.
Answer. C

Question. 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. 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

Question. 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. A 5% solution of cane sugar (mol. wt. = 342) is isotonic with 1% solution of a substance X. The molecular weight of X is
(a) 68.4
(b) 171.2
(c) 34.2
(d) 136.8 
Answer. A

Question. The vapour pressure of a solvent decreased by 10 mm 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 the decrease in the vapour pressure is to be 20 mm of mercury?
(a) 0.4
(b) 0.6
(c) 0.8
(d) 0.2
Answer. B

Question. The vapour pressure of CCl₄ at 25°C is 143 mm Hg.If 0.5 g of a non-volatile solute (mol. weight = 65) is dissolved in 100 g CCl₄, the vapour pressure of the solution will be
(a) 199.34 mm Hg
(b) 143.99 mm Hg
(c) 141.43 mm Hg
(d) 94.39 mm Hg. 
Answer. C

Question. The relationship between osmotic pressure at 273 K when 10 g glucose (p₁), 10 g urea (p₂), and 10 g sucrose (p₃) are dissolved in 250 mL of water is
(a) p₂ > p₁ > p₃
(b) p₂ > p₃ > p₁
(c) p₁ > p₂ > p₃
(d) p₃ > p₁ > p₂
Answer. A

Question. According to Raoult’s law, the relative lowering of vapour pressure for a solution is equal to
(a) mole fraction of solute
(b) mole fraction of solvent
(c) moles of solute
(d) moles of solvent.
Answer. A

Question. If 0.1 M solution of glucose and 0.1 M solution of urea are placed on two sides of the semipermeable membrane to equal heights, then it will be correct to say that
(a) there will be no net movement across the membrane
(b) glucose will flow towards glucose solution
(c) urea will flow towards glucose solution
(d) water will flow from urea solution to glucose.
Answer. A

Question. Which one is a colligative property?
(a) Boiling point
(b) Vapour pressure
(c) Osmotic pressure
(d) Freezing point 
Answer. C

Question. Blood cells retain their normal shape in solu-tion which are
(a) hypotonic to blood
(b) isotonic to blood
(c) hypertonic to blood
(d) equinormal to blood.
Answer. B

Question. The relative lowering of the vapour pressure is equal to the ratio between the number of
(a) solute molecules to the solvent molecules
(b) solute molecules to the total molecules in the solution
(c) solvent molecules to the total molecules in the solution
(d) solvent molecules to the total number of ions of the solute.
Answer. B

Question. The van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
(a) 0
(b) 1
(c) 2
(d) 3
Answer. D

Question. The boiling point of 0.2 mol kg^–1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?
(a) Molecular mass of X is less than the molecular mass of Y.
(b) Y is undergoing dissociation in water while X undergoes no change.
(c) X is undergoing dissociation in water.
(d) Molecular mass of X is greater than the molecular mass of Y.
Answer. C

Question. Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
(a) KCl
(b) C₆H₁₂O₆
(c) Al₂(SO₄)₃
(d) K₂SO₄ 
Answer. C

Question. The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively
(a) less than one and greater than one
(b) less than one and less than one
(c) greater than one and less than one
(d) greater than one and greater than one. 
Answer. C

Question. The freezing point depression constant for water is –1.86 °C m^–1. If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by –3.82 °C.Calculate the van’t Hoff factor for Na₂SO₄.
(a) 2.05
(b) 2.63
(c) 3.11
(d) 0.381 
Answer. B

Question. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be
(a) – 0.18 °C
(b) – 0.54 °C
(c) – 0.36 °C
(d) – 0.24 °C
Answer. D

Question. An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
(a) Addition of NaCl
(b) Addition of Na₂SO₄
(c) Addition of 1.00 molal KI
(d) Addition of water 
Answer. D

Question. A 0.0020 m aqueous solution of an ionic compound [Co(NH₃)₅(NO₂)]Cl freezes at –0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = –1.86 °C/m)
(a) 3
(b) 4
(c) 1
(d) 2 
Answer. D

Question. 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol^–1, the lowering in freezing point of the solution is
(a) 0.56 K
(b) 1.12 K
(c) –0.56 K
(d) –1.12 K
Answer. B

Question. Which of the following 0.10 m aqueous solution will have the lowest freezing point?
(a) KI
(b) C₁₂H₂₂O₁₁
(c) Al₂(SO₄)₃
(d) C₅H₁₀O₅ 
Answer. C

Question. Which of the following salts has the same value of van’t Hoff factor (i) as that of K₃[Fe(CN)₆]?
(a) Na₂SO₄
(b) Al(NO₃)₃
(c) Al₂(SO₄)₃
(d) NaCl 
Answer. B

Question. At 25°C, the highest osmotic pressure is exhibited by 0.1 M solution of
(a) glucose
(b) urea
(c) CaCl₂
(d) KCl. 
Answer. C

Question. Which of the following aqueous solution has minimum freezing point?
(a) 0.01 m NaCl
(b) 0.005 m C₂H₅OH
(c) 0.005 m MgI₂
(d) 0.005 m MgSO₄
Answer. A

 

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