CBSE Class 12 Chemistry Solutions MCQs Set H

Question. A 0.5 molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point constant of water be 1.86°C per mole, the mixture shall freeze at
(a) 0.93°C
(b) –0.93°C
(c) 1.86°C
(d) –1.86°C

Answer : B

Question. A solution of urea (mol. mass 56 g mol-1) boils at 100.180C 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. The freezing point of 1% solution of lead nitrate in water will be
(a) 2°C
(b) 1°C
(c) 0°C
(d) below 0°C

Answer : D

Question. Most of the processes in our body occur in
(a) solid solution
(b) liquid solution
(c) gaseous solution
(d) colloidal solution

Answer : B

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. Which is an application of Henry’s law?
(a) Spray paint
(b) Bottled water
(c) Filling up atire
(d) Soft drinks (soda)

Answer : D

Question. A solution containing 1.8 g of a compound (empirical formula CH₂O) in 40 g of water is observed to freeze at –0.465° C.The molecular formula of the compound is (K_f of water = 1.86 kg K mol^–1)
(a) C₂H₄O₂
(b) C₃H₆O₃
(c) C₄H₈O₄
(d) C₆H₁₂O₆ 

Answer : D

Question. Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon __________.
(a) Temperature
(b) Nature of solute
(c) Pressure
(d) Nature of solvent

Answer : C

Question. 12 g of a nonvolatile solute dissolved in 108 g of water produces the relative lowering of vapour pressure of 0.1.The molecular mass of the solute is
(a) 80
(b) 60
(c) 20
(d) 40

Answer : C

Question. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to _________.
(a) low temperature
(b) low atmospheric pressure
(c) high atmospheric pressure
(d) both low temperature and high atmospheric pressure

Answer : B

Question. As a result of osmosis, the volume of more concentrated solution
(a) gradually decreases
(b) gradually increases
(c) is not affected
(d) suddenly increases

Answer : B

Question. Osmotic pressure of a solution at a given temperature
(a) increases with concentration
(b) decreases with concentration
(c) remains same
(d) initially increases and then decreases

Answer : A

Question. Van’t Hoff factor is given by the expression ________.
(a) i= Normalmolar mass/Abnormal molar mass
(b) i= Abnormal molar mass/Normalmolar mass
(c) i= Observed colligativeproperty/Calculated colligativeproperty
(d) Both (a) and (c)

Answer : D

Question. Negative deviation from Raoult’s law is observed in which one of the following binary liquid mixtures?
(a) Ethanol and acetone
(b) Benzene and toluene
(c) Acetone and chloroform
(d) Chloroethane and bromoethane

Answer : A

Question. If α is the degree of dissociation of Na₂SO₄, the Vant Hoff’s factor (i) used for calculating the molecular mass is
(a) 1 + α
(b) 1 – α
(c) 1 + 2α
(d) 1 – 2α

Answer : C

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

Answer : A

Question. Which of the following factor do not affect solubility of solid solute in liquid ?
(a) Temperature
(b) Pressure
(c) Nature of solute
(d) All of these

Answer : B

Question. Which one of the following salts will have the same value of van’t Hoff factor (i) as that of K₄[Fe(CN)₆].
(a) Al₂(SO₄)₃
(b) NaCl
(c) Al(NO₃)₃
(d) Na₂SO₄.

Answer : A

Question. At 25°C, at 5 % aqueous solution of glucose (molecular weight = 180 g mol^–1) is isotonic with a 2% aqueous solution containing an unknown solute. What is the molecular weight of the unknown solute?
(a) 60
(b) 80
(c) 72
(d) 63

Answer : C

Question. Which one of the following statements is false?
(a) Raoult’s law states that the vapour pressure of a component over a binary solution of volatile liquids is directly proportional to its mole fraction
(b) Two sucrose solutions of the same molality prepared in different solvents will have the same depression of freezing point
(c) The correct order of osmotic pressures of 0.01 M solution of each compound is BaCl₂ > KCl > CH₃COOH > glucose
(d) In the equation osmotic pressureπ = MRT, M is the molarity of the solution

Answer : B

Question. Which of the following statements is false?
(a) Units of atmospheric pressure and osmotic pressure are the same.
(b) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region of lower concentration of solute to a region of higher concentration.
(c) The value of molal depression constant depends on nature of solvent.
(d) Relative lowering of vapour pressure, is a dimensionless quantity.

Answer : B

Question. Which of the following concentration unit is independent of temperature ?
(a) Normality
(b) Molarity
(c) Formality
(d) Molality

Answer : D

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 urea solution
(c) urea will flow towards glucose solution
(d) water will flow from urea solution to glucose

Answer : A

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. Vapour pressure of benzene at 30°C is 121.8 mm Hg. When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm Hg. The molecular weight of the solute (Mo. wt. of solvent = 78)
(a) 356.2
(b) 456.8
(c) 530.1
(d) 656.7

Answer : A

Question. We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van’t Hoff factor for these solutions will be in the order ________.
(a) i_A < i_B < i_C
(b) i_A > i_B > i_C
(c) i_A = i_B = i_C
(d) i_A < i_B > i_C 

Answer : C

Question.The value of P° for benzene is 640 mm of Hg. The vapour pressure of solution containing 2.5gm substance in 39gm. benzene is 600mm of Hg the molecular mass of X is
(a) 65.25
(b) 130
(c) 40
(d) 75

Answer : D

Question. The freezing point of equimolal aqueous solutions will be highest for
(a) C₆H₅NH₃ ⁺Cl^– (aniline hydrochloride)
(b) Ca(NO₃)₂
(c) La(NO₃)₃
(d) C₆H₁₂O₆ (glucose)

Answer : D

Question. Which one of the following is non-ideal solution
(a) Benzene + toluene
(b) n-hexane + n-heptane
(c) Ethyl bromide + ethyl iodide
(d) CCl₄ + CHCl₃

Answer : D

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

Answer : C

Question. The correct relationship between the boiling points of very dilute solutions of AlCl₃ (t₁) and CaCl₂ (t₂), having the same molar concentration is
(a) t₁ = t₂
(b) t₁ > t₂
(c) t₂ > t₁
(d) t₂ ≥ t₁

Answer : B

Question. The vapour pressure of water at 20°C is 17.5 mm Hg. 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) 17.325 mm Hg
(b) 15.750 mm Hg
(c) 16.500 mm Hg
(d) 17.500 mm Hg

Answer : A

Question. The vapour pressure of two liquids X and Y are 80 and 60 torr respectively. The total vapour pressure of the ideal solution obtained by mixing 3 moles of X and 2 moles of Y would be
(a) 68 Torr
(b) 140 Torr
(c) 48 Torr
(d) 72 Torr

Answer : D

Question. The rise in the boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is 0.1°C. The molal elevation constant of the liquid is
(a) 0.01 K/m
(b) 0.1 K/m
(c) 1 K/m
(d) 10 K/m

Answer : C

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 depression constant, Kf of benzene is 5.12 K kg mol^–1, the freezing point of benzene will be lowered by
(a) 0.3 K
(b) 0.5 K
(c) 0.4 K
(d) 0.2

Answer : C