1.Only one pair of liquid in the following set does not obey Raoult’s law, identify it:
(a) Methanol and Ethanol
(b) Benzene and toluene
(c) n-Hexane and n-heptane
(d) Ethanol and Acetone
Explanation (Answer; d)
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. Some examples of ideal solutions are methanol and ethanol, benzene, and toluene, n-hexane and n-heptane, bromoethane and chloroethane etc.
Methanol and ethanol are polar in nature with similar intermolecular forces, hence they form an ideal solution.
Both benzene and toluene and n-hexane and n-heptane are non-polar in nature with similar intermolecular forces. Hence, they form an ideal solution.
Ethanol and Acetone
When Acetone and ethanol are mixed, the weakening of hydrogen bonds in ethanol takes place. Hence, an ideal solution is not formed.
2. Identify the incorrect statement about colloidal solution:
(a) It shows Tyndall effect
(b) Its particles movement is Brownian type
(c) Its particle size is less than 1 nm
(d) Its physical appearance is translucent.
Explanation (Answer; c)
In colloidal solution, the particles is in between 1 nm to 1000 nm.
4. Which is not a colligative property?
(a) Lowering in vapours pressure
(b) Elevation in boiling point
(c) Depression in freezing point
(d) Atmospheric pressure
Explanation (Answer; d)
Atmospheric pressure is a colligative property.
4. Effect of pressure change play significant role in the solubility of:
(a) Solid into liquid
(b) Liquid into liquid
(c) Gas into liquid
(d) All of them
Explanation (Answer; c)
The solubility of gas into liquid is significantly changed by the change in pressure obeying Henry’s law accordingly the solubility of gas in liquid solvent is directly proportional to the pressure.
5. According to Raoult law the relative lowering of vapour pressure is equal to:
(a) Molality
(b) Mole fraction of solute
(c) Mole fraction of solvent
(d) Molarity
Explanation (Answer; b)
According to the third form of Raoult’s law, the relative lowering of vapour pressure is equal to the mole fraction of solute (X2).
∆P/Po = X2---------------- [Here ∆P/Po is referred as relative lowering in vapours pressure]
6. The sum of mole fractions of components of a solution is equal to:
(a) 0.0
(b) 1.0
(c) 10
(d) 100
Explanation (Answer; b)
The sum of mole fractions of components of a solution is equal to 1.0.
7. How many mole of NaOH are present in 2dm3 of 1 molar aqueous solution of it.
(a) 0.5 mole
(b) 1 mole
(c) 1.5 mole
(d) 2 mole
Explanation (Answer; d)
1 molar aqueous solution contains 1 mole of solute per 1 dm3. Hence 1 molar aqueous solution contains 2 mole per 2 dm3.
8. A colloidal solution of liquid into liquid is known as:
(a) Gel
(b) Foam
(c) Sol
(d) Emulsion
Explanation (Answer; d)
A colloidal solution of liquid into liquid is known as Emulsion
9. An example of completely immiscible liquid pair is:
(a) Benzene to toluene
(b) Water and phenol
(c) Water and Benzene
(d) Water & methanol
Explanation (Answer; b)
Water and phenol is an example of completely immiscible liquid pair
10. A 15% W/W KOH solution can be prepared by mixing 15g KOH in:
(a) 15g water
(b) 85g water
(c) 100g water
(d) 115g water
Explanation (Answer; b)
The numerical value of percent concentration indicates amount of solute. The amount of solvent can be calculated by subtracting amount of solute from 100 (only in case of W/W % or V/V%). Hence 15% W/W KOH solution can be prepared by mixing 15g KOH in 85 g (100 – 15) water.
11. Particles size of solute and solvent in solution (atoms, molecules or ions) are extremely small having a diameter of approximately less than.
(a) 1 nm
(b) 10 nm
(c) 1000 nm
(d) 0.1 nm
11. Explanation (Answer; b)
The particles size in solution is less than 0.1-1nm.
12. Tyndall effect is not shown by
(a) Solution
(b) suspension
(c) Colloids
(d) All of them
12. Explanation (Answer; a)
Solution does show exhibit Tyndall effect due to small size of their particles.
13.When a solution is formed, attraction between solute-solvent molecules becomes ………… than solute-solute and solvent-solvent molecules.
(a) equal or greater
(b) lesser or equal
(c) lesser
(d) All of them
13. Explanation (Answer; a)
A solution is only formed when attraction between solute-solvent molecules becomes equal or greater than solute-solute and solvent-solvent molecules.
14. Generally, dissolution process is
(a) endothermic
(b) exothermic
(c) both of them
(d) None of them
14. Explanation (Answer; c)
Generally, dissolution process is either exothermic or endothermic depending the values of two types of energy changes i.e. lattice energy and solvation energy.
Lattice energy is the energy required to separate solute particles from solid crystal into solution. Its value is positive for bond breaking. solvation energy which released when interaction of solute and solvent particles takes place. The Ions with greater charge and smaller size have high solvation energy hence their dissolution in exothermic. On the other hand many solid substances have strong particle binding in their crystal lattice. They must absorb energy from the solution and make the process endothermic.
15.Which of the following energies involved in dissolution process?
(a) Lattice energy
(b) solvation energy
(c) both of them
(d) None of them
15. Explanation (Answer; c)
Generally, dissolution process is either exothermic or endothermic depending the values of two types of energy changes i.e. lattice energy and solvation energy.
16.Lattice energy is always:
(a) Positive
(b) Negative
(c) both of them
(d) None of them
16. Explanation (Answer; b)
Lattice energy is the energy required to separate solute particles from solid crystal into solution. Its value is positive for bond breaking.
17. Solvation energy is always:
(a) Positive
(b) Negative
(c) both of them
(d) None of them
17. Explanation (Answer; b)
solvation energy which released when interaction of solute and solvent particles takes place.
18.The difference lattice energy and solvation energy is known as
(a) heat of solution
(b) Heat of hydration
(c) heat of atomization
(d) None of them
18. Explanation (Answer; a)
The difference lattice energy and solvation energy is known as heat of solution.
19. The change in enthalpy when a substance dissolves in solvent at constant pressure is called
(a) Heat of solution
(b) Heat of hydration
(c) Heat of atomization
(d) None of them
19. Explanation (Answer; a)
heat of solution is the change in enthalpy when a substance dissolves in solvent at constant pressure
20. The solute substances whose particles (ions or molecules) are strongly associated with water are termed as …………. molecules.
(a) Hydrophilic
(b) Hydrophobic
(c) Amphoteric
(d) None of them
20. Explanation (Answer; a)
The solute substances whose particles (ions or molecules) are strongly associated with water are termed as hydrophilic molecules.
21.The hydrophilic molecules are …………….. in nature.
(a) Polar
(b) Non-polar
(c) Both of them
(d) None of them
21. Explanation (Answer; a)
The hydrophilic molecules are polar in nature.
22.Which one of the following is hydrophilic molecules?
(a) ethanol
(b) amino acids
(c) acetone
(d) All of them
22. Explanation (Answer; d)
All the given molecules being polar are hydrophilic molecules.
23.The hydrophilic molecules dissolves in water due to formation of
(a) hydrogen bond
(b) Dipole-dipole force
(c) London forces
(d) All of them
23. Explanation (Answer; a)
The hydrophilic molecules dissolves in water due to formation of hydrogen bond.
24.The solute substances whose particles (ions or molecules) do not mix into water and do not form aqueous solution due to their non-polar nature are termed as
(a) hydrophilic
(b) hydrophobic
(c) Amphoteric
(d) None of them
24. Explanation (Answer; b)
The solute substances whose particles (ions or molecules) do not mix into water and do not form aqueous solution due to their non-polar nature are termed as hydrophobic.
25.The hydrophobic molecules are …………….. in nature.
(a) Polar
(b) Non-polar
(c) Both of them
(d) None of them
25. Explanation (Answer; b)
The hydrophobic molecules are Non-polar in nature.
26.Which one of the following is hydrophobic molecules?
(a) petrol
(b) benzene
(c) toluene
(d) All of them
26. Explanation (Answer; d)
All the given molecules being non-polar are hydrophobic molecules.
27.Particles size of the components of suspension are above
(a) 1 nm
(b) 10 nm
(c) 103nm
(d) 0.1 nm
27. Explanation (Answer; c)
Particles size of the components of suspension are above 1000 or 103nm.
28. The translucent heterogeneous mixture of the particles of 1nm to 1000 nm size whose dispersed particles may be seen microscopically dispersed into a dispersion medium which cannot be settled down but coagulate on heating or adding an electrolyte is called
(a) Solution
(b) suspension
(c) Colloids
(d) All of them
28. Explanation (Answer; c)
Colloids or collodial solution is the translucent heterogeneous mixture of the particles of 1nm to 1000 nm size whose dispersed particles may be seen microscopically dispersed into a dispersion medium which cannot be settled down but coagulate on heating or adding an electrolyte.
29. The size range of colloidal particles is from.
(a) 1 to 1000 nm
(b) 100 to 1000 nm
(c) 1000 to 1200 nm
(d) None of them
29. Explanation (Answer; a)
The size range of colloidal particles is from 1 to 1000 nm.
30. A colloidal solution of gas into liquid or solid is known as:
(a) Gel
(b) Foam
(c) Aerosol
(d) Emulsion
30. Explanation (Answer; b)
Foam is a solution of a gas in a liquid. The substance being dispersed would be the gas, triggering the fluid to become frothy and foamy. A sample of this would be shaving cream.
31.A colloidal solution of liquid into gas is known as:
(a) Gel
(b) Foam
(c) Aerosol
(d) Emulsion
31. Explanation (Answer; b)
A colloidal solution gas or air as the dispersion medium into liquid or solid is known as foam
32. A colloidal solution of liquid into solid is known as:
(a) Gel
(b) Foam
(c) Aerosol
(d) Emulsion
32. Explanation (Answer; a)
A colloidal solution of liquid into solid is known as Gel.
33. Fog and spray are the examples of
(a) Gel
(b) Foam
(c) Aerosol
(d) Emulsion
33. Explanation (Answer; c)
Fog, mist, Clouds, sprays, steam etc. are the examples aerosol or liquid-gas colloidal solution
34. Smoke is the examples of
(a) Gel
(b) Foam
(c) Aerosol
(d) Emulsion
34. Explanation (Answer; c)
Smoke, ice cloud, dust, automobile exhaust, atmospheric particulate matter etc are examples of aerosol or (Solid) Aerosol which are solid-gas colloidal solutions.
35. Shaving cream is the examples of
(a) Gel
(b) Foam
(c) Aerosol
(d) Emulsion
35. Explanation (Answer; b)
Shaving cream, Froth, Whipped Cream, soaps suds etc. are examples of foam which are gas-liquid colloidal solutions.
36. Paint is the example of
(a) Gel
(b) Sols
(c) Aerosol
(d) Emulsion
36. Explanation (Answer; b)
Paints, Gum, Cell Fluids, pigmented ink, sediments, sewage etc. are examples of sols which are solid-liquid colloidal solutions.
37. Milk is the example of
(a) Gel
(b) Foam
(c) Aerosol
(d) Emulsion
37. Explanation (Answer; d)
Milk, oil in water, mayonnaise, hand cream, latex etc. are examples of emulsion which are liquid-liquid colloidal solution.
38. The process of precipitation of colloidal particles which can be accomplished by heating or by adding electrolyte is known as
(a) Coagulation
(b) Tyndall effect
(c) Electrophoresis
(d) None of them
38. Explanation (Answer; )
On heating, colloidal particles strike to each other many times with high energy and aggregate into each other to form large particle and thus settle out.
Dispersed particles of colloidal solution possesses either positive or negative charge. However the charge of all dispersed particles remain the same that is why they repel each other and keep them self suspended in the dispersion medium instead of combining of form a large molecule. Electrophoresis the movement of colloidal particles towards particular electrode in an applied electric field.
39. The dissolution of these solids in liquids solvent is exothermic:
(a) AlCl3
(b) Na2SO4
(c) lipids
(d) All of them
39. Explanation (Answer; d)
All the given solids dissolution is exothermic.
40. The dissolution of gases in liquids solvent is:
(a) exothermic
(b) endothermic
(c) both of them
(d) None of them
40. Explanation (Answer; a)
The dissolution of gases in liquids solvent is exothermic. Hence solubility of gases decreases with increasing temperature.
41. Solubility of those substance whose dissolution is exothermic ……………..with the rise of temperature.
(a) decreases
(b) increases
(c) both of them
(d) None of them
41. Explanation (Answer; a)
The dissolution of some solids (AlCl3, Na2SO4 etc.) and many as well as gases in liquid solvent is exothermic and releases heat.
42. The solubility of gases …………. with the increasing pressure.
(a) decreases
(b) increases
(c) both of them
(d) None of them
42. Explanation (Answer; b)
The solubility of gases increases with the increasing pressure. This is known as Henry’s law.
43. The solubility of gases increases with the increasing pressure. This is known as
(a) Raoult’s law
(b) Henry’s law
(c) Le-Chatelier’s principle
(d) None of them
43. Explanation (Answer; b)
The solubility of gases increases with the increasing pressure. This is known as Henry’s law.
44. An alloy is type of…….. solution:
(a) Solid-gas
(b) Solid-solid
(c) Liquid-liquid
(d) Gas-gas
44. Explanation (Answer; b)
Alloys are solid-solid solution.
45. In which mode of expression does the concentration of solution remains independent of temperature?
(a) % concentration
(b) Molarity
(c) Molality
(d) Normality
45. Explanation (Answer; c)
Molality is independent of temperature.
46. Isotonic solutions must have the same:
(a) Density
(b) Normality
(c) Volume
(d) osmotic pressure
46. Explanation (Answer; d)
Isotonic solutions must have same osmotic pressure at a given temperature hence must have same volume and number of moles i.e., same molar concentration. Thus, the isotonic solutions have same elevation in boiling point, and depression in freezing point.
47. A solution of known strength or concentration is called:
(a) Standard solution
(b) Normal solution
(c) Molal solution
(d) Suspension
47. Explanation (Answer; a)
Standard solution is a solution of known strength or concentration.
48. The number of gram equivalents of solute dissolved per dm3 of solution is called:
(a) Mole fraction
(b) Normality
(c) Molality
(d) % concentration
48. Explanation (Answer; b)
Normality is the number of gram equivalents of solute dissolved per dm3 of solution.
49. The number of moles of solute dissolved per kilogram of solvent is called:
(a) Mole fraction
(b) Normality
(c) Molality
(d) % concentration
49. Explanation (Answer; c)
Molality is the number of moles of solute dissolved per kilogram of solvent.
50. The number of moles of solute dissolved in 1 liter of solution is called:
(a)molarity
(b) Normality
(c) Molality
(d) % concentration
50. Explanation (Answer; a)
molarity is the number of moles of solute dissolved in 1 liter of solution
51. A solution which contains 1 mole of solute dissolved in 1000 cm3 of water is designated by:
(a) 1M
(b) 1m
(c) 1N
(d) 10%
51. Explanation (Answer; a)
1M or molar solution is a solution which contains 1 mole (molar mass) of solute dissolved in 1000 cm3 or 1 dm3 of water
52. A solution which contains 1 mole of NaCl dissolved in 1 kg of water is designated by:
(a) 1M NaCl
(b) 1m NaCl
(c) 1N NaCl
(d) 10% NaCl
52. Explanation (Answer; b)
1m or molal solution is a solution which contains 1 mole of solute dissolved in 1 kg (or 1000 g) of water.
53. 1N H2SO4 solution contains:
(a) 98 g/dm3
(b) 9.8 g/dm3
(c) 980 g/dm3
(d) 49 g/dm3
53. Explanation (Answer; d)
1N H2SO4 solution contains its 1 gram equivalent which is equal to 49 g/dm3 (gram equivalent of acid = molar mass/basicity).
54. 1N NaOH solution contains:
(a)4 g/dm3
(b) 20 g/dm3
(c) 80 g/dm3
(d) 40 g/dm3
54. Explanation (Answer; d)
1N NaOH solution contains its 1 gram equivalent which is equal to 40 g/dm3 (gram equivalent of base = molar mass/acidity).
55. The decimolar solution of H2SO4 solution contains:
(a) 98 g/dm3
(b) 9.8 g/dm3
(c) 980 g/dm3
(d) 4.9 g/dm3
55. Explanation (Answer; b)
1/10 solution is decimolar solution which contains deci mole of solute per dm3 of solution. 1 mole of H2SO4 is equal to 98 g/mol, so its deci mole will be equal to 9.8 g. Hence The decimolar solution of H2SO4 solution contains 9.8 g/dm3
56. 56 g of KOH is dissolved in 1 liter of water. The solution obtained is said to be:
(a) Molar
(b) Semimolar
(c) Decimolar
(d) decamolar
56. Explanation (Answer; a)
56 g of KOH constitute its 1 mole. A solution which contains 1 mole (molar mass) of solute dissolved in 1000 cm3 or 1 dm3 of water is called 1M or molar solution.
57. The semimolar solution of KOH contains:
(a) 56 g/dm3
(b) 5.6 g/dm3
(c) 28 g/dm3
(d) 112 g/dm3
57. Explanation (Answer; c)
½ molar solution is semimolar or hemimolar solution which contains hemi or half mole of solute per dm3 of solution. 1 mole of KOH4 is equal to 56 g/mol, so its half mole will be equal to 28 g. Hence The semimolar solution of KOH solution contains 28 g/dm3
58 5 liters of 0.5M HNO3 solution contains:
(a) 2.5 moles
(b) 3moles
(c) 3.5 moles
(d) 4 moles
58. Explanation (Answer; a)
No of moles = M x Vdm3 = 0.5 x 5 = 2.5 moles
59. 2 liters of 0.25M NaCl solution contains:
(a) 29.25g
(b) 58.5g
(c) 5.85g
(d) 585g
59. Explanation (Answer; a)
No of moles = M x Vdm3 = 0.25 x 2 = 0.5 moles
Mass = n x M = 0.5 x 58.5 = 29.25 g
60. How many gram of Na2CO3 is required to prepare its 0.5M solution of 250ml?
(a) 132.5 g
(b) 0.1325 g
(c) 1.325 g
(d) 13.25 g
60. Explanation (Answer; d)
Mass = M x Molar mass x Vdm3 = 0.5 x 106 x (250/1000) = 13.25 g
61. The molarity of solution containing 45g of H2C2O4 in 250ml of solution is:
(a) 0.25 M
(b) 2.5 M
(c) 0.025 M
(d) 2 M
61. Explanation (Answer; d)
Molar mass of = 90 g/mol
V in liter = 250/1000 = 0.25 liter
Molarity = (mass/M) x (1/Vdm3) = (45/90) x (1/0.25) = 2M
62. A 10M solution stands for:
(a) Normal solution
(b) Decimolar solution
(c) Decamolar solution
(d) Molar solution
62. Explanation (Answer; c)
The prefix deca stands for 10.
63.A N/10 solution stands for:
(a) Normal solution
(b) Decinormal solution
(c) Seminormal solution
(d) Molar solution
63. Explanation (Answer; b)
The prefix desi stands for one-tenth (1/10).
64. A N/2 solution stands for:
(a) Normal solution
(b) Decinormal solution
(c) Seminormal solution
(d) Molar solution
64. Explanation (Answer; c)
The prefix semi stands for half (½).
A N/2 solution stands for seminormal solution
65. The molarity of solution that contains 10g of urea in 500 ml of solution is:
(a) 0.33 M
(b) 3.34 M
(c) 0.033 M
(d) 2.5 M
65. Explanation (Answer; a)
M = (mass/molar mass) x (1000/Vml) = (10/60) x (1000/500) = (0.166) x (2) = 0.33 M
66. 1m Na2CO3 solution contains ………… g of in 1000g of water.
(a) 1.06 g
(b) 10.6 g
(c) 106 g
(d) 53 g
66. Explanation (Answer; c)
A molal (1m) solution contains 1 mole of solute in 1kg or 1000 g of solvent. The mass of 1 mole of Na2CO3 is 106 g.
67. How many gram of NaCl are to be taken to prepare 100 ml of 10% salt solution?
(a) 4 g
(b) 40 g
(c) 80 g
(d) 10 g
67. Explanation (Answer; d)
10% solution contains 10 g of solute per 100 ml of soliton.
Mass of solute = % concentration x volume of solution/ 100 = 10 x 100/100 = 10 g
68. How many gram of HCl are in 1000 g of a 10 % solution?
(a) 100 g
(b) 10 g
(c) 9.8 g
(d) 98 g
68. Explanation (Answer; a)
10% solution contains 10 g of solute per 100 ml of soliton.
1000 g solution contains 100 g of solute.
Mass of solute = % concentration x mass of solution/ 100 = 10 x 1000/100 = 100 g
69. The sum of the mole fractions of solute and solvent is equal to:
(a) 0
(b) 1
(c) 2
(d) 3
69. Explanation (Answer; b)
The sum of the mole fractions of solute and solvent is equal to 1.
70. The suspended particles in suspensions are generally of the size:
(a) 10 nm
(b) 100 nm
(c) 1200 nm
(d) 1 nm
70. Explanation (Answer; c)
The suspended particles in suspensions are generally of the size greater than 1000 nm.
71. A mixture having constituent particles of size of 1 nm is called
(a) Suspension
(b) Colloids
(c) Colloidal suspension
(d) Solution
71. Explanation (Answer; d)
A mixture having constituent particles of size of 1 nm is called solution.
72. The size of dispersed particles in colloids is:
(a) 1nm
(b) 10nm
(c) 1000nm
(d) 2-1000nm
72. Explanation (Answer; d)
The size of dispersed particles in colloids is 2-1000nm.
73. The intermediate type of mixture is:
(a) Solution
(b) Suspension
(c) Colloids
(d) None of them
73. Explanation (Answer; c)
Colloids are the intermediate type of mixture.
74. Which of the following is a colligative property?
(a) Sublimation temperature
(b) Freezing point
(c) Osmatic pressure
(d) Melting point
74. Explanation (Answer; c)
Osmotic pressure is colligative property as it depends upon the concentration of solute molecules or ions, but not upon the identity of the solute.
The properties that depend upon the ratio of the number of solute molecules and total molecules not upon the nature of solute molecules named as colligative properties.
Example- Osmotic pressure, elevation of boiling point, depression in freezing point and relative lowering of vapour pressure.
75. Saturated solution of NaCl on heating becomes
(a) Super-saturated
(b) Unsaturated
(c) Remains saturated
(d) None of them
75. Explanation (Answer; b)
Saturated solution of NaCl on heating becomes unsaturated as more solute or NaCl can be dissolved.
76. The movement of solvent molecules through a semipermeable membrane is called
(a) Electrolysis
(b) Cataphoresis
(c) Osmosis
(d) Electrophoresis
76. Explanation (Answer; c)
Osmosis is the movement of solvent molecules through a semipermeable membrane.
Osmosis is the spontaneous net movement or diffusion of solvent molecules through a selectively-permeable membrane separating two solutions of different concentrations from a region of high water potential (region of lower solute concentration) to a region of low water potential (region of higher solute concentration), in the direction that tends to equalize the solute concentrations on the two sides
Osmotic pressure is defined as the external pressure required to be applied so that there is no net movement of solvent across the membrane. Osmotic pressure is a colligative property, meaning that the osmotic pressure depends on the molar concentration of the solute but not on its identity.
77. The concentration units independent of temperature would be
(a) Molarity
(b) Normality
(c) Percent concentration
(d) Molality
77. Explanation (Answer; d)
Molality and mole fraction are two concentration units that are independent of temperature.
78. Isotonic solutions have same
(a) Molar concentration
(b) Molality
(c) Normality
(d) None of them
78. Explanation (Answer; a)
Isotonic solutions have same molar concentration and hence have same osmotic pressure.
79. If 5.85 g of NaCl is dissolved in 90 g of water, the mole fraction of NaCl is
(a) 0.1
(b) 0.01
(c) 0.2
(d) 0.0196
79. Explanation (Answer; d)
No. of moles of NaCl = mass/molar mass = 5.85/58.5 = 0.1 mol
No. of moles of water = mass/molar mass = 90/18 = 5 mol
Mole fraction of NaCl = nNaCl/nT = 0.1/ (0.1+5) = 0.1/5.1 = 0.0196
80. The molarity of pure water is
(a) 50
(b) 100
(c) 18
(d) 55.5
80. Explanation (Answer; d)
Pure water does not have a solution in it, it is just water. So, let’s take 1 liter of water. We know waters density, which is 1 gram per ml. So, our liter has 1,000 grams of mass. Thus density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02 = 0.055 mol/g). Therefore, the molar concentration of water is = 1000 g/L / (18.02 g/mol) = 55.5 mol/L.
81. If 18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose, it is said to be
(a) Molal
(b) Decimolal
(c) Molar
(d) Semimolal
81. Explanation (Answer; b)
18 g of glucose constitutes its decimole (0.1 mol). 1000 g corresponds to 1kg. A solution containing decimole dissolved per kg is called Semimolal solution.
82.How many gram of a dibasic acid (molecular weight = 200) should be present in 100 ml of the aqueous solution to give 0.1 normality?
(a) 1 g
(b) 2 g
(c) 10 g
(d) 20 g
82. Explanation (Answer; a)
Equivalent weight of dibasic acid = molecular weight/basicity of acid = 200/2 =100
Normality of solution (strength) = 0.1 N
Volume of solution = 100 ml = 100/1000 = 0.1 liter
Mass of dibasic acid (m) = ?
Normality = Mass of solute in gram/(Equivalent weight x Volume of solution in liter)
Mass of solute in gram = Normality x Equivalent weight x Volume of solution in liter
= 0.1 x 100 x0.1 = 1 g
83. Solutions with same osmotic pressures are called
(a) Hypertonic
(b) Hypotonic
(c) Isotonic
(d) Normal
83. Explanation (Answer; c)
Isotonic solutions have same molar concentration and hence have same osmotic pressure.
84. How much of NaOH is required to neutralize 1500 cm3 of 0.1 N HCl?
(a) 60 g
(b) 40 g
(c) 6 g
(d) 4 g
84. Explanation (Answer; c)
Normality = Mass of solute in gram/(Equivalent weight x Volume of solution in liter)
Mass of solute in gram = Normality x Equivalent weight x Volume of solution in liter = 0.1 x 40 x 1.5 = 6 g
85. The solubility of a gas in water depends upon
(a) Nature of the gas
(b) Pressure of the gas
(c) Temperature
(d) All of them
85. Explanation (Answer; d)
The solubility of a gas in water depends upon nature of gas, pressure of the gas (direct relation) and temperature (inverse relation).
86. The volume of water which should be added to 300 ml of 0.5 M NaOH solution so as to get a solution of 0.2 M is
(a)550 ml
(b) 350 ml
(c)450 ml
(d) 750 ml
86. Explanation (Answer; d)
M1V1 = M2V2
300 x 0.5 = 0.2 x V2
V2 = 750 ml
87. 300 ml of 0.1 M HCl and 200 ml of 0.3 M H2SO4 are mixed. The normality of the resulting mixture is:
(a)0.4 N
(b) 0.1 N
(c)0.3 N
(d) 0.084 N
87. Explanation (Answer; d)
For HCl solution;
V1 = 300 ml
N1 = M x Basicity = 0.1 x 1 = 0.1
For H2SO4 solution;
V2 = 200 ml
N2 = M x Basicity = 0.3 x 2 = 0.6
Normality of the mixture, N = V1N1+V2N2/V1+V2 = 300 x 0 .1 x 200 x 0. 6/500 = 30+12/500 = 0.084
88. What is the molarity of H2SO4 solution that has the density of 1.84 g/cc at 35ºC and contains 98% by weight?
(a) 4.18 M
(b) 8.14 M
(c) 18.4 M
(d) 18 M
88. Explanation (Answer; c)
As the given H2SO4 is 98% by weight, hence 100g solution contains 98g H2SO4 by mass. now
d=m/V and V = m/d
Here
d = 1.84 g/c.c., m =100 g, v=?
Volume = 100/1.84 = 54.34 mL
Molarity = (Mass/molar mass) x (1000/Volume in mL)
here
weight taken m = 98 g,
molecular weight of H2SO4 (M) = 98 g/mol,
volume (V) = 54.34 ml
putting these values we get
Molarity = (Mass/molar mass) x (1000/Volume in mL) = (98/98) x (1000/54.34) = 18.4 M
89. The mole fraction of a solution containing 3.0 g of urea per 250 g of water would be
(a)0.00357
(b) 0.00643
(c)0.99643
(d) None of these
89. Explanation (Answer; a)
No. of moles of urea; (NH2)2CO = mass/molar mass = 3.0/60 = 0.05 mol
No. of moles of water = mass/molar mass = 250/18 = 13.89 mol
Mole fraction of urea= nurea/nT = 0.05/ (0.05+13.89) = 0.05/13.94 = 0.00358
90. Five liters of 0.5 M HNO3 solution contains
(a) 2.5 moles
(b) 3.0 moles
(c) 3.5 moles
(d) 4.0 moles
90. Explanation (Answer; a)
No. of moles (n) = M x V = 0.5 x 5 = 2.5 mol
91. How many gram of NaCl are to be taken to prepare 50 ml of 5% w/v salt solution?
(a) 2.5 g
(b) 10 g
(c) 25 g
(d) 80 g
91. Explanation (Answer; a)
% concentration = (mass or volume of solute)/(mass or volume of solution) x 100
Mass or volume of solute = % concentration x mass or volume of solution/100
Mass of NaCl = 5 x 50/100 = 2.5 g
92. A solution of glucose is 10%. The volume in which 1 gram-mole of it is dissolved will be
(a) 1 dm3
(b) 200 cm3
(c) 900 cm3
(d) 1.8 dm3
92. Explanation (Answer; d)
strength of solution = 10% or 100 gL−1
Molar mass of glucose =180 g/mol
10% glucose means 10 g
= 10/180 moles in 100 c.c i.e. 0.1 litre
Hence
1 mole of glucose will present in
0.1 x 180/10 = 1.8 L
93. If it took 10 ml of 1 M HCl to titrate 20 ml of NaOH solution of unknown concentration to its end point, what was the concentration of NaOH?
(a) 0.5 M
(b) 1.5 M
(c) 2.0 M
(d) 2.5 M
93. Explanation (Answer; d)
M1V1 = M2V2
1 x 10 = M2x 20
M2 = 0.5 M
94. There are ………… possible relationship is percentage composition.
(a) 4
(b) 5
(c) 6
(d) 8
94. Explanation (Answer; a)
There are 4 possible relationship is percentage composition i.e.
m/m %, m/V%, V/m% and V/V%
95. A solution which contains one gram equivalent of a solute per liter is called
(a) One molar
(b) One malal
(c) One normal
(d) Semimolar
95. Explanation (Answer; d)
One normal or 1N solution contains one gram equivalent of a solute per liter.
96. As compared to molar solution, in the molal solution the quantity of solvent is
(a) Comparatively lesser
(b) Comparatively greater
(c) More or less equal
(d) Very large
96. Explanation (Answer; b)
1 molar aqueous solution is more concentrated than 1 molal aqueous solution because 1 molar solution contain 1 mole of solute in 1 litre of the solution which include both solute and solvent. So, the mass of solvent (i.e. water) is less than 1000 gram. Therefore 1 molar aqueous solution contains 1 mole of solute in less than 1000 gram of solvent whereas 1 molal solution has 1 mole of solute in 1000 gram of solvent. Hence concentration will be more in 1 molar aqueous solution.
suppose for example 1M KCl = 74.5 g
volume = 1 L
mass of KCl in 1L = 74.5 g
mass of water in 1 liter of total volume solution = 1000g − 74.5g =0 .925g = 0.925 kg
mass of solvent = 0.925 kg
molality = 1/.925 =1.08
thus in molality, weight of the solvent is less than total volume of the solution in which both are in denominator because weight of the solute has to be subtracted to get the molality.
97. A solution of sucrose is 34.2%. The volume of solution containing one mole of solute is
(a) 342 cm3
(b) 1000 cm3
(c) 500 cm3
(d) 242 cm3
97. Explanation (Answer; b)
34.2% sucrose solution = 34.2 g sucrose in 100 cm3 of solution
34.2 g sucrose is present in 100 cm3 of solution
1 g sucrose is present in 100/34.2 of solution
342 g (1mole) sucrose is present in (100/34.2) x 342 = 1000 cm3
98. 10 g of NaOH has been dissolved per kg of solvent, the molality of solution is
(a) 0.25 m
(b) 1.5 m
(c) 0.5 m
(d) 2.5 m
98. Explanation (Answer; a)
Molality = mass/molar mass x kg of solvent = 10/40 x1 = 0.25 m
99. The sum of mole percent of all the components of solution is always equal to
(a) Less than 100
(b) One
(c) 100
(d) Ten
99. Explanation (Answer; c)
The sum of mole percent of all the components of solution is always equal to 100.
100. 5 g of glucose is dissolved for 100 cm3 of solution. The percentage of solution is
(a) 5% v/w
(b) 5% v/v
(c) 5% w/v
(d) 5% w/w
100. Explanation (Answer; c)
A solution containing 5 g of solute per 100 cm3 of solution will have percentage concentration 5%w/v.
101. The molarity of 2% w/v NaOH solution is
(a) 0.25
(b) 0.1
(c) 0.05
(d) 0.5
101. Explanation (Answer; d)
The weight by volume percentage of NaOH solution = 2%
2% w/v NaOH solution means that 2g of NaOH is dissolved in 100ml of solution.
Mass of NaOH = 2g
Volume of NaOH solution = 100ml = 100/1000 = 0.1L
Molar mass of NaOH = 40 g/mol
Molarity = (mass/M x Vliter) = (2/40x 0.1) = 0.5 M
102. The amount of solute present in a given amount of solvent or solution is called:
(a) Concentration
(b) Molarity
(c) Molality
(d) Normality
102. Explanation (Answer; a)
This is the standard definition of concentration.
103. Alloys are the examples of:
(a) Compounds
(b) Elements
(c) Solutions
(d) Colloids
103. Explanation (Answer; c)
Alloys are the examples of solid-solid solutions.
104. What is the molarity of a 15 mL, 2M aqueous solution when 285 mL water is added to it?
(a) 0.400 M
(b) 0.105 M
(c) 0.111M
(d) 0.100 M
104. Explanation (Answer; b)
M1V1 (before mixing) = M2V2 (after mixing)
15 x 2 = M2 x 285
M2 = 0.150 M
105. The semimolar solution of NaOH contains:
(a) 4 g/dm3
(b) 20 g/dm3
(c) 40 g/dm3
(d) 0.4 g/dm3
105. Explanation (Answer; b)
Semi stands for half (0.5).
The semimolar solution of NaOH contains half mole or 20 g/dm3
106. When 19.6 g of sulphuric acid are dissolved in 500 ml of water, the resulting solution is
(a) 0.16 M
(b) 0.40 M
(c) 0.48 M
(d) 0.65 M
106. Explanation (Answer; b)
M = mass/(molar mass x Vliter) = 19.6/(98x 0.5) = 0.40 M
107. The unit of molality is
(a) mol dm-3
(b) mol kg-1
(c) g dm-3
(d) None
107. Explanation (Answer; b)
The unit of molality is mol kg-1
108. The pH of 10-3 moldm-3 of an aqueous solution of H2SO4 is
(a) 3.0
(b) 2.7
(c) 2.0
(d) 1.5
108. Explanation (Answer; b)
H2SO4 ⇌ 2H+ + SO42−
1 M 2M
10−3M 2 x 10−3 M
pH = - log [H+] = - log 2 x 10−3 = 2.698 ≈ 2.7
109. 9.8 g of H2SO4 is present in 100 mL solution what is the w/v%?
(a) 98
(b) 9.8
(c) 10
(d) 0.01
109. Explanation (Answer; b)
Amount of solute in 100 volume of solution is w/v%. Hence the correct option is b.
110. How much water is needed to dilute 10cm3 of decamolar hydrochloric acid to make it exactly decimolar solution?
(a) 100 ml
(b) 1010 ml
(c) 1000 ml
(d) 990 ml
110. Explanation (Answer; d)
M1V1 = M2V2 ⇒ 10 x 10 =10-1 x V2 ⇒ V2 = 1000 ml solution
Volume of water added = V2 – V1 = 1000 – 10 = 990 ml
111. The molality of an aqueous solution of sugar (C12H22O11) is 1.62 m. What is the mole fractions of sugar?
(a) 0.0284
(b) 0.284
(c) 0.0815
(d) 0.815
111. Explanation (Answer; a)
Molality is moles solute / kg of solvent. Therefore we know our solution is:
1.62 mol C12H22O11 in 1 kg of water
1.00 kg = 1000 g of water
moles of water = 1000 g / 18.0152 g/mol = 55.50868 mol
mole fraction of the sugar = 1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three significant figures)
112. Solution are the examples of:
(a) Compounds
(b) Homogenous mixture
(c) Heterogeneous mixture
(d) Colloids
112. Explanation (Answer; b)
Solution are the examples of homogenous mixture.
113. Molarity of 0.2 N H2SO4 is:
(a) 0.1
(b) 0.2
(c) 0.4
(d) None of these
113. Explanation (Answer; a)
n-factor = number of H+ ions or OH− ions given by an acid or base in solution
n-factor for H2SO4 = 2
Molarity = Normality/n-factor = 0.2/2 = 0.1 M
114. A solution which does not obey Raoult’s law is called a.
(a) non-ideal solution
(b) Ideal solution
(c) Both of them
(d) Binary solution
114. Explanation (Answer; a)
A non-ideal solution is the solution in which solute and solvent molecules interact with one another with a different force than the forces of interaction between the molecules of the pure components. Non-Ideal solutions do not obey Raoult’s law.
115. In how many ways can the concentration of a solution be expressed?
(a) 1
(b) 3
(c) 5
(d) 8
115. Explanation (Answer; d)
The concentration of a solution can be expressed in 5 ways: Mass %, Volume %, Mass by volume percentage, Parts per million, Molarity, Molality and Normality, Mole fraction,
116. Which of the following does not dissolve in benzene?
(a) Naphthalene
(b) Anthracene
(c) C6H12O6
(d) All of the above
116. Explanation (Answer; c)
C6H12O6 being polar is in soluble in non-polar benzene.
117. The solubility of a substance in a solvent depends on
(a) Temperature
(b) Pressure
(c) Nature of solute and solvent
(d) All of the above
117. Explanation (Answer; d)
All factors affect on the solubility of a substance in a solvent.
118. Which law explained solubility of gases in a liquid?
(a) Charles law
(b) Henry’s law
(c) Raoult’s law
(d) Boyle’s law
118. Explanation (Answer; b)
The solubility of gases in liquids is governed by Henry’s law.
119. What does Henry’s constant depend upon?
(a) Nature of gas
(b) Nature of solvent
(c) Temperature
(d) All of the above
119. Explanation (Answer; d)
Henry’s constant depend upon nature of gas, nature of solvent and temperature.
120. How is Henry’s constant dependent on temperature?
(a) Directly proportional
(b) Inversely proportional
(c) Varies exponentially
(d) None of the above
120. Explanation (Answer; d)
Henry’s constant Directly proportional To Temperature.
121. Dissolution of gas in a liquid is
(a) Endothermic
(b) Exothermic
(c) No heat change
(d) No change in temperature
121.Explanation (Answer; b)
Dissolution of gas in a liquid is exothermic.
122. The pair of miscible liquids among the following is
(a) Oil and water
(b) Kerosene and water
(c) Vegetable oil and corn syrup
(d) Ethanol and water
122.Explanation (Answer; d)
Ethanol and water is the pair of completely miscible liquids.
123. The example of a colloidal solution is
(a) Air
(b) Milk
(c) Alcohol
(d) Urea
123. Explanation (Answer; b)
Milk is a colloidal solution.
124. The example of a suspension is
(a) Milk
(b) Alcohol
(c) Urea
(d) Mixture of water and chalk
124. Explanation (Answer; d)
Mixture of water and chalk is a suspension.
125. The 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
Explanation; (Answer; c)
The maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent is called solubility. Solubility of solid solute is independent of pressure.
126. Which of the following units is useful in relating the concentration of a solution with its vapour pressure?
(a) mole fraction
(b) parts per million
(c) mass percentage
(d) molality
Explanation; (Answer; a)
Vapour pressure of a solution depends upon the mole fraction.
127. Which of the following aqueous solutions should have the highest boiling point?
(a) 1.0 M NaOH
(b) 1.0 M Na2SO4
(c) 1.0 M NH4NO3
(d) 1.0 M KNO3
Explanation; (Answer; b)
greater the value of van't Hoff factor (i) for same molar concentration, higher will be the elevation in boiling point and hence higher will be the boiling point of solution.
van't Hoff factor (i) for 1.0 M NaOH = 2
van't Hoff factor (i) for 1.0 M Na2SO4 = 3
van't Hoff factor (i) for 1.0 M NH4NO3 = 3
van't Hoff factor (i) for 1.0 M KNO3 = 3
Hence, 1.0 M Na2SO4 has highest value of boiling point.
The formula for elevation in boiling point is
ΔTb = i × Kb × m
Where,
‘i' = is the Van’t Hoff factor
Kb = ebullioscopic constant/molal boiling constant
‘m’ = molality of the solute
Here molality or molarity is same for all compounds i.e. 1M. Hence, the value of elevation in boiling point depends only on Van’t Hoff factor (i). Greater the value of Van’t Hoff factor, higher will be the elevation in boiling point.
Van’t Hoff factor (i) for NaOH (Na+ + OH−) is 2.
Van’t Hoff factor (i) for Na2SO4 (2Na+ + SO42−) is 3.
Van’t Hoff factor (i) for NH4NO3 (NH4+ + NO3−) is 2.
Van’t Hoff factor (i) for KNO3 (K+ + NO3−) is 2.
Since the value of Van’t Hoff factor (i) is greatest for Na2SO4, its aqueous solution will have highest boiling point.
128. Which has the least freezing point?
(a) 1% sucrose
(b) 1% NaCl
(c) 1% CaCl2
(d) 1% glucose
Explanation (Answer; c)
CaCl2 gives maximum number of ions on ionization i.e. 3.
129. In which of the following condition reverse osmosis takes place?
(a) Eext > osmotic pressure
(b) Eext = osmotic pressure
(c) Eext < osmotic pressure
(d) None of these
129. Explanation (Answer; a)
When Eext > osmotic pressure, reverse osmosis takes place.
130. Molarity of a given orthophosphoric acid solution is 3M. Its normality is
(a) 0.3 N
(b) 9M
(c) 1N
(d) 3N
130. Explanation (Answer; b)
n-factor = number of H+ ions or OH− ions given by an acid or base in solution
n-factor for H3PO4 = 3
Molarity = Normality/n-factor
Normality = molarity x n-factor = 3 x 3 = 9
131.The osmotic pressure of 0.1 M aqueous solution of NaCl is ………….. osmotic pressure of 0.1 M aqueous solution of glucose.
(a) equal to
(b) less than
(c) half of
(d) nearly double than
131. Explanation (Answer; d)
The osmotic pressure π = iCRT, where ′i′ is the vant Hoff's factor, C is molar concentration, R is ideal gas constant and T is absolute temperature. Since, C,R and T have same values for different solutions, the solution having higher value of ′i′ will have higher value of the osmotic pressure. ′i′ represents the number of ions obtained from 1 solute molecule. The ionization 0.1 sodium chloride gives 2 ions/particles. Glucose is non-electrolyte and does not undergo dissociation.
0.1 M aqueous solution of NaCl has more osmotic pressure nearly double than 0.1 M aqueous solution of glucose because Na2SO4 gives 3 ions.
132. Which of the following 0.1 M aqueous solution is likely to have the highest boiling point?
(a) K2SO4
(b) KCl
(c) Glucose
(d) Urea
132. Explanation (Answer; a)
For same molar concentration, solute with greater number of ions will have the highest boiling point. K2SO4 gives 3 ions in aqueous solution and hence it shows the highest boiling point. Glucose and urea are non-electrolyte and do not undergo dissociation
133.Increasing the temperature of an aqueous solution will cause decrease in
(a) molarity
(b) Molality
(c) mole fraction
(d) % w/w
133. Explanation (Answer; a)
An increase in temperature increases the volume of the solution which in turn decreases its molarity as molarity is moles per unit volume.
134. The amount of H2SO4 present in 1200 ml of 0.2 N solution is
(a)10.76 g
(b) 11.76 g
(c)12.76 g
(d)14.76 g
134. Explanation (Answer; b)
Volume of solution in liter = 1200 ml/1000 = 1.2 liter
gram equivalent mass of H2SO4 = 49 g/mol
Mass of solute = N x gram equivalent mass x Vdm3 = 0.2 x 49 x 1.2 = 11.76 g
135. The molarity of solution containing 26.5 g of Na2CO3 in 500 ml of solution is
(a) 1 M
(b) 0.25 M
(c) 0.5 M
(d) 0.75 M
135. Explanation (Answer; c)
Volume of solution in liter = 500 ml/1000 = 0.5 liter
Molar mass of Na2CO3 = 106 g/mol
Molarity = (mass/M x Vliter) = (26.5/106 x 0.5) = 0.5 M
136. If a solution of ethanol is 2 molal, how many gram of ethanol are dissolved in 2000 g of water?
(a)184 g
(b) 733.45 g
(c) 87.24 g
(d) 92.35 g
136. Explanation (Answer; a)
Mass = molality x molar mass x mass of solvent in kg = 2 x 46 x (2000/1000) = 184 g
137. A beaker contains a solution of a substance ‘A’. Precipitation of substance ‘A’ takes place when a small amount of ‘A’ is added to the solution. The solution is _________.
(a) saturated
(b) supersaturated
(c) unsaturated
(d) concentrated
137. Explanation; (Answer; b)
The formation of precipitate in an indication of supersaturated solution.
138. Which law specifically governs the relative lowering of vapour pressures in solutions?
(a) van’t Hoff law
(b) Raoult’s law
(c) Gay-Lussac’s law
(d) Amagat’s law
138. Explanation; (Answer; b)
Raoult’s law quantifies relative lowering in vapour pressure.
139. Which of the following aqueous solutions has the highest vapour pressure at 300 K?
(a) 1 M Na3PO4
(b) 1 M CaCl2
(c) 1 M KNO3
(d) 1 M C6H12O6
139. Explanation; (Answer; d)
The Van’t Hoff Factor (i) quantifies the effect of a solute on various colligative properties of solutions. An electrolytic solute’s Van’t Hoff factor is always equal to the number of ions in which it is ionized. It is equal to one for a non-electrolytic solute.
Upon ionization of electrolytic solutes Na3PO4, CaCl2 and KNO3, the value of the van’t hoff factor(i) comes out to be 4, 3 and 2 respectively. Because C6H12O6 is a non-electrolytic solute, it has i=1.
The greater the value of the Van’t Hoff factor (i), the lower the vapour pressure, and vice versa. As a result of the lower van’t Hoff factor(i) value, C6H12O6 has a higher vapour pressure.
140. Which of the following has equal boiling point?
(a) 0.1 M Na2SO3
(b) 0.1 M Al(NO3)3
(c) 0.1 MgCl2
(d) both a and c
140. Explanation; (Answer; d)
Boiling point will be the same for those solutes which will have the same value for Van’t Hoff factor.
Van’t Hoff factor(i) for Na2SO3 = 3
Van’t Hoff factor(i) for C12H22O11 = 1
Van’t Hoff factor(i) for Al(NO3)3 = 4
Van’t Hoff factor(i) for MgCl2 = 3
Hence, Na2SO3 and MgCl2 will have the same value for Van’t Hoff factor(i).
In the rest of the cases, it will be equal to the number of ions it produces.
141. Molarity of 4% (w/v) solution of NaOH is:
(a) 0.1
(b) 0.5
(c) 0.001
(d) 1.0
141. Explanation; (Answer; d)
4%(w/v) solution means 4 g of NaOH is present in 100 mL of the solution.
Number of moles of NaOH = mass /molar mass = 4/40 = 0.1 mol
Volume of solution in L = 100 mL = 100/1000 = 0.1 L
Molarity = n/Vliter = 0.1/0.1 = 1.0 M
142. 250 mL of a 2% (w/v) NaOH solution is diluted up to 600 mL. Find the resultant molarity of the solution.
(a) 4.0 M
(b) 0.166 M
(c) 0.33 M
(d) 0.21 M
142. Explanation; (Answer; d)
2%(w/v) solution means 2 g of NaOH is present in 100 mL of the solution.
Number of moles of NaOH = mass /molar mass =2/40 = 0.05 mol
Volume of solution in L = 100 mL = 100/1000 = 0.1 L
Molarity = n/Vliter = 0.05/0.1 = 0.5 M
Conserving moles before and after dilution:
M1V1=M2V2
Where, M1= 0.5 M and V1= 250 mL
M2 and V2 are the molarity and volume of the resultant solution.
(0.5 × 250) = (M×600)
M = 0.21 M
Molarity of the resultant solution is 0.21 M.
143. Which condition holds for the ideal solution?
(a) Change is volume is zero
(b) Change in volume is non-zero
(c) Change is enthalpy is non-zero
(d) None of the above
143. Explanation (Answer; a)
Four Distinct Properties of An Ideal Solution
1. It obeys Raoult’s law at all temperature and concentration.
2. The molecular attraction of liquids A – A, B – B and A – B remains almost constant.
3. No heat absorb or evolve in the formation of ideal solution (∆H = 0).
4. No change in volume takes place in the formation of ideal solution (∆V = 0).
144. Which condition holds for a non-ideal solution?
(a) Change is volume is zero
(b) Change in volume is non-zero
(c) Change is enthalpy is zero
(d) None of the above
144. Explanation (Answer; b)
For non-ideal solution, change is volume is non-zero since the volume of blending is not zero. Either volume change occurs.
145. From the below options, choose the correct example for gaseous solutions.
(a) Oxygen dissolved in water
(b) Camphor in nitrogen gas
(c) Carbon dioxide dissolved in water
(d) Hydrogen in palladium
145. Explanation (Answer; c)
A gaseous solution is a solution in which the solvent is a gas.
146. Which among the following is an example of a solid solution?
(a) Copper dissolved in gold
(b) Ethanol dissolved in water
(c) Glucose dissolved in water
(d) Sodium chloride dissolved in water
146. Explanation (Answer; a)
Alloys are solid-solid solution.
147. What is Raoult's Law?
(a) Psolution = (Xsolvent) × (Posolvent)
(b) Psolvent = (Xsolvent) × (Posolution)
(c) Psolution = (Xsolvent)/(Posolvent)
(d) Psolvent = (Xsolvent)/(Posolution)
147. Explanation (Answer; a)
The Raoult's Law is: Psolution = (Xsolvent) × (Posolvent)
The vapour pressure of a solution of non-volatile solute is the product of vapour pressure of pure solvent and mole fraction of solvent.
148. How does the solubility of gases vary with pressure?
(a) Increases with pressure
(b) Decreases with pressure
(c) First increases and then decreases
(d) No effect
148. Explanation (Answer; a)
Solubility of gases is dreictly proportional to the pressure. This is known as Henry’s law.
149. Which of the following is not an example of an Ideal solution?
(a) Benzene + Toluene
(b) n-Hexane + n-Heptane
(c) Ethyl alcohol + Water
(d) Ethyl bromide + Ethyl chloride
149. Explanation (Answer; c)
An ideal solution may be defined as the solution in which no volume change and no enthalpy change take place on mixing the solute and the solvent in any proportion. Ethyl alcohol + Water is a Non-Ideal solution.
150. Which of the following is false regarding Non-Ideal solutions?
(a) They do not obey Raoult’s law
(b) ΔVmixing ≠ 0
(c) ΔHmixing = 0
(d) They form azeotropes
150. Explanation (Answer; c)
Non-Ideal solutions do not obey Raoult’s law. For Non-Ideal solutions, ΔVmixing ≠ 0, ΔHmixing ≠ 0. Non-ideal solutions form azeotropes or constant boiling mixtures, i.e., they have the same concentration in the vapour phase and the liquid phase.
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