NEET Chemistry: Solutions, Redox, Electrochemistry and Kinetics · Chapter 1

Solutions

What to remember

Molality (m) does not change with temperature; molarity (M) does, because volume changes with temperature but mass does not.
Raoult's law: p₁ = x₁p₁°. For a non-volatile solute, the relative lowering of vapour pressure (p° − p)/p° equals the mole fraction of the solute. Henry's law: p = K_H x, and a larger K_H means a lower solubility of the gas.
Colligative properties depend only on the number of solute particles: ΔT_b = K_b m, ΔT_f = K_f m, π = CRT. For electrolytes and for associating solutes, multiply by the van't Hoff factor i (i > 1 for dissociation, i < 1 for association).
Positive deviation: weaker A–B attractions, ΔmixH > 0, ΔmixV > 0, minimum boiling azeotrope. Negative deviation: stronger A–B attractions, ΔmixH < 0, ΔmixV < 0, maximum boiling azeotrope.

Types of solutions

A solution is a homogeneous mixture of two or more components. The component present in the larger amount is the solvent; the other is the solute. The physical state of the solution is the physical state of the solvent. A binary solution has two components.

Solute	Solvent	Example
Gas	Gas	Mixture of oxygen and nitrogen (air)
Gas	Liquid	Oxygen dissolved in water
Gas	Solid	Hydrogen in palladium
Liquid	Liquid	Ethanol in water
Liquid	Solid	Amalgam of mercury with sodium
Solid	Liquid	Glucose in water
Solid	Solid	Copper in gold

Expressing concentration

Mass percentage = (mass of component / total mass of solution) × 100.
Volume percentage = (volume of component / total volume of solution) × 100.
ppm = (parts of the component / total parts of the solution) × 10⁶. It is used for very small amounts, such as pollutants in water.
Mole fraction of a component, x₁ = n₁ / (n₁ + n₂). The sum of all mole fractions is 1.
Molarity, M = moles of solute / volume of solution in litres. Unit: mol L⁻¹.
Molality, m = moles of solute / mass of solvent in kg. Unit: mol kg⁻¹.
Normality is used only when needed: N = gram equivalents of solute / volume of solution in litres.

Unit	Depends on temperature?	Reason
Mass %, ppm, mole fraction, molality	No	Based on mass or moles only
Volume %, molarity, normality	Yes	Based on volume, which changes with temperature

Worked example 1. A solution contains 46 g of ethanol (molar mass 46 g mol⁻¹) and 54 g of water (molar mass 18 g mol⁻¹). Moles of ethanol = 1; moles of water = 3. Mole fraction of ethanol = 1/(1 + 3) = 0.25.

Solubility

Solubility is the maximum amount of solute that dissolves in a given amount of solvent at a given temperature and pressure. A solution that cannot dissolve more solute is saturated.

Solids in liquids. "Like dissolves like": polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. If dissolution is endothermic, solubility rises with temperature; if exothermic, it falls. Pressure has almost no effect on the solubility of a solid.

Gases in liquids. Solubility increases with pressure and decreases with rise in temperature, because dissolution of a gas is exothermic.

Henry's law: the partial pressure of a gas in the vapour phase is proportional to its mole fraction in the solution: p = K_H x. A higher K_H means a lower solubility at the same pressure. K_H increases with temperature.

Gas (in water, 298 K)	K_H (kbar)
Helium	144.97
Hydrogen	69.16
Nitrogen	76.48
Oxygen	34.86

Applications of Henry's law:

Soda water and soft drinks are packed under high pressure of carbon dioxide; the gas fizzes out when the bottle is opened.
Deep-sea divers use air diluted with helium, because helium is much less soluble in blood than nitrogen at high pressure. This avoids the bends.
At high altitude the partial pressure of oxygen is low, so climbers have a low oxygen level in blood (anoxia).
Aquatic animals are more comfortable in cold water, which holds more dissolved oxygen.

Vapour pressure and Raoult's law

Vapour pressure is the pressure of the vapour in equilibrium with its liquid at a given temperature.

Two volatile liquids: p₁ = x₁p₁° and p₂ = x₂p₂°. Total pressure p_total = p₁° x₁ + p₂° x₂ = p₁° + (p₂° − p₁°) x₂. The vapour phase is richer in the more volatile component.

Non-volatile solute: only the solvent contributes to the vapour pressure, so p = x₁p₁°. Hence (p₁° − p)/p₁° = x₂. This relative lowering of vapour pressure depends only on the mole fraction of the solute.

Raoult's law as a special case of Henry's law: in p = K_H x, if K_H equals the vapour pressure of the pure component p°, Henry's law becomes Raoult's law.

Ideal and non-ideal solutions

Property	Ideal	Positive deviation	Negative deviation
Obeys Raoult's law	At all concentrations	No; p is higher than expected	No; p is lower than expected
A–B attraction compared with A–A and B–B	Equal	Weaker	Stronger
ΔmixH	0	Positive	Negative
ΔmixV	0	Positive	Negative
Example	n-Hexane and n-heptane; benzene and toluene; chlorobenzene and bromobenzene	Ethanol and acetone; ethanol and water; carbon disulphide and acetone	Chloroform and acetone; phenol and aniline; nitric acid and water

Azeotropes are liquid mixtures that boil at a constant temperature and distil without change in composition, so they cannot be separated by fractional distillation.

Minimum boiling azeotrope: formed by solutions with large positive deviation. Example: ethanol and water, about 95% ethanol by volume.
Maximum boiling azeotrope: formed by solutions with large negative deviation. Example: nitric acid and water, about 68% nitric acid by mass.

Colligative properties

These properties depend on the number of solute particles, not on their nature: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

Elevation of boiling point: ΔT_b = K_b m. For molar mass: M₂ = K_b × w₂ × 1000 / (ΔT_b × w₁), where w₂ is the mass of solute in g and w₁ the mass of solvent in g. K_b is the molal elevation constant (K kg mol⁻¹); for water it is 0.52.

Depression of freezing point: ΔT_f = K_f m. M₂ = K_f × w₂ × 1000 / (ΔT_f × w₁). For water K_f = 1.86 K kg mol⁻¹.

Osmosis is the flow of solvent through a semipermeable membrane from the dilute solution (or pure solvent) to the concentrated solution. The osmotic pressure π is the extra pressure that must be applied to the solution to stop osmosis. For a dilute solution: π = CRT, where C is molarity. With π = (n₂/V)RT, the molar mass is M₂ = w₂RT / (πV). R = 0.0821 L atm K⁻¹ mol⁻¹ = 8.314 J K⁻¹ mol⁻¹. Osmotic pressure is measured at room temperature and is the preferred method for polymers and proteins, because even a dilute solution gives a measurable value.

Isotonic solutions have the same osmotic pressure at the same temperature. A 0.9% (mass/volume) sodium chloride solution is isotonic with blood cells.
Hypertonic solution: higher osmotic pressure than the cell fluid; the cell loses water and shrinks.
Hypotonic solution: lower osmotic pressure; the cell takes in water and swells.
Reverse osmosis: a pressure larger than the osmotic pressure is applied to the solution side, and the solvent flows from the concentrated solution to the pure solvent. It is used for desalination of sea water.

Worked example 2. 6 g of urea (molar mass 60 g mol⁻¹) is dissolved in 200 g of water. Moles of urea = 0.1. Molality = 0.1 / 0.200 = 0.5 mol kg⁻¹. ΔT_f = 1.86 × 0.5 = 0.93 K. The freezing point is −0.93 °C.

Worked example 3. A solute (3 g) in 100 g of water lowers the freezing point by 0.93 K. M₂ = 1.86 × 3 × 1000 / (0.93 × 100) = 60 g mol⁻¹.

Abnormal molar masses and van't Hoff factor

Association or dissociation of solute changes the number of particles, so the observed colligative property differs from the calculated one.

i = (observed colligative property) / (calculated colligative property) = (normal molar mass) / (observed molar mass) = (total moles of particles after association or dissociation) / (moles before).

Case	Effect on particle number	Value of i	Example
Dissociation	Increases	i > 1	NaCl (i ≈ 2), K₂SO₄ (i ≈ 3), K₄[Fe(CN)₆] (i ≈ 5)
Association	Decreases	i < 1	Ethanoic acid (acetic acid) dimer in benzene (i ≈ 0.5 for complete dimerisation)
No change	Same	i = 1	Glucose, urea

Modified equations: ΔT_b = i K_b m; ΔT_f = i K_f m; π = i CRT; (p° − p)/p° = i x₂.

For dissociation of a compound giving n ions with degree of dissociation α: i = 1 + (n − 1)α. For association of n molecules into one with degree of association α: i = 1 − α(1 − 1/n).

Worked example 4. A binary electrolyte AB (n = 2) at 0.1 m depresses the freezing point by 0.3348 K (K_f = 1.86). i = 0.3348 / (1.86 × 0.1) = 1.8. Then α = i − 1 = 0.8.

Formula summary

Quantity	Formula	Unit
Molarity	moles of solute / litres of solution	mol L⁻¹
Molality	moles of solute / kg of solvent	mol kg⁻¹
Boiling point elevation	ΔT_b = i K_b m	K
Freezing point depression	ΔT_f = i K_f m	K
Osmotic pressure	π = i CRT	atm or bar
Relative lowering of vapour pressure	(p° − p)/p° = i x₂	no unit
Henry's law	p = K_H x	bar or kbar

Constants used: R = 0.0821 L atm K⁻¹ mol⁻¹ (also 0.082 for quick calculations), R = 8.314 J K⁻¹ mol⁻¹, and 1 bar is about 0.987 atm. Remember that a change of 1 K equals a change of 1 °C, so ΔT values may be quoted in either unit.

Steps for a colligative numerical: first find the moles of solute from its mass and molar mass; then divide by the mass of solvent in kg to get the molality; then multiply by the constant K_b or K_f, and by i if the solute ionises or associates. The new boiling point is the boiling point of the pure solvent plus ΔT_b; the new freezing point is the freezing point of the pure solvent minus ΔT_f. For osmotic pressure, convert the volume to litres and the temperature to kelvin (K = °C + 273).

A useful comparison: for the same molal concentration and complete dissociation, the order of colligative effect follows the value of i. For 0.1 m solutions, glucose (i = 1) < NaCl (i = 2) < K₂SO₄ (i = 3) < K₄[Fe(CN)₆] (i = 5). So the last solution has the lowest freezing point, the highest boiling point and the highest osmotic pressure.

NEET traps

1. Molarity changes with temperature; molality does not. Do not mix them up.
2. Molality uses kg of solvent; molarity uses litres of solution.
3. A higher K_H means lower solubility, not higher.
4. Positive deviation gives a minimum boiling azeotrope; negative deviation gives a maximum boiling azeotrope.
5. Acetone with chloroform shows negative deviation (hydrogen bonding); acetone with carbon disulphide or ethanol shows positive deviation.
6. In positive deviation ΔmixH and ΔmixV are positive; in negative deviation both are negative.
7. 0.1 M NaCl is not isotonic with 0.1 M glucose; its particle concentration is nearly double.
8. Association lowers i below 1, while dissociation raises i above 1.
9. Freezing point is lowered and boiling point is raised by a solute; do not reverse the signs.
10. In reverse osmosis, pressure above π is applied to the concentrated side; in ordinary osmosis no external pressure is applied.

One-liners

1. A solution is a homogeneous mixture.
2. Mole fraction has no unit.
3. ppm = parts per million = mass of component / mass of solution × 10⁶.
4. Molality is independent of temperature.
5. Gas solubility falls as temperature rises.
6. Henry's law: p = K_H x.
7. Helium-diluted air is used by deep-sea divers.
8. Raoult's law for a non-volatile solute gives (p° − p)/p° = x₂.
9. n-Hexane and n-heptane form an ideal solution.
10. Ethanol and water form a minimum boiling azeotrope.
11. Nitric acid and water form a maximum boiling azeotrope.
12. K_f of water is 1.86 K kg mol⁻¹; K_b of water is 0.52 K kg mol⁻¹.
13. π = CRT.
14. Isotonic solutions have equal osmotic pressure.
15. Reverse osmosis is used for desalination of sea water.

Practice questions

A solution is prepared by dissolving 5.85 g of NaCl (molar mass 58.5 g mol⁻¹) in water to make 500 mL of solution. What is its molarity?
1. 0.2 M
2. 0.5 M
3. 0.1 M
4. 1.0 M
Answer
A. 0.2 M
Moles = 5.85/58.5 = 0.1; M = 0.1/0.5 L = 0.2 M.
What is the mole fraction of ethanol in a solution containing 46 g of ethanol (molar mass 46 g mol⁻¹) and 54 g of water (molar mass 18 g mol⁻¹)?
1. 0.75
2. 0.33
3. 0.25
4. 0.50
Answer
C. 0.25
Ethanol = 1 mol, water = 3 mol; x = 1/(1+3) = 0.25.
What is the molality of a solution of 4 g of NaOH (molar mass 40 g mol⁻¹) in 250 g of water?
1. 0.1 mol kg⁻¹
2. 0.8 mol kg⁻¹
3. 0.2 mol kg⁻¹
4. 0.4 mol kg⁻¹
Answer
D. 0.4 mol kg⁻¹
Moles = 0.1; mass of solvent = 0.25 kg; m = 0.1/0.25 = 0.4 mol kg⁻¹.
Which of the following concentration terms does not change with temperature?
1. Volume percentage
2. Molality
3. Molarity
4. Normality
Answer
B. Molality
Molality uses masses only. Molarity, normality and volume percentage use volume, which changes with temperature.
At 298 K the Henry's law constants in water are: helium 144.97 kbar, hydrogen 69.16 kbar, nitrogen 76.48 kbar, oxygen 34.86 kbar. Which gas is least soluble at the same partial pressure?
1. Helium
2. Nitrogen
3. Oxygen
4. Hydrogen
Answer
A. Helium
Solubility is inversely related to K_H, and helium has the highest K_H.
The vapour pressure of a pure solvent is 200 mm Hg. One mole of a non-volatile solute is dissolved in nine moles of this solvent. What is the vapour pressure of the solution?
1. 20 mm Hg
2. 190 mm Hg
3. 200 mm Hg
4. 180 mm Hg
Answer
D. 180 mm Hg
x(solvent) = 9/10 = 0.9; p = 0.9 × 200 = 180 mm Hg.
Which pair of liquids forms an ideal solution?
1. Ethanol and acetone
2. Phenol and aniline
3. n-Hexane and n-heptane
4. Chloroform and acetone
Answer
C. n-Hexane and n-heptane
Chemically similar liquids such as n-hexane and n-heptane have A–B attractions equal to A–A and B–B attractions. The other pairs show deviations.
Which of the following mixtures forms a maximum boiling azeotrope?
1. n-Hexane and n-heptane
2. Nitric acid and water
3. Benzene and toluene
4. Ethanol and water
Answer
B. Nitric acid and water
Nitric acid and water show a large negative deviation and form a maximum boiling azeotrope (about 68% nitric acid). Ethanol and water form a minimum boiling azeotrope.
For water, K_b = 0.52 K kg mol⁻¹. What is the elevation of boiling point when 18 g of glucose (molar mass 180 g mol⁻¹) is dissolved in 500 g of water?
1. 0.104 K
2. 0.052 K
3. 0.52 K
4. 1.04 K
Answer
A. 0.104 K
m = 0.1/0.5 = 0.2 mol kg⁻¹; ΔT_b = 0.52 × 0.2 = 0.104 K.
For water, K_f = 1.86 K kg mol⁻¹. What is the freezing point of a solution of 6 g of urea (molar mass 60 g mol⁻¹) in 200 g of water?
1. −0.465 °C
2. −3.72 °C
3. −1.86 °C
4. −0.93 °C
Answer
D. −0.93 °C
m = 0.1/0.2 = 0.5; ΔT_f = 1.86 × 0.5 = 0.93 K; freezing point = 0 − 0.93 = −0.93 °C.
What is the osmotic pressure of a 0.1 M glucose solution at 300 K? (R = 0.082 L atm K⁻¹ mol⁻¹)
1. 24.6 atm
2. 4.92 atm
3. 2.46 atm
4. 0.246 atm
Answer
C. 2.46 atm
π = CRT = 0.1 × 0.082 × 300 = 2.46 atm.
A non-electrolyte (3 g) dissolved in 100 g of water lowers the freezing point by 0.93 K. (K_f of water = 1.86 K kg mol⁻¹.) What is its molar mass?
1. 30 g mol⁻¹
2. 60 g mol⁻¹
3. 180 g mol⁻¹
4. 120 g mol⁻¹
Answer
B. 60 g mol⁻¹
M = K_f × w₂ × 1000/(ΔT_f × w₁) = 1.86 × 3 × 1000/(0.93 × 100) = 60 g mol⁻¹. Check: m = 0.93/1.86 = 0.5, so 0.05 mol in 3 g gives 60.
Assuming complete dissociation, which 0.1 m aqueous solution has the lowest freezing point?
1. Glucose
2. K₂SO₄
3. NaCl
4. K₄[Fe(CN)₆]
Answer
D. K₄[Fe(CN)₆]
The depression of freezing point is proportional to i. The values of i are 1, 2, 3 and 5 respectively, so K₄[Fe(CN)₆] is lowest.
What is the van't Hoff factor for a solute that undergoes complete dimerisation in a solvent?
1. 1
2. 0.5
3. 2
4. 0.25
Answer
B. 0.5
Two molecules form one dimer, so the number of particles halves and i = 1/2 = 0.5.
An electrolyte A₂B dissociates in water into 2A⁺ and B²⁻ with degree of dissociation 0.5. What is the van't Hoff factor?
1. 2.0
2. 3.0
3. 1.5
4. 2.5
Answer
A. 2.0
i = 1 + (n − 1)α = 1 + 2 × 0.5 = 2.0. Check: 0.5 undissociated + 0.5 × 3 particles = 2.0.
Applying a pressure larger than the osmotic pressure on the solution side of a semipermeable membrane is used for
1. making a solution isotonic with blood
2. separating an ethanol–water azeotrope by distillation
3. desalination of sea water
4. increasing the solubility of oxygen in water
Answer
C. desalination of sea water
This is reverse osmosis: pure solvent passes from the concentrated solution to the solvent side, which is used to desalinate sea water.
Given below are two statements: Statement I: Molality of a solution does not change with temperature. Statement II: Molarity of a solution changes with temperature because the volume of the solution changes.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Molality uses mass of solvent; molarity uses volume of solution. Both statements are correct.
Given below are two statements: Statement I: Raoult's law becomes a special case of Henry's law when K_H equals the vapour pressure of the pure component. Statement II: In Henry's law, K_H is always equal to the vapour pressure of the pure gas.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Statement II is wrong: K_H is a separate constant that differs from p° (except in the special case).
Given below are two statements: Statement I: In a solution showing positive deviation, the A–B attractions are stronger than the A–A and B–B attractions. Statement II: A solution showing positive deviation has a negative enthalpy of mixing.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Positive deviation: A–B attractions are weaker, and ΔmixH and ΔmixV are positive. Both statements are incorrect.
Given below are two statements: Statement I: Solutions showing negative deviation from Raoult's law form minimum boiling azeotropes. Statement II: Solutions showing negative deviation have a negative volume change on mixing.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Negative deviation gives maximum boiling azeotropes (Statement I incorrect) and ΔmixV < 0 (Statement II correct).
Given below are two statements: Statement I: Osmotic pressure is a colligative property. Statement II: The osmotic pressure of a solution is independent of temperature.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
π = CRT, so π increases with temperature. Statement II is incorrect.
Given below are two statements: Statement I: 0.1 M glucose solution and 0.1 M urea solution are isotonic at the same temperature. Statement II: 0.1 M glucose solution and 0.1 M NaCl solution are isotonic at the same temperature.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Glucose and urea do not ionise, so both have the same particle concentration. NaCl gives about twice the particles, so Statement II is incorrect.
Given below are two statements: Statement I: Association of solute molecules in a solvent gives a van't Hoff factor greater than 1. Statement II: Dissociation of solute molecules in a solvent gives a van't Hoff factor greater than 1.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Association reduces the number of particles (i < 1); dissociation increases it (i > 1).
Which of the following statements are correct? 1. In osmosis, solvent flows through a semipermeable membrane from the dilute solution to the concentrated solution. 2. The osmotic pressure of a given solution increases when the temperature increases. 3. In reverse osmosis, the solvent flows from the dilute solution to the concentrated solution.
1. 1 and 3 only
2. 1 and 2 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 2 only
In reverse osmosis the solvent flows from the concentrated solution to the dilute one, so 3 is wrong.
Which of the following statements are correct? 1. K_b depends on the solvent only. 2. K_f of water is 0.52 K kg mol⁻¹. 3. At the same molality and with complete dissociation, 0.1 m NaCl shows a larger depression of freezing point than 0.1 m glucose.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
K_f of water is 1.86 K kg mol⁻¹ (0.52 is K_b), so 2 is wrong. NaCl has i ≈ 2, so 3 is correct.
Given below are two statements: Statement I: Acetone and chloroform form a solution that shows negative deviation because of hydrogen bonding between them. Statement II: Such solutions form a maximum boiling azeotrope.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both correct: the A–B attractions are stronger, which lowers vapour pressure and raises the boiling point at the azeotropic composition.
Assertion (A): Aquatic species are more comfortable in cold water than in warm water. Reason (R): The solubility of oxygen in water decreases with rise in temperature.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Dissolution of gases is exothermic, so less oxygen dissolves in warm water; R explains A.
Assertion (A): 0.1 m NaCl shows a greater depression of freezing point than 0.1 m glucose. Reason (R): Depression of freezing point is a colligative property.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true, but the reason for the greater value is i ≈ 2 for NaCl, not the fact that the property is colligative.
Assertion (A): Reverse osmosis can be used for desalination of sea water. Reason (R): In reverse osmosis a pressure smaller than the osmotic pressure is applied on the solution side.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
A is true; R is false, because the applied pressure must be larger than the osmotic pressure.
Assertion (A): Acetic acid dissolved in benzene shows a van't Hoff factor greater than 1. Reason (R): Acetic acid associates to form dimers in benzene.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A is false (i ≈ 0.5); R is true.
Assertion (A): Deep-sea divers use air diluted with helium. Reason (R): Helium is much less soluble in blood than nitrogen at high pressure.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Less dissolved gas avoids the bends; R explains A.
Assertion (A): 1 M glucose solution and 1 M NaCl solution have the same osmotic pressure at the same temperature. Reason (R): Osmotic pressure is a colligative property.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A is false: NaCl gives about twice the particles. R is true.
Match List I with List II. List I: a. Molarity b. Molality c. Mole fraction d. ppm List II: i. Moles of solute per kg of solvent ii. Parts of component per million parts of solution iii. Moles of solute per litre of solution iv. Moles of a component per total moles of all components
1. a-iii, b-i, c-ii, d-iv
2. a-iii, b-i, c-iv, d-ii
3. a-i, b-iii, c-iv, d-ii
4. a-iv, b-i, c-iii, d-ii
Answer
B. a-iii, b-i, c-iv, d-ii
Molarity is per litre of solution; molality per kg of solvent; mole fraction is a ratio of moles; ppm is per 10⁶ parts.
Match List I with List II. List I: a. Ideal solution b. Positive deviation c. Negative deviation d. Azeotrope with about 68% of one component by mass List II: i. Nitric acid and water ii. Benzene and toluene iii. Carbon disulphide and acetone iv. Phenol and aniline
1. a-iii, b-ii, c-iv, d-i
2. a-ii, b-iv, c-iii, d-i
3. a-ii, b-iii, c-i, d-iv
4. a-ii, b-iii, c-iv, d-i
Answer
D. a-ii, b-iii, c-iv, d-i
Benzene–toluene is ideal; CS₂–acetone shows positive deviation; phenol–aniline shows negative deviation; nitric acid–water forms the 68% maximum boiling azeotrope.
Match List I with List II. List I: a. Elevation of boiling point b. Depression of freezing point c. Osmotic pressure d. Relative lowering of vapour pressure List II: i. CRT ii. K_f m iii. Mole fraction of the solute iv. K_b m
1. a-iv, b-ii, c-i, d-iii
2. a-ii, b-iv, c-i, d-iii
3. a-i, b-ii, c-iv, d-iii
4. a-iv, b-ii, c-iii, d-i
Answer
A. a-iv, b-ii, c-i, d-iii
ΔT_b = K_b m; ΔT_f = K_f m; π = CRT; (p° − p)/p° = x₂.
Match List I with List II. List I: a. Glucose b. NaCl c. K₂SO₄ d. K₄[Fe(CN)₆] List II: i. 5 ii. 1 iii. 2 iv. 3
1. a-ii, b-iii, c-i, d-iv
2. a-ii, b-iii, c-iv, d-i
3. a-ii, b-iv, c-iii, d-i
4. a-iii, b-ii, c-iv, d-i
Answer
B. a-ii, b-iii, c-iv, d-i
Particles per formula unit: glucose 1; NaCl 2; K₂SO₄ 3; K₄[Fe(CN)₆] 5.
Match List I with List II. List I: a. Henry's law b. Raoult's law for two volatile liquids c. Van't Hoff factor d. Reverse osmosis List II: i. Desalination of sea water ii. p = K_H x iii. Ratio of observed to calculated colligative property iv. p(total) = p₁°x₁ + p₂°x₂
1. a-ii, b-iv, c-iii, d-i
2. a-iii, b-iv, c-ii, d-i
3. a-ii, b-iv, c-i, d-iii
4. a-iv, b-ii, c-iii, d-i
Answer
A. a-ii, b-iv, c-iii, d-i
Match each law or term with its equation or use.
Match List I with List II. List I: a. Gas in liquid b. Liquid in solid c. Gas in solid d. Solid in solid List II: i. Hydrogen in palladium ii. Copper in gold iii. Oxygen in water iv. Mercury in sodium
1. a-iii, b-iv, c-ii, d-i
2. a-iii, b-iv, c-i, d-ii
3. a-iii, b-i, c-iv, d-ii
4. a-iv, b-iii, c-i, d-ii
Answer
B. a-iii, b-iv, c-i, d-ii
Oxygen in water (gas in liquid); sodium amalgam (liquid in solid); hydrogen in palladium (gas in solid); copper in gold (solid in solid).
At 1 bar partial pressure, the mole fraction of a gas dissolved in water is 5 × 10⁻⁵. At the same temperature, what partial pressure gives a mole fraction of 2 × 10⁻⁴?
1. 2 bar
2. 10 bar
3. 4 bar
4. 0.25 bar
Answer
C. 4 bar
p = K_H x, so p is proportional to x: (2 × 10⁻⁴/5 × 10⁻⁵) × 1 bar = 4 bar.
Liquids A and B form an ideal solution. At a given temperature p_A° = 300 mm Hg and p_B° = 100 mm Hg. What is the total vapour pressure of a solution with mole fraction of A equal to 0.4?
1. 160 mm Hg
2. 120 mm Hg
3. 200 mm Hg
4. 180 mm Hg
Answer
D. 180 mm Hg
p = 0.4 × 300 + 0.6 × 100 = 120 + 60 = 180 mm Hg.
A water sample contains 3.6 mg of a pollutant in 1.2 kg of solution. What is the concentration of the pollutant in ppm?
1. 300 ppm
2. 30 ppm
3. 0.3 ppm
4. 3 ppm
Answer
D. 3 ppm
ppm = (3.6 × 10⁻³ g/1200 g) × 10⁶ = 3. Check: 3.6 mg per 1.2 kg = 3 mg per kg.
0.6 g of a non-electrolyte is dissolved in water to make 100 mL of solution. The osmotic pressure at 300 K is 2.46 atm. (R = 0.082 L atm K⁻¹ mol⁻¹.) What is the molar mass of the solute?
1. 6 g mol⁻¹
2. 120 g mol⁻¹
3. 60 g mol⁻¹
4. 600 g mol⁻¹
Answer
C. 60 g mol⁻¹
C = π/RT = 2.46/(0.082 × 300) = 0.1 M; moles in 0.1 L = 0.01; M = 0.6/0.01 = 60. Direct: M = wRT/(πV) = 60.
0.1 mol of NaCl (assume complete dissociation) is dissolved in 500 g of water. (K_b of water = 0.52 K kg mol⁻¹.) What is the elevation of boiling point?
1. 0.416 K
2. 0.208 K
3. 0.520 K
4. 0.104 K
Answer
B. 0.208 K
m = 0.1/0.5 = 0.2; i = 2; ΔT_b = 2 × 0.52 × 0.2 = 0.208 K.
A 0.1 molal solution of a binary electrolyte AB shows a freezing point depression of 0.3348 K. (K_f = 1.86 K kg mol⁻¹.) What is its degree of dissociation?
1. 0.8
2. 0.6
3. 0.4
4. 1.0
Answer
A. 0.8
i = 0.3348/(1.86 × 0.1) = 1.8; for n = 2, i = 1 + α, so α = 0.8. Check: 0.8 × 2 + 0.2 = 1.8 particles per mole.
Assuming 100% dissociation of NaCl, what concentration of glucose solution is isotonic with 0.1 M NaCl at the same temperature?
1. 0.2 M
2. 0.3 M
3. 0.1 M
4. 0.05 M
Answer
A. 0.2 M
0.1 M NaCl gives 0.2 mol of particles per litre (i = 2). Isotonic means equal particle concentration, so glucose = 0.2 M.

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