Chapter 1: Solutions Case Study Questions | chemistry Class 12 CBSE

ByteNotes

Chapter 1: Solutions Case Study Questions | chemistry Class 12 CBSE | ByteNotes.in

Case Study

Passage with linked questions

Case Set

Case Set 1

Case Analysis

Passage

A nurse in a hospital is preparing intravenous (IV) fluids for patients. She knows that blood plasma has a specific ionic concentration, and any IV solution must match this osmotic pressure to prevent harm to blood cells. She prepares a normal saline solution containing 0.9% (mass/volume) NaCl in water. When a patient receives a hypotonic solution (less than 0.9% NaCl), water flows into the blood cells, causing them to swell and potentially burst. On the other hand, a hypertonic solution (more than 0.9% NaCl) causes water to exit the cells, making them shrink. The nurse must also calculate the molarity and osmotic pressure of the IV fluid before administration.

Question 1: What is the term for two solutions that have the same osmotic pressure at a given temperature?

Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions.
No osmosis occurs when isotonic solutions are separated by a semipermeable membrane.

Question 2: Explain why blood cells placed in a solution containing less than 0.9% (mass/volume) NaCl swell, while those in more than 0.9% NaCl shrink.

In a hypotonic solution (less than 0.9% NaCl), the osmotic pressure outside the cell is lower than inside. Water flows into the cell by osmosis, causing it to swell.
In a hypertonic solution (more than 0.9% NaCl), osmotic pressure outside the cell is higher than inside. Water flows out of the cell by osmosis, causing it to shrink.

Question 3: Calculate the osmotic pressure of normal saline (0.9% mass/volume NaCl) at 37°C. Given: Molar mass of NaCl = 58.5 g/mol, R = 0.083 L bar mol⁻¹ K⁻¹, assume complete dissociation (i = 2).

Concentration: 0.9 g per 100 mL = 9 g per 1000 mL = 9 g/L
Moles of NaCl per litre = 9 / 58.5 = 0.1538 mol/L
Temperature T = 37 + 273 = 310 K
Osmotic pressure Π = i × C × R × T = 2 × 0.1538 × 0.083 × 310 = 7.93 bar (approximately 7.9 bar)

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 2

Case Analysis

Passage

A scuba diver is preparing for a deep-sea dive. At high pressures underwater, atmospheric gases dissolve more readily in blood. Nitrogen, being the major component of air, can accumulate in dangerous amounts in the bloodstream at depth. When the diver ascends too quickly, the sudden decrease in pressure causes dissolved nitrogen to come out of solution rapidly, forming bubbles in the blood and tissues — a condition called 'decompression sickness' or 'bends.' To prevent this, divers use tanks filled with a helium-nitrogen-oxygen mix (trimix): approximately 11.7% helium, 56.2% nitrogen, and 32.1% oxygen. Henry's law is central to understanding gas solubility under pressure.

Question 1: State Henry's law and write its mathematical expression.

Henry's law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Mathematically: p = K_H × x, where p is partial pressure, K_H is Henry's law constant, and x is mole fraction of gas in solution.

Question 2: Why does nitrogen form dangerous bubbles when a diver ascends rapidly? Explain using Henry's law.

According to Henry's law, solubility of a gas is directly proportional to its partial pressure. At depth, high pressure increases nitrogen's partial pressure, so more nitrogen dissolves in the blood.
When the diver ascends rapidly, pressure decreases suddenly. The dissolved nitrogen becomes supersaturated and comes out of solution as bubbles, blocking capillaries and causing 'bends.'

Question 3: The KH for N₂ in water at 293 K is 76.48 kbar. If N₂ exerts a partial pressure of 0.987 bar over water, calculate the mole fraction of N₂ and the number of millimoles of N₂ dissolved in 1 litre of water.

Mole fraction of N₂: x = p / K_H = 0.987 / 76480 = 1.29 × 10⁻⁵
1 litre of water contains 1000/18 = 55.5 mol
Using x = n / (n + 55.5) ≈ n / 55.5 (since n << 55.5): n = 1.29 × 10⁻⁵ × 55.5 = 7.16 × 10⁻⁴ mol = 0.716 mmol

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 3

Case Analysis

Passage

A chemistry teacher is demonstrating antifreeze solutions to students. In cold countries, the radiator water in cars can freeze, causing engine damage. Ethylene glycol (C₂H₆O₂) is mixed with water to lower its freezing point and raise its boiling point. A 35% (v/v) solution of ethylene glycol lowers the freezing point of water to approximately 255.4 K (–17.6°C). This is a classic application of the colligative property known as depression of freezing point. The teacher explains that the extent of freezing point depression depends on the number of solute particles and is independent of the nature of the solute.

Question 1: What are colligative properties? Name any two colligative properties.

Colligative properties are properties of solutions that depend on the number of solute particles present, irrespective of their nature (chemical identity).
Examples: depression of freezing point, elevation of boiling point (also osmotic pressure and relative lowering of vapour pressure).

Question 2: Write the mathematical relation for depression in freezing point and explain how it can be used to determine the molar mass of the solute.

ΔTf = Kf × m, where Kf is the molal freezing point depression constant and m is molality of solution.
Molar mass M₂ = (Kf × w₂ × 1000) / (ΔTf × w₁), where w₂ is mass of solute and w₁ is mass of solvent in grams. By measuring ΔTf experimentally, M₂ can be calculated.

Question 3: 45 g of ethylene glycol (C₂H₆O₂, molar mass = 62 g/mol) is mixed with 600 g of water. Calculate (a) the molality of the solution, (b) the depression in freezing point, and (c) the freezing point of the solution. (Kf for water = 1.86 K kg mol⁻¹)

Moles of ethylene glycol = 45/62 = 0.726 mol; Mass of water = 600 g = 0.6 kg
Molality m = 0.726 / 0.6 = 1.21 mol kg⁻¹
ΔTf = Kf × m = 1.86 × 1.21 = 2.25 K
Freezing point = 273.15 – 2.25 = 270.90 K

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 4

Case Analysis

Passage

A student is studying the vapour pressure behaviour of a binary solution of benzene and toluene — a classic example of an ideal solution. When benzene (vapour pressure = 50.71 mm Hg at 300 K) and toluene (vapour pressure = 32.06 mm Hg at 300 K) are mixed, they obey Raoult's law over the entire range of composition. The student plots the partial vapour pressures of both components and the total vapour pressure as a function of mole fraction. The experiment confirms that both components behave ideally: the enthalpy of mixing is zero and the volume of mixing is also zero.

Question 1: State Raoult's law for a solution of volatile liquids.

Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction present in the solution.
Mathematically, p₁ = x₁ × p₁⁰, where p₁⁰ is the vapour pressure of the pure component.

Question 2: What are ideal solutions? State the conditions at molecular level for a solution to behave ideally.

Ideal solutions are solutions that obey Raoult's law over the entire range of concentration with ΔmixH = 0 and ΔmixV = 0.
At molecular level, the intermolecular attractive forces between A-A and B-B molecules are nearly equal to those between A-B molecules. Examples: benzene-toluene, n-hexane–n-heptane.

Question 3: 80 g of benzene (M = 78 g/mol) is mixed with 100 g of toluene (M = 92 g/mol). Calculate (a) the total vapour pressure of the solution and (b) the mole fraction of benzene in the vapour phase. (p⁰benzene = 50.71 mm Hg, p⁰toluene = 32.06 mm Hg at 300 K)

Moles: benzene = 80/78 = 1.026 mol; toluene = 100/92 = 1.087 mol; Total = 2.113 mol
x(benzene) = 1.026/2.113 = 0.4854; x(toluene) = 1 – 0.4854 = 0.5146
p(benzene) = 0.4854 × 50.71 = 24.61 mm Hg; p(toluene) = 0.5146 × 32.06 = 16.50 mm Hg
p(total) = 24.61 + 16.50 = 41.11 mm Hg; y(benzene) = 24.61/41.11 = 0.599

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 5

Case Analysis

Passage

A pharmaceutical company is determining the molar mass of a newly synthesized protein. Since proteins are large, thermally sensitive biomolecules, the boiling point elevation or freezing point depression methods are unsuitable — the temperature changes are too small and proteins denature at higher temperatures. The company instead uses osmotic pressure measurement. A solution of 1.26 g of protein in 200 mL of water at 300 K shows an osmotic pressure of 2.57 × 10⁻³ bar. The osmotic pressure method is preferred because even very dilute solutions give measurably large osmotic pressures at room temperature.

Question 1: Why is the osmotic pressure method preferred over other colligative property methods for determining the molar mass of proteins?

Osmotic pressure measurement is done at room temperature, so thermally sensitive biomolecules like proteins are not denatured.
The osmotic pressure is large even for very dilute solutions, making it measurable with high accuracy. Molarity is used instead of molality, which is more convenient.

Question 2: Write the equation relating osmotic pressure (Π) to the molar mass of a solute and explain each symbol.

Π V = w₂RT / M₂, therefore M₂ = w₂RT / (ΠV)
Where Π = osmotic pressure, V = volume of solution in litres, w₂ = mass of solute, R = gas constant (0.083 L bar mol⁻¹ K⁻¹), T = temperature in kelvin, M₂ = molar mass of solute.

Question 3: Calculate the molar mass of the protein from the data given: 1.26 g of protein in 200 mL water, osmotic pressure = 2.57 × 10⁻³ bar at 300 K. (R = 0.083 L bar mol⁻¹ K⁻¹). Also comment on why osmotic pressure is more suitable than boiling point elevation for such biomolecules.

V = 200 mL = 0.200 L; Π = 2.57 × 10⁻³ bar; T = 300 K; w₂ = 1.26 g
M₂ = (1.26 × 0.083 × 300) / (2.57 × 10⁻³ × 0.200) = 31.374 / 5.14 × 10⁻⁴ = 61,035 g/mol ≈ 61,022 g/mol
This large molar mass would cause an immeasurably small boiling point elevation (ΔTb), but gives a measurable osmotic pressure at room temperature, making it safer and more accurate for biomolecules.

ByteNotes

Premium Access

Unlock all Case Study sets

You are viewing 5 free case study sets from this chapter. Upgrade to Premium to unlock the remaining 10 sets.

Upgrade to Premium10 questions locked

Chapter sections

Chapter 1: Solutions | Case Study Questions

Passage

Question 1: What is the term for two solutions that have the same osmotic pressure at a given temperature?

Question 2: Explain why blood cells placed in a solution containing less than 0.9% (mass/volume) NaCl swell, while those in more than 0.9% NaCl shrink.

Question 3: Calculate the osmotic pressure of normal saline (0.9% mass/volume NaCl) at 37°C. Given: Molar mass of NaCl = 58.5 g/mol, R = 0.083 L bar mol⁻¹ K⁻¹, assume complete dissociation (i = 2).

Passage

Question 1: State Henry's law and write its mathematical expression.

Question 2: Why does nitrogen form dangerous bubbles when a diver ascends rapidly? Explain using Henry's law.

Question 3: The KH for N₂ in water at 293 K is 76.48 kbar. If N₂ exerts a partial pressure of 0.987 bar over water, calculate the mole fraction of N₂ and the number of millimoles of N₂ dissolved in 1 litre of water.

Passage

Question 1: What are colligative properties? Name any two colligative properties.

Question 2: Write the mathematical relation for depression in freezing point and explain how it can be used to determine the molar mass of the solute.

Question 3: 45 g of ethylene glycol (C₂H₆O₂, molar mass = 62 g/mol) is mixed with 600 g of water. Calculate (a) the molality of the solution, (b) the depression in freezing point, and (c) the freezing point of the solution. (Kf for water = 1.86 K kg mol⁻¹)

Passage

Question 1: State Raoult's law for a solution of volatile liquids.

Question 2: What are ideal solutions? State the conditions at molecular level for a solution to behave ideally.

Question 3: 80 g of benzene (M = 78 g/mol) is mixed with 100 g of toluene (M = 92 g/mol). Calculate (a) the total vapour pressure of the solution and (b) the mole fraction of benzene in the vapour phase. (p⁰benzene = 50.71 mm Hg, p⁰toluene = 32.06 mm Hg at 300 K)

Passage

Question 1: Why is the osmotic pressure method preferred over other colligative property methods for determining the molar mass of proteins?

Question 2: Write the equation relating osmotic pressure (Π) to the molar mass of a solute and explain each symbol.

Question 3: Calculate the molar mass of the protein from the data given: 1.26 g of protein in 200 mL water, osmotic pressure = 2.57 × 10⁻³ bar at 300 K. (R = 0.083 L bar mol⁻¹ K⁻¹). Also comment on why osmotic pressure is more suitable than boiling point elevation for such biomolecules.

Unlock all Case Study sets