SOLUTIONS: COLLIGATIVE PROPERTY

 5 important numericals based on all four colligative properties including van’t Hoff factor (i) — aligned with CBSE PYQs and exam style:

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1. Relative Lowering of Vapour Pressure (PYQ Style)

Q: 5.85 g of NaCl is dissolved in 100 g of water. Calculate the relative lowering of vapour pressure. (Given: Molar mass of NaCl = 58.5 g/mol, i = 2)

Concepts Involved:

• Mole calculation
• Raoult’s Law
• Use of van’t Hoff factor

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2. Elevation of Boiling Point

Q: 3.42 g of sucrose (C₁₂H₂₂O₁₁) is dissolved in 180 g of water. Calculate the elevation in boiling point. (Kb = 0.52 K·kg/mol, Molar mass of sucrose = 342 g/mol)

Concepts Involved:

• Molality
• Non-electrolyte (i = 1)
• ΔTb = i·Kb·m

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3. Depression in Freezing Point (CBSE 2020)

Q: 1.8 g of glucose is dissolved in 100 g of water. Calculate depression in freezing point. (Kf = 1.86 K·kg/mol, Molar mass of glucose = 180 g/mol)

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4. Osmotic Pressure

Q: Calculate the osmotic pressure of a solution containing 0.5 mol NaCl in 2 L solution at 27°C. (R = 0.0821 L·atm/mol·K, i = 2)

Concepts Involved:

• π = i·M·R·T
• NaCl dissociation

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5. Van’t Hoff Factor (Abnormal Mol. Mass)

Q: 0.1 mol of K₂SO₄ is dissolved in 1 kg water. The freezing point is found to be depressed by 0.488 K. Calculate the van’t Hoff factor (i) and degree of dissociation. (Kf = 1.86 K·kg/mol)

Concepts Involved:

• Use i = ΔTf / (Kf·m)
• K₂SO₄ dissociates into 3 ions → Theoretical i = 3
• Use of degree of dissociation formula

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ANSWERS:

1. Relative Lowering of Vapour Pressure

Q: 5.85 g of NaCl is dissolved in 100 g of water. Calculate the relative lowering of vapour pressure.
(Given: Molar mass of NaCl = 58.5 g/mol, i = 2)

Solution:

• Moles of NaCl (solute)

  $$n_{\text{solute}} = \frac{5.85}{58.5} = 0.1 \, \text{mol}$$

• Moles of water (solvent)

  $$n_{\text{solvent}} = \frac{100}{18} \approx 5.56 \, \text{mol}$$

• Relative lowering of vapour pressure:

  $$\frac{\Delta P}{P_0} = i \cdot \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$$

• Substitute values:

  $$\frac{\Delta P}{P_0} = 2 \cdot \frac{0.1}{0.1 + 5.56} = 2 \cdot \frac{0.1}{5.66} \approx 0.0353$$

✅ Answer: Relative lowering of vapour pressure ≈ 0.0353

🔷 Question 2: Elevation of Boiling Point

Q: 3.42 g of sucrose (C₁₂H₂₂O₁₁) is dissolved in 180 g of water. Calculate the elevation in boiling point.
Given:

• Molar mass of sucrose = 342 g/mol

• Kb (for water) = 0.52 K·kg/mol

• Mass of water = 180 g = 0.180 kg

• Sucrose is a non-electrolyte, so van’t Hoff factor i = 1

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✅ Step-by-Step Solution

🔹 Step 1: Calculate moles of solute (sucrose)

$$\text{Moles of sucrose} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{3.42}{342} = 0.01 \, \text{mol}$$

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🔹 Step 2: Calculate molality (m)

Molality is defined as moles of solute per kg of solvent:

$$m = \frac{0.01 \, \text{mol}}{0.180 \, \text{kg}} \approx 0.0556 \, \text{mol/kg}$$

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🔹 Step 3: Use the formula for elevation in boiling point

$$\Delta T_b = i \cdot K_b \cdot m$$

Substitute the known values:

$$\Delta T_b = 1 \cdot 0.52 \cdot 0.0556 = 0.0289 \, \text{K}$$

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✅ Final Answer:

$$\boxed{\Delta T_b \approx 0.0289 \, \text{K}}$$

So, the boiling point of the solution is raised by approximately 0.0289 K compared to pure water.

🔷 Question 3: Depression in Freezing Point (CBSE 2020 Style)

Q: 1.8 g of glucose is dissolved in 100 g of water. Calculate the depression in freezing point.
Given:

• Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol

• Kf (for water) = 1.86 K·kg/mol

• Mass of water = 100 g = 0.100 kg

• Glucose is a non-electrolyte → van’t Hoff factor (i) = 1

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✅ Step-by-Step Solution

🔹 Step 1: Calculate moles of solute (glucose)

$$\text{Moles of glucose} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{1.8}{180} = 0.01 \, \text{mol}$$

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🔹 Step 2: Calculate molality (m)

$$\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.01}{0.100} = 0.1 \, \text{mol/kg}$$

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🔹 Step 3: Apply the formula for depression in freezing point

$$\Delta T_f = i \cdot K_f \cdot m$$

Substitute the values:

$$\Delta T_f = 1 \cdot 1.86 \cdot 0.1 = 0.186 \, \text{K}$$

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✅ Final Answer:

$$\boxed{\Delta T_f = 0.186 \, \text{K}}$$

So, the freezing point of the solution is lowered by 0.186 K compared to pure water.

🔷 Question 4: Osmotic Pressure

Q: Calculate the osmotic pressure of a solution containing 0.5 mol NaCl in 2 L solution at 27°C.
Given:

• $n = 0.5 \, \text{mol}$

• $V = 2 \, \text{L}$

• $T = 27^\circ C = 300 \, \text{K}$

• $R=0.0821\text{L.atm / mole K}$

• NaCl is an electrolyte → it dissociates into Na⁺ and Cl⁻

• So, van’t Hoff factor (i) = 2

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✅ Concept Used:

$$\pi = i \cdot M \cdot R \cdot T$$

Where:

• $\pi$ = osmotic pressure

• $i$ = van’t Hoff factor

• $M$ = molarity = $\frac{n}{V}$

• $R$ = gas constant

• $T$ = temperature in Kelvin

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✅ Step-by-Step Calculation

🔹 Step 1: Calculate molarity (M)

$$M = \frac{0.5 \, \text{mol}}{2 \, \text{L}} = 0.25 \, \text{mol/L}$$

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🔹 Step 2: Use the formula for osmotic pressure

$$\pi = i \cdot M \cdot R \cdot T$$

Substitute values:

$$\pi = 2 \cdot 0.25 \cdot 0.0821 \cdot 300$$ $$\pi = 2 \cdot 0.25 \cdot 24.63 = 2 \cdot 6.1575 = 12.315 \, \text{atm}$$

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✅ Final Answer:

$$\boxed{\pi = 12.32 \, \text{atm}}$$

So, the osmotic pressure of the solution is approximately 12.32 atm.

ANSWER 5:🧪 Given:

• Moles of K₂SO₄ = 0.1 mol

• Mass of water = 1 kg

• $K_f = 1.86 \, \text{K·kg/mol}$

• $\Delta T_f = 0.488 \, \text{K}$

• Molality $m = \frac{0.1}{1} = 0.1 \, \text{mol/kg}$

• Dissociation:

  $$\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \Rightarrow i_{\text{theoretical}} = 3$$

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✅ (a) Calculate van’t Hoff factor (i):

$$i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.488}{1.86 \times 0.1} = \frac{0.488}{0.186} \approx \boxed{2.62}$$

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✅ (b) Calculate Degree of Dissociation (α):

We use the relation:

$$i = 1 + 2\alpha$$

Substitute $i = 2.62$:

$$2.62 = 1 + 2\alpha \Rightarrow 2\alpha = 1.62 \Rightarrow \alpha = \frac{1.62}{2} = \boxed{0.81}$$

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🔚 Final Answers (for ΔTf = 0.488 K):

• ✅ Van’t Hoff factor (i) = $\boxed{2.62}$

• ✅ Degree of dissociation (α) = $\boxed{0.81 \text{ or } 81\%}$