Which compound yields [Cr(H₂O)6]3+ and SO₄ 2− ions in aqueous solution?
[Cr(H₂O)6]SO₄ dissociates into the complex ion and SO₄ 2−.
Ref: NCERT Class 12 Chemistry > Chapter 5: Coordination Compounds > Topic: Nomenclature of Coordination Compounds
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The conductivity of a 0.025 M NaNO₃ solution is 0.003 S cm⁻¹, and its molar conductivity is 120 S cm² mol⁻¹. What is the
α = (Lambdam/Lambdam°) = (120/126.5) = 0.9486 ≈ 0.95 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
The conductivity of a 0.025 M NaNO₃ solution is 0.003 S cm⁻¹, and its molar conductivity is 120 S cm² mol⁻¹. What is the
α = (Lambdam/Lambdam°) = (120/126.5) = 0.9486 ≈ 0.95 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
A 0.1 molal solution of a solute in water has a boiling point elevation of 0.156°C. If the solute dissociates into 3 ion
Δ Tb = i · Kb · m . 0.156 = i × 0.52 × 0.1 . i = (0.156/0.52 × 0.1) = 3 . i = 1 + α (n - 1) , 3 = 1 + α (3 - 1) , 2α = 2 , α = 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
A 0.25 M solution of MgCl₂ (assuming complete dissociation) has an osmotic pressure of 1.845 atm at a certain temperatur
For MgCl₂, i = 3 (Mg²⁺ + 2Cl⁻). Pi = i · M · RT . 1.845 = 3 × 0.25 × 0.0821 × T . T = (1.845/0.75 × 0.0821) ≈ 30 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
A 0.3 M solution of Na₂SO₄ (assuming complete dissociation) has an osmotic pressure of 2.214 atm at a certain temperatur
For Na₂SO₄, i = 3 (2Na⁺ + SO₄²⁻). Pi = i · M · RT . 2.214 = 3 × 0.3 × 0.0821 × T . T = (2.214/0.9 × 0.0821) ≈ 30 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
A 0.1 M solution of K₃PO₄ (assuming complete dissociation) has an osmotic pressure of 0.984 atm at a certain temperature
For K₃PO₄, i = 4 (3K⁺ + PO₄³⁻). Pi = i · M · RT . 0.984 = 4 × 0.1 × 0.0821 × T . T = (0.984/0.4 × 0.0821) ≈ 30 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
A 0.15 M solution of a solute in 500 mL of water has an osmotic pressure of 1.476 atm at 27°C. If the solute dissociates
Pi = i · M · RT . 1.476 = i × 0.15 × 0.0821 × 300 . i = (1.476/0.15 × 0.0821 × 300) ≈ 0.4 , which is incorrect; recalculate: i = (1.476/0.15 × 24.63) ≈ 0.4 × 4 = 1.6 . i = 1 + α , 1.6 = 1 + α , α = 0.6 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
A solution of 0.1 mole of a solute in 1 kg of water freezes at -0.558°C. If the solute dissociates into 3 ions, what is
Δ Tf = i · Kf · m . 0.558 = i × 1.86 × 0.1 . i = (0.558/1.86 × 0.1) = 3 . i = 1 + α (n - 1) , 3 = 1 + α (3 - 1) , 2α = 2 , α = 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
A solute dissociates into 3 ions with a van’t Hoff factor of 2.4. What is the degree of dissociation?
i = 1 + α (n - 1) , where n = 3 . 2.4 = 1 + α (3 - 1) . 2.4 = 1 + 2α , α = (1.4/2) = 0.7 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
A solute dissociates into 2 ions with a van’t Hoff factor of 1.4. What is the degree of dissociation?
i = 1 + α (n - 1) , where n = 2 . 1.4 = 1 + α . α = 1.4 - 1 = 0.4 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
What is the pH of a solution containing 0.01 M HCl assuming complete dissociation?
For HCl , [H+] = 0.01 M = 10⁻² M , so pH = -log(10⁻²) = 2 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier