#chemistry problems

Temperature Dependence - Arrhenius Equation and Activation Energy

A first-order reaction has a half-life of 25 minutes. What percentage of the reactant remains after 75 minutes?

3 half-lives → fraction remaining = 1/8 = 12.5%.

Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Temperature Dependence - Arrhenius Equation and Activation Energy

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Temperature Dependence - Arrhenius Equation and Activation Energy

A reaction’s rate increases by 25 times when the concentration of the reactant is increased by 5 times. What is the orde

5ⁿ = 25 → n = 2.

Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Temperature Dependence - Arrhenius Equation and Activation Energy

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Temperature Dependence - Arrhenius Equation and Activation Energy

A reaction’s rate increases by 2.25 times when the concentration of the reactant is increased by 1.5 times. What is the

1.5ⁿ = 2.25 → n log 1.5 = log 2.25 → n ≈ 2.

Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Temperature Dependence - Arrhenius Equation and Activation Energy

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Types of Solutions and Expressing Concentration

The vapor pressure of pure toluene is 40 mm Hg at a certain temperature. If a solution with a non-volatile solute has a

(p⁰ - p/p⁰) = xsolute . (40 - 38/40) = (2/40) = 0.05 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration

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Types of Solutions and Expressing Concentration

What is the mole fraction of chloroform (molar mass = 119.5 g/mol) in a solution containing 59.75 g of chloroform and 36

Moles of chloroform = (59.75/119.5) = 0.5 mol . Moles of water = (36/18) = 2 mol . Total moles = 0.5 + 2 = 2.5 . Mole fraction = (0.5/2.5) = 0.2 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration

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Solubility - of Solids and Gases in Liquids Henry's Law

A 0.2 molal solution of a non-electrolyte in water has its boiling point elevated by 0.104°C. What is the value of Kb fo

Δ Tb = Kb · m . 0.104 = Kb × 0.2 . Kb = (0.104/0.2) = 0.52 K kg mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law

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Solubility - of Solids and Gases in Liquids Henry's Law

A gas has a partial pressure of 5 bar and a Henry’s law constant of 500 bar. What is the mole fraction in the solution?

p = KH · x . x = (5/500) = 0.01 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law

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Solubility - of Solids and Gases in Liquids Henry's Law

A solution of a non-volatile solute in water boils at 100.52°C at 1 atm. What is the molality of the solution? ( Kb = 0.

Δ Tb = Kb · m . 100.52 - 100 = 0.52 · m . m = (0.52/0.52) = 1 mol/kg .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law

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Solubility - of Solids and Gases in Liquids Henry's Law

The solubility of a gas in water follows Henry’s law with a constant of 150 bar. If the gas’s partial pressure decreases

Henry’s law: p = KH · x . Initial x₁ = (6/150) = 0.04 , final x₂ = (3/150) = 0.02 . Factor = (x₂/x₁) = (0.02/0.04) = 0.5 (decreases by a factor of 2).

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

What is the molar mass of a solute if 4 g of it in 500 mL of solution produces an osmotic pressure of 0.492 atm at 27°C?

Pi = (w/M V) RT . 0.492 = (4/M × 0.5) × 0.0821 × 300 . M = (4 × 0.0821 × 300/0.492 × 0.5) ≈ 200 g/mol .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

The van’t Hoff factor of a 0.05 m NaCl solution is 1.9. What is the osmotic pressure at 27°C? ( R = 0.0821 L atm mol⁻¹ K

Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.05 M . Pi = 1.9 × 0.05 × 0.0821 × 300 ≈ 2.339 atm .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

What is the molar mass of a solute if 5 g of it in 250 mL of solution produces an osmotic pressure of 0.984 atm at 27°C?

Pi = (w/M V) RT . 0.984 = (5/M × 0.25) × 0.0821 × 300 . M = (5 × 0.0821 × 300/0.984 × 0.25) ≈ 50 g/mol .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis

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