#dilution calculation

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Colligative Properties - Depression of Freezing Point

A solution contains 7 g of a solute (molar mass = 140 g/mol) in 350 mL of solution. What is the molarity?

Moles of solute = (7/140) = 0.05 mol . Volume = 350 mL = 0.35 L. Molarity = (0.05/0.35) ≈ 0.1429 M .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Stoichiometry and Stoichiometric Calculations and Limiting Reagent

A 0.2 M NaCl solution is diluted by adding 150 mL of water to 50 mL of the original solution. What is the final molarity

M₁V₁ = M₂V₂. 0.2 × 50 = M₂ × (50 + 150). M₂ = (0.2 × 50)/200 = 0.05 M.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent

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Stoichiometry and Stoichiometric Calculations and Limiting Reagent

A 0.25 M NaOH solution is diluted by adding 300 mL of water to 200 mL of the original solution. What is the final molari

M₁V₁ = M₂V₂. 0.25 × 200 = M₂ × (200 + 300). M₂ = (0.25 × 200)/500 = 0.1 M.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent

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#dilution calculation

No questions with this tag

A solution contains 7 g of a solute (molar mass = 140 g/mol) in 350 mL of solution. What is the molarity?

A 0.2 M NaCl solution is diluted by adding 150 mL of water to 50 mL of the original solution. What is the final molarity

A 0.25 M NaOH solution is diluted by adding 300 mL of water to 200 mL of the original solution. What is the final molari

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