What is the mass of solute required to prepare 250 g of a 12% by mass solution?
Mass of solute = (12/100) × 250 = 30 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
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The vapor pressure of pure water is 28 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor
(p⁰ - p/p⁰) = xsolute . (28 - 26.6/28) = (1.4/28) = 0.05 . Moles of water = (360/18) = 20 . xsolute = (nsolute/nsolute + 20) = 0.05 . nsolute = 0.05 (nsolute + 20) , 0.95 nsolute = 1 , nsolute ≈ 1.0526 . Mass = 1.0526 × 50 ≈ 52.63 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
A solution contains 6 g of a solute (molar mass = 120 g/mol) in 300 mL of solution. What is the molarity?
Moles of solute = (6/120) = 0.05 mol . Volume = 300 mL = 0.3 L. Molarity = (0.05/0.3) ≈ 0.1667 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
What is the mass percentage of a solution prepared by dissolving 40 g of salt in 160 g of water?
Total mass = 40 + 160 = 200 g . Mass % = (40/200) × 100 = 20% .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
A solution contains 25 g of a solute in 75 g of water. If the mass percentage of the solute is reduced to 20% by adding
Initial mass % = (25/25 + 75) × 100 = 25% . New mass % = 20% = (25/25 + 75 + w) × 100 . 0.2 (100 + w) = 25 , 100 + w = 125 , w = 25 g . Total mass = 100 + 25 = 125 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
What is the mass percentage of a solution if 25 g of a solute is dissolved in 225 g of water?
Total mass = 25 + 225 = 250 g . Mass % = (25/250) × 100 = 10% .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
A solution is made by mixing 30 g of a solute (molar mass = 60 g/mol) with water. If the molality is 0.5 mol/kg, what is
Moles of solute = (30/60) = 0.5 mol . Molality = (Moles/Mass of solvent in kg) , 0.5 = (0.5/w) . w = 1 kg = 1000 g . Total mass = 30 + 1000 = 1030 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
A 0.5 M solution of K₂SO₄ (molar mass = 174 g/mol) has a density of 1.15 g/mL. What is the mass of the solute in 300 mL
Moles = 0.5 × 0.3 = 0.15 mol . Mass of K₂SO₄ = 0.15 × 174 = 26.1 g . (Density is extra info, not needed for molarity-based calculation.)
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
A 0.5 M solution of Na₂SO₄ (molar mass = 142 g/mol) has a density of 1.2 g/mL. What is the mass of the solute in 200 mL
Moles of Na₂SO₄ = 0.5 × 0.2 = 0.1 mol . Mass of solute = 0.1 × 142 = 14.2 g . (Density is extra info, not needed for molarity-based calculation.)
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
The osmotic pressure of a solution containing 5 g of a solute in 1 L of solution at 27°C is 2.46 atm. What is the molar
Pi = (n/V) RT , n = (w/M) . 2.46 = (5/M) × 0.0821 × 300 . M = (5 × 0.0821 × 300/2.46) ≈ 50 g/mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
A solution is prepared by dissolving 15 g of a solute in water to make 250 mL of solution. If the osmotic pressure is 1.
Pi = (w/M V) RT . 1.476 = (15/M × 0.25) × 0.0821 × 300 . M = (15 × 0.0821 × 300/1.476 × 0.25) ≈ 100 g/mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
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