57 public questions tagged with this topic.
Moles of NaOH = (4/40) = 0.1 mol . Molarity = (Moles/Volume in L) , so 0.5 = (0.1/V) . Volume = (0.1/0.5) = 0.2 L = 200 mL . Since total volume is 200 mL, water volume = 200 mL (assuming solute volume is negligible).
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Moles of CaCl₂ = 0.2 × 0.6 = 0.12 mol . Mass = 0.12 × 111 = 13.32 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Molarity = (Moles of solute/Volume in liters) . Moles = Molarity × Volume = 0.2 mol/L × 0.5 L = 0.1 mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Moles of NaOH = (4/40) = 0.1 mol . Volume = 250 mL = 0.25 L. Molarity = (0.1/0.25) = 0.4 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Moles of KCl = 0.4 × 0.25 = 0.1 mol . Mass = 0.1 × 74.5 = 7.45 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Moles of solute = (8/80) = 0.1 mol . Volume = 400 mL = 0.4 L. Molarity = (0.1/0.4) = 0.25 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Moles of solute = (9/90) = 0.1 mol . Volume = 450 mL = 0.45 L. Molarity = (0.1/0.45) ≈ 0.2222 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Moles = M × V = 0.5 × 0.4 = 0.2 mol . Molar mass = (Mass/Moles) = (20/0.2) = 100 g/mol . (Density is extra info, not needed here.)
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = M · RT . 0.984 = M × 0.0821 × 300 . M = (0.984/0.0821 × 300) ≈ 0.04 M . Cross-check: Moles = 0.04 × 0.25 = 0.01 , but given 0.02 mole, i = 2 , adjust assumption if needed.
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Moles of glucose = (9/180) = 0.05 mol . Volume = 500 mL = 0.5 L. Molarity = (0.05/0.5) = 0.1 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
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
Moles = 0.6 × 0.2 = 0.12 mol . Mass of NaCl = 0.12 × 58.5 = 7.02 g . Total mass = 200 × 1.05 = 210 g . Mass of water = 210 - 7.02 = 202.98 g = 0.20298 kg . Molality = (0.12/0.20298) ≈ 0.591 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
