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1 Solutions 2 Electrochemistry 3 Chemical Kinetics 4 The d-and f-Block Elements 5 Coordination Compounds 6 Haloalkanes and Haloarenes 7 Alcohols, Phenols and Ethers 8 Aldehydes, Ketones and Carboxylic Acids 9 Amines 10 Biomolecules

1 Solutions

1.1 Types of Solutions 1.2 Expressing Concentration of Solutions 1.3 Solubility 1.4 Vapour Pressure of Liquid Solutions 1.5 Ideal and Non-ideal Solutions 1.6 Colligative Properties and Determination of Molar Mass 1.7 Abnormal Molar Masses 2.1 Electrochemical Cells 2.2 Galvanic Cells 2.3 Nernst Equation 2.4 Conductance of Electrolytic Solutions 2.5 Electrolytic Cells and Electrolysis 2.6 Batteries 2.7 Fuel Cells 2.8 Corrosion

Expressing Concentration of Solutions

NCERT Reference:Chapter 1 – Solutions – Page 2

Quick Notes

  • Concentration: The amount of solute present in a given quantity of solution or solvent.
  • Mass % (w/w) = (mass of solute / mass of solution) × 100
  • Volume % (v/v) = (volume of solute / volume of solution) × 100
  • Mass by volume % = (mass of solute / volume of solution in mL) × 100
  • Parts per million (ppm) = (mass of component / total mass of solution) × 106
  • Mole fraction (χ) = moles of component / total moles of all components
  • Molarity (M) = moles of solute / volume of solution in litres
  • Molality (m) = moles of solute / mass of solvent in kg

Full Notes

The concentration of a solution refers to the quantity of solute present in a given amount of solvent or solution.

There are several standard ways of expressing the concentration of a solution. These are chosen based on the nature of the experiment, accuracy required, and the physical state of the solute and solvent.

Mass Percentage (w/w)

Used when both solute and solvent are solids or when mass measurement is convenient.

Formula:

NCERT 12 Chemistry formula image showing mass percentage equals mass of solute divided by mass of solution times 100.

Volume Percentage (v/v)

Used when both solute and solvent are liquids.

Formula:

NCERT 12 Chemistry formula image showing volume percentage equals volume of solute divided by volume of solution times 100.

Mass by Volume Percentage (w/v)

Frequently used in medical or laboratory settings.

Mass of solute dissolved in 100 mL of solution

Example: A 5% w/v NaCl solution contains 5 g NaCl in 100 mL solution.

Parts per Million (ppm)

Used for expressing very low concentrations, especially in environmental chemistry.

Formula:

NCERT 12 Chemistry formula image showing parts per million equals mass of component divided by total mass of solution times 10^6.

Example: 1 ppm = 1 mg solute per kg solution = 1 mg/L for water.

Mole Fraction (χ)

A ratio of moles, useful in thermodynamics and phase equilibrium.

Formula:

NCERT 12 Chemistry formula image defining mole fraction as moles of a component divided by total moles of all components.

Mole fraction of component A:

Where nA and nB are moles of A and B respectively. The sum of all mole fractions in a solution is always 1.

Molarity (M)

Defined as the number of moles of solute per litre of solution.

Formula:

NCERT 12 Chemistry formula image showing molarity equals moles of solute divided by volume of solution in litres.

Note: Molarity is temperature dependent because volume changes with temperature.

Molality (m)

Number of moles of solute per kg of solvent.

Formula:

NCERT 12 Chemistry formula image showing molality equals moles of solute divided by mass of solvent in kilograms.

m = n / mass of solvent (kg) (1.9)

Advantage: Molality is temperature independent as it is based on mass, not volume.

Comparison of Molarity and Molality

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Matt’s exam tip

Make sure you know the difference between molarity and molality — both are used in chemistry and it is easy to get the two confused.

Aspect Molarity (M) Molality (m)
Definition moles per litre moles per kg of solvent
Temperature Dependent Independent
Use case Solutions with fixed volume Thermodynamic calculations
Worked Example

Q: A solution contains 20 g of NaOH in 500 g of water. What is its molality?

  1. Molar mass of NaOH = 40 g/mol
  2. Moles of NaOH = 20 / 40 = 0.5 mol
  3. Mass of solvent (water) = 500 g = 0.5 kg
  4. Molality = 0.5 mol / 0.5 kg = 1 m

Summary