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AQA GCSE Chemistry Notes

3.3.1 Concentration Measurements in Chemistry

Understanding Concentration in Solutions

Concentration is a measure of how much of a given substance (the solute) is mixed with another substance (the solvent). It's a fundamental concept in chemistry, crucial for preparing solutions, conducting experiments, and understanding chemical reactions.

Units of Concentration

Grams per Decimetre Cubed (g/dm³)

  • Definition: It represents the mass of the solute in grams present in one cubic decimetre (1 dm³) of solution.
  • Usage: This unit is common when the exact molecular composition of the solute is unknown or when dealing with non-ionic substances.

Moles per Decimetre Cubed (mol/dm³)

  • Definition: This unit, also known as molarity, measures the number of moles of a substance in one cubic decimetre of solution.
  • Significance: Molarity is widely used in chemical stoichiometry as it directly relates to the number of particles in a solution, which is essential for reaction equations.
Formula of molarity

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Calculating Concentration

Method 1: Concentration in g/dm³

  • Formula: Concentration (g/dm³) = Mass of Solute (g) / Volume of Solution (dm³)
  • Example Calculation:
    • Suppose we dissolve 50g of sugar in 250cm³ of water. First, convert the volume from cm³ to dm³ (250cm³ = 0.25dm³).
    • Calculate the concentration: 50g / 0.25dm³ = 200g/dm³.
  • Practical Considerations:
    • Ensure the mass of the solute is accurately measured using a balance.
    • Volume conversions are essential; remember that 1000cm³ is equivalent to 1dm³.

Method 2: Concentration in mol/dm³

  • Formula: Concentration (mol/dm³) = Number of Moles of Solute / Volume of Solution (dm³)
  • Moles Calculation: Moles = Mass (g) / Molar Mass (g/mol)
  • Example Calculation:
    • Consider dissolving 117g of NaCl (molar mass = 58.5g/mol) in 2dm³ of water.
    • Calculate moles of NaCl: 117g / 58.5g/mol = 2 moles.
    • Calculate concentration: 2 moles / 2dm³ = 1mol/dm³.
  • Key Points:
    • Molar mass must be determined accurately. It can be found using the periodic table and adding up the atomic masses of the constituent elements.
    • When measuring volume, use a volumetric flask for precision.

Practical Application and Importance

Preparing Solutions

  • Accuracy: Accurate measurement is essential for preparing solutions with the desired concentration.
  • Equipment: Use a digital balance for solute mass and a volumetric flask or graduated cylinder for solvent volume.
Chemistry laboratory equipments

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Importance in Chemical Reactions

  • Reproducibility: Correct concentration ensures that experiments can be replicated with consistent results.
  • Stoichiometry: In chemical equations, reactants and products are often expressed in moles, making molarity a key factor in calculations.

Common Mistakes and Tips

  • Volume Conversion: Be vigilant about converting cm³ to dm³ in calculations.
  • Measurement Accuracy: Use calibrated equipment for both mass and volume measurements.
  • Temperature Effects: Remember that the volume of liquids can change with temperature, affecting concentration.

Advanced Concepts

Dilution of Solutions

  • Concept: Dilution involves adding more solvent to a solution, decreasing its concentration.
  • Formula: C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume after dilution.
  • Application: This is particularly useful in laboratories for preparing solutions of desired concentrations from more concentrated stock solutions.
Dilution of Solutions- C1V1 = C2V2

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Concentration in Gases

  • Volume at RTP: The concentration of gases can also be expressed in terms of volume, especially at Room Temperature and Pressure (RTP).
  • Molar Volume: At RTP, one mole of any gas occupies 24dm³. This is a crucial factor in calculations involving gaseous reactants or products.

In conclusion, understanding and accurately measuring concentration in various units is a cornerstone in IGCSE Chemistry. It underpins many laboratory techniques and theoretical concepts. Students should practice these calculations regularly and be mindful of the common pitfalls to gain proficiency in this fundamental area of chemistry. Remember, accuracy in concentration measurements is not just about getting the right answer; it's about understanding the relationships between different chemical quantities.

FAQ

The presence of a solute affects the boiling point of a solution through a phenomenon known as boiling point elevation. When a non-volatile solute is added to a solvent, it causes the boiling point of the solvent to increase. This happens because the solute particles disrupt the solvent molecules' ability to escape into the gas phase, which requires a higher temperature to achieve.

The degree of boiling point elevation depends on the number of solute particles in the solution, not their identity. This is a colligative property, meaning it is affected by the concentration of particles in the solution. For every mole of solute particles dissolved in one kilogram of solvent, the boiling point is elevated by a certain amount, known as the ebullioscopic constant, which is specific to each solvent. Understanding this concept is important in fields like cooking (e.g., boiling point of saltwater), industrial processes (e.g., antifreeze solutions), and research (e.g., molecular weight determination of macromolecules).

Considering the dissociation of ionic compounds in solutions is crucial when calculating concentration in mol/dm³ because these compounds dissociate into ions when dissolved. For example, NaCl dissociates into Na⁺ and Cl⁻ ions. This dissociation affects the number of particles in the solution, which in turn impacts the solution's properties, such as conductivity and boiling point.

When calculating molarity, if the solute is an ionic compound, you must account for all the ions produced. For instance, if 1 mole of NaCl is dissolved, it produces 1 mole of Na⁺ and 1 mole of Cl⁻, resulting in a total of 2 moles of particles in solution. Neglecting this fact can lead to underestimating the solution’s osmotic pressure, electrical conductivity, and other colligative properties. This understanding is vital in both theoretical calculations and practical applications, such as in determining the correct dosage of electrolytes in medical solutions or the salt concentration in a chemical reaction.

To convert concentration from ppm (parts per million) to g/dm³, it's important to understand what ppm represents. Ppm is a way of expressing very dilute concentrations of substances. One ppm implies one part of the solute in one million parts of the solution. To convert ppm to g/dm³, you use the fact that 1 ppm is equivalent to 1 mg of solute in 1 dm³ (1000 L) of solution. Therefore, to convert, simply divide the ppm value by 1000. For instance, a concentration of 50 ppm is equivalent to 50 mg/dm³, which is the same as 0.050 g/dm³. This conversion is especially useful in environmental chemistry, where pollutant concentrations are often given in ppm or ppb (parts per billion) and need to be converted into more standard units for regulatory or comparative purposes.

Molarity and molality are both measures of concentration, but they differ in how they are calculated. Molarity, denoted as M, is the number of moles of solute per litre of solution. It's a volume-based measure and is affected by temperature changes, as the volume of solutions can expand or contract with temperature. Molarity is commonly used in stoichiometry, reactions in solutions, and titrations because of its direct relation to volume, which is easy to measure in a laboratory setting.

On the other hand, molality, denoted as m, is the number of moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by temperature since it's based on mass, not volume. This makes it particularly useful in situations where temperature varies significantly, such as in boiling point elevation or freezing point depression studies. Molality is less commonly used than molarity but provides more accurate results in thermochemical calculations where temperature variance is a factor.

Weight/volume percentage concentration (w/v%) is often used in situations where ease of preparation and practicality are more important than extreme precision. This method of concentration measurement expresses the mass of solute in grams per 100 ml of solution. It is straightforward and useful in everyday laboratory practices, especially in biology and medicine, where solutions are often prepared by directly weighing the solute and adding a specific volume of solvent.

Chemists might choose w/v% over molarity in scenarios where the exact number of moles is less critical, or when working with solid solutes and liquid solvents. For example, in preparing media for cell cultures, pharmaceutical formulations, or when dealing with very dilute or very concentrated solutions. It's also commonly used in industrial processes and product formulations, where ease and speed of preparation are crucial, and the effects of temperature on volume are not a significant concern. Additionally, w/v% is more intuitive for non-chemists, making it a preferred choice in interdisciplinary fields.

Practice Questions

Calculate the concentration in g/dm³ of a solution prepared by dissolving 20g of sodium chloride in 250cm³ of water.

To calculate the concentration in g/dm³, we first convert the volume of the solution from cm³ to dm³. Since 1 dm³ is equivalent to 1000 cm³, 250 cm³ converts to 0.25 dm³. The concentration in g/dm³ is calculated by dividing the mass of the solute by the volume of the solution in dm³. Thus, the concentration = ( \frac{20\, \text{g}}{0.25\, \text{dm}³} = 80\, \text{g/dm}³ ). Therefore, the concentration of the sodium chloride solution is 80 g/dm³.

A student prepares a solution by dissolving 36g of glucose (C₆H₁₂O₆) in enough water to make 300cm³ of solution. Given the molar mass of glucose is 180g/mol, calculate the concentration of the solution in mol/dm³.

First, we calculate the number of moles of glucose. The number of moles is given by the mass of the substance divided by its molar mass. Here, the number of moles of glucose = ( \frac{36\, \text{g}}{180\, \text{g/mol}} = 0.2\, \text{mol} ). Next, we convert the volume of the solution from cm³ to dm³. The volume of 300 cm³ is equivalent to 0.3 dm³. The concentration in mol/dm³ is calculated as the number of moles divided by the volume in dm³, which is ( \frac{0.2\, \text{mol}}{0.3\, \text{dm}³} = 0.67\, \text{mol/dm}³ ). Therefore, the concentration of the glucose solution is 0.67 mol/dm³.

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