Amount of Substance Revision Notes for Grade 10 NSC Matric Physical Sciences

Amount of Substance

Molar volumes of gases

When working with gases, it's helpful to know that all gases behave similarly under the same conditions. At standard temperature and pressure, every gas follows the same rule for volume.

Molar volume of gases: One mole of any gas occupies 22.4 dm³ at standard temperature and pressure.

This principle applies to any gas when it's at standard temperature and pressure conditions. Understanding this concept becomes more important as you progress through your chemistry studies, particularly when learning about gas laws in Grade 11.

infoNote

Standard temperature and pressure (S.T.P.) refers to specific conditions:

Temperature: 273.15 K (0°C)
Pressure: 0.986 atm

Practical examples of molar volume

Let's see how this works with real gases:

2 mol of hydrogen gas (H₂) will occupy 44.8 dm³ at S.T.P.
67.2 dm³ of ammonia gas (NH₃) contains 3 mol of ammonia

This relationship makes it easy to convert between moles and volume for gases at standard conditions.

Molar concentrations of liquids

Solutions are formed when we dissolve a solid substance (the solute) in a liquid (usually water, which acts as the solvent). The resulting mixture is called a solution.

The concentration tells us how much solute is dissolved in a specific volume of liquid. Understanding concentration is crucial for many chemical calculations and practical applications.

Concentration: A measure of the amount of solute dissolved in a given volume of liquid, measured in mol·dm⁻³.

Mathematically, we express concentration using the formula:

$C = \frac{n}{V}$

This formula can be rearranged using the triangle method to solve for any variable:

Where:

C = concentration (mol·dm⁻³)
n = number of moles of solute
V = volume of solution (dm³)

chatImportant

Note that volume is measured in dm³, which equals litres. Therefore, concentration units are mol·dm⁻³ (or mol/L). Always convert volume units to dm³ when calculating concentration.

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Worked Example 1: Finding concentration from mass

Question: If 3.5 g of sodium hydroxide (NaOH) is dissolved in 2.5 dm³ of water, what is the concentration of the solution in mol·dm⁻³?

Solution:

Step 1: Find the number of moles of sodium hydroxide

$n = \frac{m}{M} = \frac{3.5 \text{ g}}{40.01 { g·mol}^{-1}} = 0.0875 \text{ mol}$

Step 2: Calculate the concentration

$C = \frac{n}{V} = \frac{0.0875 { mol}}{2.5 \text{ dm}^3} = 0.035 { mol·dm}^{-3}$

The concentration of the solution is 0.035 mol·dm⁻³.

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Worked Example 2: Finding mass from concentration

Question: You have a 1 dm³ container and need to prepare a solution of potassium permanganate (KMnO₄). What mass of KMnO₄ is needed to make a solution with a concentration of 0.2 mol·dm⁻³?

Solution:

Step 1: Calculate the number of moles needed

$C = \frac{n}{V}, \text{ therefore: } n = C \times V = 0.2 { mol·dm}^{-3} \times 1 \text{ dm}^3 = 0.2 \text{ mol}$

Step 2: Find the mass required

$m = n \times M = 0.2 \text{ mol} \times 158 { g·mol}^{-1} = 31.6 \text{ g}$

The mass of KMnO₄ needed is 31.6 g.

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Worked Example 3: Volume conversion and calculations

Question: How much sodium chloride (in g) will you need to prepare 500 cm³ of solution with a concentration of 0.01 mol·dm⁻³?

Solution:

Step 1: Convert the given volume to the correct units

$V = 500 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3} = 0.5 \text{ dm}^3$

Step 2: Find the number of moles needed

$n = C \times V = 0.01 { mol·dm}^{-3} \times 0.5 \text{ dm}^3 = 0.005 \text{ mol}$

Step 3: Find the mass required

$m = n \times M = 0.005 { mol} \times 58.45 { g·mol}^{-1} = 0.29 \text{ g}$

The mass of sodium chloride needed is 0.29 g.

Introduction to stoichiometric calculations

Stoichiometry involves calculating the quantities of reactants and products in chemical reactions. This skill helps you determine how much product will form in a chemical reaction, making it essential for practical chemistry applications.

The principles you've learned about moles, concentration, and molar volumes form the foundation for more complex stoichiometric calculations that you'll encounter in advanced chemistry topics.

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Key Points to Remember:

One mole of any gas occupies 22.4 dm³ at standard temperature and pressure (S.T.P.)
Concentration formula: $C = \frac{n}{V}$ , where C is in mol·dm⁻³
Always convert volume units to dm³ when calculating concentration
The triangle method helps rearrange the concentration formula for different variables
Stoichiometry builds on these mole concepts to solve complex reaction problems

Amount of Substance (Grade 10 NSC Matric Physical Sciences): Revision Notes