Gases and Solutions (Grade 11 NSC Matric Physical Sciences): Revision Notes

Worked Example: Molar Gas Volume

Question: What is the volume of 2.3 mol of hydrogen gas at STP?

Solution:

Use the molar gas volume formula

$V_g = 22.4n_g$

$V_g = 22.4 \times 2.3$

$V_g = 51.52 dm³$

Worked Example: Volume and Gases

Question: Hydrogen and oxygen react to form water according to: $2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{g})$

If 3 dm³ of oxygen is used, what volume of water is produced?

Solution:

Step 1: Identify the stoichiometric coefficients

From the equation: $2\text{H}_2 + 1\text{O}_2 \rightarrow 2\text{H}_2\text{O}$

• Coefficient of O₂ (b) = 1
• Coefficient of H₂O (a) = 2

Step 2: Apply the volume relationship formula

$V_A = \frac{a}{b}V_B$

$V_{\text{H}_2\text{O}} = \frac{2}{1} \times V_{\text{O}_2}$

$V_{\text{H}_2\text{O}} = 2 \times 3 = 6 dm³$

Worked Example: Gas Phase Calculations

Question: What volume of oxygen at STP is needed for the complete combustion of 3.3 dm³ of propane (C₃H₈)? The products are CO₂ and H₂O.

Solution:

Step 1: Write the balanced equation

$\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{g})$

Step 2: Use the volume relationship

$V_A = \frac{a}{b}V_B$

$V_{\text{O}_2} = \frac{5}{1} \times V_{\text{C}_3\text{H}_8}$

$V_{\text{O}_2} = 5 \times 3.3 = 16.5 dm³$

Worked Example: Concentration Calculations

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

Solution:

Step 1: Convert volume to dm³

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

Step 2: Calculate moles of NaCl needed

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

$0.01 = \frac{n}{0.5}$

$n = 0.005 \text{ mol}$

Step 3: Convert moles to mass

$m = nM = 0.005 \times 58 = 0.29 g$

The mass of sodium chloride needed is 0.29 g.

Worked Example: Titration Calculation

Question: 25 cm³ of 0.2 mol·dm⁻³ hydrochloric acid solution was pipetted into a conical flask and titrated with sodium hydroxide. 15 cm³ of sodium hydroxide was needed to neutralise the acid. Calculate the concentration of the sodium hydroxide.

Equation: $\text{NaOH}(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})$

Solution: Step 1: List known information

• NaOH: $V = 15$ cm³
• HCl: $V = 25$ cm³, $C = 0.2$ mol·dm⁻³
• Equation is balanced (coefficients = 1)

Step 2: Convert volumes to dm³

$V_{\text{NaOH}} = \frac{15}{1000} = 0.015 \text{ dm}^3$

$V_{\text{HCl}} = \frac{25}{1000} = 0.025 \text{ dm}^3$

Step 3: Apply titration formula

$\frac{C_A V_A}{a} = \frac{C_B V_B}{b}$

$\frac{(0.2)(0.025)}{1} = \frac{(C_{\text{NaOH}})(0.015)}{1}$

$0.005 = 0.015 \times C_{\text{NaOH}}$

$C_{\text{NaOH}} = 0.33 mol·dm⁻³$