Here are the solutions to the questions: 1. (a) Define the term "mole of a substance" The mole is the SI unit for the amount of substance. It is defined as the amount of substance that contains exactly $6.02214076 \times 10^{23}$ elementary entities (such as atoms, molecules, ions, or electrons). This specific number is known as Avogadro's number ($N_A$). 1. (b) 1.50 g of ethyl ethanoate was hydrolysed with 50.0 cm³ of 0.5 M NaOH solution and then made up to 250 cm³ with distilled water in a volumetric flask. A: The number of moles of ethyl ethanoate in 1.50 g. (RMM: CH₃CO₂C₂H₅ = 88) Step 1: Identify the given mass and molar mass. Mass of ethyl ethanoate = $1.50 \text{ g}$ Molar mass (RMM) of ethyl ethanoate = $88 \text{ g/mol}$ Step 2: Calculate the number of moles using the formula: $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$. $$ \text{Moles of ethyl ethanoate} = \frac{1.50 \text{ g}}{88 \text{ g/mol}} $$ $$ \text{Moles of ethyl ethanoate} \approx 0.017045 \text{ mol} $$ $$ \text{Moles of ethyl ethanoate} = \boxed{0.0170 \text{ mol (3 s.f.)}} $$ B: The number of moles of NaOH in 50 cm³ of 0.5 M solution. Step 1: Identify the given concentration and volume. Concentration of NaOH = $0.5 \text{ M}$ (or $0.5 \text{ mol/L}$) Volume of NaOH solution = $50 \text{ cm}^3$ Step 2: Convert the volume from $\text{cm}^3$ to $\text{L}$. $$ \text{Volume in L} = 50 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} = 0.050 \text{ L} $$ Step 3: Calculate the number of moles using the formula: $\text{Moles} = \text{Concentration} \times \text{Volume}$. $$ \text{Moles of NaOH} = 0.5 \text{ mol/L} \times 0.050 \text{ L} $$ $$ \text{Moles of NaOH} = \boxed{0.025 \text{ mol}} $$ C: The percentage yield when 1.50 g of ethyl ethanoate reacts completely with sodium hydroxide to produce 0.63 g of ethanol. (RAM: C = 12.0, H = 1.0, O = 16.0, Na = 23.0) Step 1: Write the balanced chemical equation for the hydrolysis of ethyl ethanoate. $$ \text{CH}_3\text{COOC}_2\text{H}_5 \text{(aq)} + \text{NaOH (aq)} \rightarrow \text{CH}_3\text{COONa (aq)} + \text{C}_2\text{H}_5\text{OH (aq)} $$ The molar ratio between ethyl ethanoate and ethanol is $1:1$. Step 2: Calculate the molar mass of ethanol ($\text{C}_2\text{H}_5\text{OH}$). $$ \text{Molar mass of C}_2\text{H}_5\text{OH} = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0) $$ $$ \text{Molar mass of C}_2\text{H}_5\text{OH} = 24.0 + 6.0 + 16.0 = 46.0 \text{ g/mol} $$ Step 3: Calculate the theoretical moles of ethanol produced from 1.50 g of ethyl ethanoate. From part (b) A, moles of ethyl ethanoate = $0.017045 \text{ mol}$. Since the ratio is $1:1$, theoretical moles of ethanol = $0.017045 \text{ mol}$. Step 4: Calculate the theoretical mass of ethanol. $$ \text{Theoretical mass of ethanol} = \text{Moles} \times \text{Molar Mass} $$ $$ \text{Theoretical mass of ethanol} = 0.017045 \text{ mol} \times 46.0 \text{ g/mol} $$ $$ \text{Theoretical mass of ethanol} \approx 0.78407 \text{ g} $$ Step 5: Calculate the percentage yield. Actual yield of ethanol = $0.63 \text{ g}$ (given) $$ \text{Percentage yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% $$ $$ \text{Percentage yield} = \frac{0.63 \text{ g}}{0.78407 \text{ g}} \times 100\% $$ $$ \text{Percentage yield} \approx 80.35 \% $$ $$ \text{Percentage yield} = \boxed{80.4\% \text{ (3 s.f.)}} $$