This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions for question 1(b): 1. (b) 1.50 g of ethyl ethanoate was hydrolysed with 50.0 cm^3 of 0.5 M NaOH solution and the mixture was made up to 250 cm^3 with distilled water in a volumetric flask. A: The number of moles of ethyl ethanoate in 1.50 g. (RMM: CH_3COOC_2H_5 = 88) Step 1: Use the formula for moles. Moles = MassRelative Molecular Mass (RMM) Step 2: Substitute the given values. Moles of ethyl ethanoate = 1.50 g88 g/mol Moles of ethyl ethanoate = 0.017045... mol Step 3: Round to an appropriate number of significant figures (e.g., 3 s.f.). Moles of ethyl ethanoate ≈ 0.0170 mol The number of moles of ethyl ethanoate is 0.0170 mol. B: The number of moles of NaOH in 50 cm^3 of 0.5 M solution. Step 1: Convert the volume from cm^3 to dm^3. Volume in dm^3 = 50.0 cm^31000 cm^3/dm^3 = 0.050 dm^3 Step 2: Use the formula for moles in solution. Moles = Concentration × Volume (in dm^3) Step 3: Substitute the given values. Moles of NaOH = 0.5 mol/dm^3 × 0.050 dm^3 Moles of NaOH = 0.025 mol The number of moles of NaOH is 0.025 mol. C: The percentage yield when 1.50 g of ethyl ethanoate reacts completely with sodium hydroxide to give 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. CH_3COOC_2H_5 (aq) + NaOH (aq) → CH_3COONa (aq) + C_2H_5OH (aq) From the equation, 1 mole of ethyl ethanoate produces 1 mole of ethanol. Step 2: Calculate the molar mass of ethanol (C_2H_5OH). RMM of C_2H_5OH = (2 × 12.0) + (6 × 1.0) + (1 × 16.0) RMM of C_2H_5OH = 24.0 + 6.0 + 16.0 = 46.0 g/mol Step 3: Determine the theoretical moles of ethanol produced from 1.50 g of ethyl ethanoate. From part A, moles of ethyl ethanoate = 0.017045 mol. Since the mole ratio is 1:1, theoretical moles of ethanol = 0.017045 mol. Step 4: Calculate the theoretical mass of ethanol. Theoretical mass of ethanol = Theoretical moles × RMM of ethanol Theoretical mass of ethanol = 0.017045 mol × 46.0 g/mol Theoretical mass of ethanol = 0.78407 g Step 5: Calculate the percentage yield. Percentage yield = Actual yieldTheoretical yield × 100\% Given actual yield of ethanol = 0.63 g. Percentage yield = 0.63 g0.78407 g × 100\% Percentage yield = 80.35\% Step 6: Round to an appropriate number of significant figures (e.g., 2 s.f. based on 0.63 g). Percentage yield ≈ 80\% The percentage yield is 80%.