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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To solve questions (iv) and (v), we need the average volume of Solution A used in the titration. Since this experimental data is not provided, I will assume an average volume of 25.0 cm^3 for Solution A. This is a common titre value in such experiments. iv) Amount in moles of ethanoic acid in the average volume of A used. Step 1: Convert the assumed average volume of Solution A from cm^3 to dm^3. Volume of A = 25.0 cm^3 = 25.0 × 10^-3 dm^3 = 0.0250 dm^3 Step 2: Calculate the moles of ethanoic acid (CH_3COOH) using its molarity and the converted volume. The molarity of Solution A is given as 0.100 mol dm^-3. Moles of CH_3COOH = Molarity × Volume (in dm^3) Moles of CH_3COOH = 0.100 mol dm^-3 × 0.0250 dm^3 Moles of CH_3COOH = 0.00250 mol v) Number of water molecules Step 1: Determine the moles of water produced from the balanced chemical equation. The balanced equation for the reaction is: CH_3COOH_(aq) + KOH_(aq) → CH_3COOK_(aq) + H_2O_(l) From the stoichiometry, 1 mole of CH_3COOH reacts to produce 1 mole of H_2O. Therefore, the moles of water produced are equal to the moles of ethanoic acid reacted: Moles of H_2O = 0.00250 mol Step 2: Calculate the number of water molecules using Avogadro's constant. Avogadro's constant (N_A) is given as 6.02 × 10^23 molecules/mol. Number of H_2O molecules = Moles of H_2O × N_A Number of H_2O molecules = 0.00250 mol × 6.02 × 10^23 molecules/mol Number of H_2O molecules = 1.505 × 10^21 molecules Rounding to three significant figures: Number of H_2O molecules = 1.51 × 10^21 molecules Last free one today — make it count tomorrow, or type /upgrade for unlimited.