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.
ScanSolve AI Answer
To solve these questions, I need some information that was likely provided in earlier parts of the exam paper, specifically the formula of the hydrated salt B and the concentration of acid A. Since these are not in the current image, I will make reasonable assumptions based on typical WASSCE Chemistry practicals to demonstrate the calculation method. Assumptions: For part (ii), I will assume that the hydrated salt B is sodium carbonate decahydrate (Na_2CO_3 · 10H_2O), as its anhydrous form (Na_2CO_3) has a molar mass of 106 g mol^-1. For part (iii), I will use the average titre value of acid A as 24.65 cm^3 (from our previous hypothetical calculation). I will also assume that acid A is a monoprotic acid (e.g., HCl) and its concentration is 0.100 mol dm^-3, which is a common concentration in such experiments. --- (ii) percentage of water of hydration in B; Step 1: Determine the molar mass of water of hydration. The molar mass of water (H_2O) is 2(1.0) + 16.0 = 18.0 g mol^-1. For Na_2CO_3 · 10H_2O, the molar mass of water of hydration is 10 × 18.0 g mol^-1 = 180 g mol^-1. Step 2: Determine the molar mass of the hydrated salt B. The molar mass of anhydrous salt B (Na_2CO_3) is given as 106 g mol^-1. The molar mass of the hydrated salt B (Na_2CO_3 · 10H_2O) is the sum of the molar mass of the anhydrous salt and the molar mass of the water of hydration. Molar mass of hydrated B = 106 g mol^-1 + 180 g mol^-1 = 286 g mol^-1 Step 3: Calculate the percentage of water of hydration. Percentage of water of hydration = Molar mass of water of hydrationMolar mass of hydrated salt × 100\% Percentage of water of hydration = 180 g mol^-1286 g mol^-1 × 100\% Percentage of water of hydration = 0.62937 × 100\% Percentage of water of hydration = 62.94\% --- (iii) number of moles of hydrogen ions in the average titre value. Step 1: Convert the average titre volume to dm^3. The average titre value of acid A is 24.65 cm^3. Volume of acid A = 24.65 cm^3 × 1 dm^31000 cm^3 = 0.02465 dm^3 Step 2: Calculate the number of moles of acid A. Using the assumed concentration of acid A as $0.100 mol