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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Fresh day Hurry, let's solve. Here are the solutions to your chemistry questions. Titration Problem: Assumption: Since the average titre value for solution A (HCl) is not provided, I will assume an average titre volume of 20.00 cm^3 for the calculations. Given Information: Solution A: HCl, prepared by dissolving 7.3 g of HCl gas in 1 dm^3 of solution. Solution B: Na_2CO_3, contains 2.65 g in 250 cm^3 of solution. Volume of B used for titration = 20 cm^3. Reaction: Na_2CO_3(aq) + 2HCl(aq) → 2NaCl(aq) + H_2O(l) + CO_2(g) Relative atomic masses: H=1.0, C=12.0, O=16.0, Na=23.0, Cl=35.5. Molar volume of gas at S.T.P. = 22.4 dm^3. Molar Masses: Molar mass of Na_2CO_3 = (2 × 23.0) + 12.0 + (3 × 16.0) = 46.0 + 12.0 + 48.0 = 106.0 g/mol. Molar mass of HCl = 1.0 + 35.5 = 36.5 g/mol. Calculations: 1. (i) Concentration of B in mol dm^-3 Step 1: Calculate the number of moles of Na_2CO_3 in 250 cm^3. Moles of Na_2CO_3 = MassMolar mass = 2.65 g106.0 g/mol = 0.025 mol Step 2: Calculate the concentration of Na_2CO_3 in mol dm^-3. Concentration of B = MolesVolume in dm^3 = 0.025 mol250 cm^3 × 1 dm^31000 cm^3 = 0.025 mol0.250 dm^3 = 0.100 mol dm^-3 The concentration of B is 0.100 mol dm^-3. 1. (ii) Concentration of A in mol dm^-3 Step 1: Calculate the number of moles of Na_2CO_3 in the 20 cm^3 portion used for titration. Moles of Na_2CO_3 in 20 cm^3 = Concentration × Volume in dm^3 = 0.100 mol dm^-3 × (20 cm^3 × 1 dm^31000 cm^3) = 0.100 mol dm^-3 × 0.020 dm^3 = 0.002 mol Step 2: Use the mole ratio from the balanced equation: 1 mol of Na_2CO_3 reacts with 2 mol of HCl. Moles of HCl = 2 × Moles of Na_2CO_3 = 2 × 0.002 mol = 0.004 mol Step 3: Calculate the concentration of A (HCl) using the assumed average titre volume of 20.00 cm^3. Concentration of A = Moles of HClVolume of A in dm^3 = 0.004 mol20.00 cm^3 × 1 dm^31000 cm^3 = 0.004 mol0.020 dm^3 = 0.200 mol dm^-3 The concentration of A is 0.200 mol dm^-3. 1. (iii) Concentration of A in g dm^-3 Step 1: Convert the molar concentration of A (HCl) to mass concentration using its molar mass. Concentration of A in g dm^-3 = Concentration in mol dm^-3 × Molar mass of HCl = 0.200 mol dm^-3 × 36.5 g/mol = 7.30 g dm^-3 The concentration of A is 7.30 g dm^-3. 1. (iv) Volume of hydrogen chloride gas dissolved in 1 dm^3 of solution A at S.T.P. Step 1: The problem states that solution A is prepared by dissolving 7.3 g of hydrogen chloride gas in 1 dm^3 of solution. We need to find the volume this mass of gas would occupy at S.T.P. Moles of HCl gas = MassMolar mass = 7.3 g36.5 g/mol = 0.20 mol Step 2: Use the molar volume of gas at S.T.P. (22.4 dm^3/mol) to find the volume. Volume of HCl gas = Moles × Molar volume at S.T.P. = 0.20 mol × 22.4 dm^3/mol = 4