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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Welcome back — been a while! Let's pick up where you left off. Here are the calculations for the concentrations of solutions 'An' and 'Bn' based on the provided instructions. a) Solution 'An' (H₂SO₄) Step 1: Calculate the mass of the concentrated H₂SO₄. The relative density (R.D.) of concentrated H₂SO₄ is 1.84. This means its density is 1.84 g/cm^3. The volume of concentrated H₂SO₄ per dm^3 of solution is 5.60 cm^3. Mass of concentrated H_2SO_4 = Density × Volume Mass of concentrated H_2SO_4 = 1.84 g/cm^3 × 5.60 cm^3 = 10.304 g Step 2: Calculate the mass of pure H₂SO₄. The purity of the concentrated H₂SO₄ is 98\%. Mass of pure H_2SO_4 = Mass of concentrated H_2SO_4 × Purity Mass of pure H_2SO_4 = 10.304 g × 0.98 = 10.09792 g Step 3: Calculate the moles of pure H₂SO₄. The molar mass of H₂SO₄ is 98.0 g/mol. Moles of H_2SO_4 = Mass of pure H_2SO_4Molar mass of H_2SO_4 Moles of H_2SO_4 = 10.09792 g98.0 g/mol = 0.10304 mol Step 4: Calculate the molarity of solution 'An'. The problem states this amount of H₂SO₄ is present per dm^3 of solution. Molarity of solution 'An' = Moles of H_2SO_4Volume of solution in dm^3 Molarity of solution 'An' = 0.10304 mol1 dm^3 = 0.10304 M Rounding to three significant figures: Molarity of solution 'An' = 0.103 M b) Solution 'Bn' (NaOH) Step 1: Calculate the moles of NaOH. The mass of NaOH per dm^3 of solution is 4.0 g. The molar mass of NaOH is 40.0 g/mol. Moles of NaOH = Mass of NaOHMolar mass of NaOH Moles of NaOH = 4.0 g40.0 g/mol = 0.10 mol Step 2: Calculate the molarity of solution 'Bn'. The problem states this amount of NaOH is present per dm^3 of solution. Molarity of solution 'Bn' = Moles of NaOHVolume of solution in dm^3 Molarity of solution 'Bn' = 0.10 mol1 dm^3 = 0.10 M Rounding to two significant figures: Molarity of solution 'Bn' = 0.10 M Send me the next one 📸