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
Okay, Ruddy, let's present the titration calculations in a clear, step-by-step format. Titration Calculations Step 1: Calculate the average volume of hydrochloric acid (HCl) used. The consistent readings are 23.70 cm^3 and 18.80 cm^3. Average volume of HCl = 23.70 cm^3 + 18.80 cm^32 Average volume of HCl = 42.50 cm^32 Average volume of HCl = 21.25 cm^3 Step 2: Calculate the number of moles of hydrochloric acid (HCl) used. Given concentration of HCl = 0.100 mol/dm^3. Volume of HCl = 21.25 cm^3 = 0.02125 dm^3. Moles of HCl = Concentration × Volume Moles of HCl = 0.100 mol/dm^3 × 0.02125 dm^3 Moles of HCl = 0.002125 mol Step 3: Write the balanced chemical equation for the reaction. Assuming solution A is sodium carbonate (Na_2CO_3): Na_2CO_3(aq) + 2HCl(aq) → 2NaCl(aq) + H_2O(l) + CO_2(g) Step 4: Calculate the number of moles of solution A (Na_2CO_3) in 25.0 cm^3. From the balanced equation, the mole ratio of Na_2CO_3 to HCl is 1:2. Moles of Na_2CO_3 = Moles of HCl2 Moles of Na_2CO_3 = 0.002125 mol2 Moles of Na_2CO_3 = 0.0010625 mol Step 5: Calculate the concentration of solution A (Na_2CO_3) in mol/dm^3. Volume of solution A = 25.0 cm^3 = 0.0250 dm^3. Concentration of Na_2CO_3 = MolesVolume Concentration of Na_2CO_3 = 0.0010625 mol0.0250 dm^3 Concentration of Na_2CO_3 = 0.0425 mol/dm^3 Step 6: Calculate the molar mass of Na_2CO_3. Atomic masses: Na = 22.99, C = 12.01, O = 16.00. Molar mass of Na_2CO_3 = (2 × 22.99) + 12.01 + (3 × 16.00) Molar mass of Na_2CO_3 = 45.98 + 12.01 + 48.00 Molar mass of Na_2CO_3 = 105.99 g/mol Step 7: Calculate the concentration of solution A (Na_2CO_3) in g/dm^3. Concentration (g/dm^3) = Concentration (mol/dm^3) × Molar mass Concentration (g/dm^3) = 0.0425 mol/dm^3 × 105.99 g/mol Concentration (g/dm^3) = 4.50 g/dm^3 (to 3 significant figures) Step 8: Calculate the percentage purity of sample A. Original concentration of sample A = 5.00 g/dm^3. Percentage purity = Concentration of pure Na_2CO_3Original concentration of sample A × 100\% Percentage purity = 4.50 g/dm^35.00 g/dm^3 × 100\% Percentage purity = 0.900 × 100\% Percentage purity = 90.0\% Last free one today — make it count tomorrow, or type /upgrade for unlimited.