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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Step 1: Calculate the volume of solution K used for each trial. Trial I: Volume used = Final reading - Initial reading = 5.0 \, cm^3 - 0.0 \, cm^3 = 5.0 \, cm^3 Trial II: Volume used = Final reading - Initial reading = 10.1 \, cm^3 - 5.0 \, cm^3 = 5.1 \, cm^3 Trial III: Volume used = Final reading - Initial reading = 15.3 \, cm^3 - 10.1 \, cm^3 = 5.2 \, cm^3 Step 2: Calculate the average volume of solution K used. The concordant titres are 5.1 \, cm^3 and 5.2 \, cm^3. We average these values. Average volume of K = 5.1 \, cm^3 + 5.2 \, cm^32 Average volume of K = 10.3 \, cm^32 Average volume of K = 5.15 \, cm^3 The average volume of solution K used is 5.15 cm^3. b) Determine the concentration of solution L in moles per litre. Step 1: Calculate the molar mass of sodium hydroxide (NaOH). Given atomic masses: Na = 23, O = 16, H = 1. M_r(NaOH) = 23 + 16 + 1 = 40 \, g/mol Step 2: Calculate the concentration of solution L (NaOH) in moles per litre. Solution L contains 2 \, g of NaOH per litre. Concentration of L = Mass concentrationMolar mass Concentration of L = 2 \, g/L40 \, g/mol Concentration of L = 0.05 \, mol/L The concentration of solution L is 0.05 mol/L. c) Determine the number of moles of solution L that reacted with solution K. Step 1: Convert the volume of solution L from cm^3 to litres. The volume of solution L pipetted was 25 \, cm^3. Volume of L = 25 \, cm^3 = (25)/(1000) \, L = 0.025 \, L Step 2: Calculate the number of moles of solution L. Moles of L = Concentration of L × Volume of L Moles of L