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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Hey Abiodun😍😍😍, good to see you again. Here are the calculations based on the provided information: a) The question asks to record the pipette volume, tabulate burette readings, and calculate the average volume of F used. Since no experimental data is provided, we will use the information given in the problem and the teacher's hint. The volume of the pipette used for solution G (NaOH) is assumed to be 25 cm^3. The average volume of F (H₂X) used from the burette, as suggested by the teacher's hint, is 24.50 cm^3. b) i) Concentration of G in mol dm^-3. Solution G contains 1.00 g of sodium hydroxide (NaOH) in 250 cm^3 of solution. The atomic masses are H = 1, O = 16, Na = 23. Step 1: Calculate the molar mass of NaOH. Molar mass of NaOH = Na + O + H = 23 + 16 + 1 = 40 g/mol Step 2: Calculate the number of moles of NaOH. Moles of NaOH = Mass of NaOHMolar mass of NaOH = 1.00 g40 g/mol = 0.025 mol Step 3: Calculate the concentration of NaOH in mol dm^-3. The volume of the solution is 250 cm^3. Convert this to dm^3: 250 cm^3 = (250)/(1000) dm^3 = 0.250 dm^3 Concentration of G (NaOH) = Moles of NaOHVolume of solution in dm^3 = 0.025 mol0.250 dm^3 = 0.100 mol dm^-3 The concentration of G is 0.100 mol dm^-3. ii) Concentration of F in mol dm^-3. The balanced chemical equation is: H_2X(aq) + 2NaOH(aq) → Na_2X(aq) + 2H_2O(l) From the equation, the mole ratio of H₂X to NaOH is 1:2. Given values: Concentration of base, C_b = Concentration of G = 0.100 mol dm^-3 Volume of base, V_b = Pipette volume of G = 25 cm^3 = 0.025 dm^3 Volume of acid, V_a = Average volume of F = 24.50 cm^3 = 0.0245 dm^3 Moles of acid from equation, n_a = 1 Moles of base from equation, n_b = 2 Using the titration formula: (C_a V_a)/(C_b V_b) = (n_a)/(n_b) Rearrange to solve for C_a: C_a = (n_a)/(n_b) × (C_b V_b)/(V_a) Substitute the values: C_a = (1)/(2) × (0.100 mol dm^-3) × (0.025 dm^3)0.0245 dm^3 C_a = (1)/(2) × (0.0025)/(0.0245) C_a = (1)/(2) × 0.1020408... C_a = 0.0510204... mol dm^-3 Rounding to three significant figures: C_a = 0.0510 mol dm^-3 The concentration of F is 0.0510 mol dm^-3. iii) Molar mass of the acid H₂X. We are given that 100 cm^3 of solution F contained 0.485 g of acid. From part (b)(ii), the concentration of F is 0.0510 mol dm^-3. Step 1: Calculate the number of moles of H₂X in 100 cm^3 of solution F. Convert 100 cm^3 to dm^3: 100 cm^3 = (100)/(1000) dm^3 = 0.100 dm^3 Moles of H_2X = Concentration × Volume in dm^3 Moles of H_2X = 0.0510204 mol dm^-3 × 0.100 dm^3 = 0.00510204 mol Step 2: Calculate the molar mass of H₂X. Molar mass of H_2X = Mass of H_2XMoles of H_2X Molar mass of H_2X = 0.485 g0.00510204 mol = 95.059 g/mol Rounding to three significant figures: Molar mass of H_2X = 95.1 g/mol The molar mass of the acid H₂X is 95.1 g/mol. Send me the next one 📸