lunous dak set is automatically saved. 12. At the direction of your instructor,
ID: 304321 • Letter: L
Question
lunous dak set is automatically saved. 12. At the direction of your instructor, conduct a third trial. DATA TABLE Trial 1 Trial 2 Trial 3 Volume of H2O2 solution Volume of MnO, solution used at equivalence point (mL) DATA ANALYSIS 1. Calculate the moles of Mn04 used to reach the equivalence point of the reaction for each 2. Use your answer to question 1, along with the balanced redox equation in the introduction, to 3. Calculate the molar concentration of the H202 solution for each trial. trial. calculate the moles of H202 in the sample of solution for each trial. 4. The hydrogen peroxide solution that you tested is a commercial product with a concentration, as described on the label of the container, as 3%. As stated in the introductory remarks, a 3% H202 solution converts to a molarity of 0.88 M. Compare your experimentally determined molarity of H202 to the label description. 32-8 Advanced Chemistry with VerniExplanation / Answer
1. Trial 1:
M1V1 = M2V2, Where M1 is the molarity of H2O2 solution, M2 is the molarity of MnO4- solution.
i.e. M1*10 mL= M2*15.1 mL
i.e. M1/M2 = 15.1/10
Therefore, M1 = 15.1 M and M2 = 10 M
Hence, the no. of moles of MnO4- used = M2V2 = 10 mol/L * 15.1*10-3 L = 0.151 mol (Since 1 L = 103 mL)
Trial 2:
By repeating the calculations shown for Trial 1, you will get
M1 = 15.5 M and M2 = 10 M
Therefore, the no. of moles of MnO4- used = M2V2 = 10 mol/L * 15.5*10-3 L = 0.155 mol
2. At equivalence point, the no. of moles of H2O2 = no. of moles of MnO4-.
Therefore, the no. of moles of H2O2 used in trial 1 = 0.151 mol
And the no. of moles of H2O2 used in trial 1 = 0.155 mol
3. The molar concentration of H2O2 for trial 1 = 15.1 M
The molar concentration of H2O2 for trial 1 = 15.5 M
4. For trial 1, 3% of 15.1 M H2O2 = 3*15.1/100 = 0.453 M
And for trial 2, 3% of 15.5 M H2O2 = 3*15.5/100 = 0.465 M
Therefore, the experimentally determined molarity of H2O2 is almost half of the given molarity (0.88 M).
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