Alt EXPERIMENT 14 Name Date Part A: Diluting the KMnO4 Stock Solution mL of 020

Question

Alt EXPERIMENT 14 Name Date Part A: Diluting the KMnO4 Stock Solution mL of 020 Molar KMnO4 that I will use for the I am going to prepare titrations in this expenment Calculation: Preparing titrating strength KMnO, from stock, more concentrated KMnO v2 ~ (o.02.0m)(250m L v, ? m Part B: Standardizing the diluted KMnO4 Solution Step 3: Calculation: Volume of H,C0, 2H,O standard that would use about 15 mL of KMno 0.2 0.02 nof Step 6 DATA: Use the area below to record the data of your Step 6 titrations Trial 2 Trial 3 Trial 1 Finap Vol. Km.? 13, ?e «L // , q2 mL 12.39L 35? at am 169

Explanation / Answer

Exact molarity of KMnO4 solution:

Trial 1

Molecular weight of KMnO4 = 158.034 g/mol

Mass = Volume x Density

Density of  KMnO4 =1.01 g/mL

Mass of KMnO4 = 13.64 mL x 1.01 g/mL

Mass of KMnO4 = 13.78 g

Moles of  KMnO4 = weight / molecular weight

= 13.76 g / 158.034 g/mol

= 0.0871 mol

Molarity of KMnO4 = (0.0871 mol / 13.64 mL) x 1000

Molarity of KMnO4 = 6.3856 M

Similarly for Trial 2

Molarity of KMnO4 = (0.0871 mol / 11.91 mL) x 1000

Molarity of KMnO4 = 7.3132 M

Trial 3

Molarity of KMnO4 = (0.0871 mol / 12.39 mL) x 1000

Molarity of KMnO4 = 7.0299 M

Calculation of Average Exact Molarity and Precision of Three Trials:

Average Exact Molarity = (6.3856 + 7.3132 + 7.0299 )/3

= 6.9096 M

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