1. Titration of an oxalate sample gave the following percentages: 70.59%, 70.51%

Question

1. Titration of an oxalate sample gave the following percentages: 70.59%, 70.51%, and 71.56%. calculate the average and the standard deviation

2. Calculate the % C2O42- in each of the following :- a. H2C2O4.2H2O b. K2[Cu (C2O4)2].2H2O

3. A solution of 0.05M KMnO4 was titrated with a solution of 0.252g H2C2O4.2H2O 100mL. What is the volume of KMnO4 required to the end point?

4. If 0.126 g of Oxalic acid dihydrate H2C2O4.2H2O, requires 21 mL of a KMnO4 solution to react to the end point, what is the molarity of the KMnO4 solution?

Explanation / Answer

Average =70.59 + 70.51 + 71.56 =212.66/3 = 70.887

SD Calculation.

Step 1:

(70.59 - 70.887)2 = 0.088

(70.51 - 70.887)2 = 0.142

(71.56 - 70.887)2 = 0.453

Step 2:

0.088 + 0.142 + 0.453 =0.683/3 = 0.227

Step 3

SD = Square root of 0.227 = 0.476

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