Question

Molarity of your standardized KMnO₄ solution : 0.0200 M

Experiment Number	#1	#2	#3	#4
Mass of sample (g)	0.134	0.152	0.129	0.143
Volume of KMnO4 required (mL) :	17.70	12.90	14.20	14.70
Volume of blank titration (mL) :	0.0500	0.0500	0.0500	0.0500

Calculate the corrected volumes for KMnO₄:

Experiment Number	#1	#2	#3	#4
Corrrected Volume (mL)

Calculate the moles of KMnO₄ added :

Experiment Number	#1	#2	#3	#4
Moles KMnO4 (moles)

Calculate the moles of C₂O₄²⁻ in the sample :

Experiment Number	#1	#2	#3	#4
Moles C2O42− (moles)

Calculate the percent of C₂O₄²⁻ in the sample :

Experiment Number	#1	#2	#3	#4
Percent C2O42− %

Answer 1

Molarity of kim Oq solution = 0.002 (M) volume of blank titration = 0.05 mL Connected volumes for kMn Og = Experimentally taken komog - Blank titration) SO, connected volumes - 1. 17,7-0.05= 17.65 mL 2. 12:9 – 0.05 = 12,85 mL 3. 14.2 - 0.05 - 14.15 mL. 4. 14.7 -0.05 = 14.65 me mo4 Iwme Moles of kamoa added the 1000 - 5 0.002 x 17.65 = 3.53x10 1000 mol 0.002 x 12.85 – 2.57x105 mol 1000 2 x 14.15 - 2.83x105 mol 0.002 4, 0.002 x 14.65 = 2.93 x 105 mol 1000 Balanced nedox reaction between klm Oy & C₂07 : 2 MnO4 + 16H+ + 56200 + 50₂ + 4H₂O . 2 mol amor = 5 mol CO2- : 1000 mL 2 (M) MnOy = 5 mol la – 1000 ml con emoa = 1500 mol 04 - I = 215x 103 mol Goa 1. 17.65 mL 0.002 (M) amba = 215x10 3 x 17.65% 0.002 - 8.825x105 mol CO2- C202- in sample 1= 8.825 x 155 mol

2. 12.85 mL 0.002 (M) kinoa = 2.5x10 X 12.858 0.002 mol n = 6.425x105 mol (209- 3.19.15 mL 0.002 (M) KMmoa = 14.15% 2.5x103 4 0.002 mol - 7.075x105 nuol 07 4. 14.65 mL 0.002 (M) KMnO 4 - 14.65% 2.5x10² +0.002 nool - 7.325 x 155 sol (2007- 1 % of CO2 in the sample: mass of 2017 in the sample x. 1000 total mass of the sample taken molen of 207 estimated & Formula weight of 40% - x 100 total mass of the sample taken. mass of IS 09 = 24 + 64 = 88. A 8.825 x 10 x 88 x 100 - 5.795 %. 0.134 6.425x105 x 88 x 100 = 3.72 % 0.152 2. 13. = 9.82% 7.075 x 10 x 88 x 100 0.129 7.325 x 105 x 88 x 100 0.143 = 4.51%