IV CALCULATIONS A. Molarity of Standard H,Cro. Solution Use the molar mass of H

Question

IV CALCULATIONS A. Molarity of Standard H,Cro. Solution Use the molar mass of H Co.. 2 HO and the volume of the volumetric flask to determine the molariry of the oxalic acid solution that you have prepared The stoichiometry of Equation 5 shows that the following using hold true at the equivalence point equation will of the titration. 5 x mol Mno. 2 x mol H,CO (6) Since moles can be expressed in terms of molarity and volume data for a given reagent, Equation 6 becomes the following equation. Using your data to provide the two volumes and the molarity of oxalic acid that you calculated in Part A, calculate the molarity of KMnor C. Percent Iron in Unknown Sample To find the percent by mass of iron you need the following equation. Fe mass of ron(II) X 100 (8) mass of sample The mass of your sample can be obtained directly from your data, while the mass of iron las iron(II)] present in the sample must come from your titration as data shown below. The stoichiometry of Equation 3 shows that the following equation will hold true at the equivalence point of the titration 5 x mol MnO (9) The moles of the KMno, solution can be expressed in terms of molarity and volume data, while the moles of the iron can be expressed in terms of mass divided by molar mass. 5x (Mv)aro mass of ron (II) molar mass of iron With the molarity of KMno, determined from Part B and the volume data from your titration in Part C, you can determine the mass of iron from Equation 10 and then use this value to find the percent ofiron using Equation 8.

Explanation / Answer

Ans:

Step 1: Calculating the Molarity of Standard Oxalic Acid solution :-

mass of oxalic acid = 0.20g

Molar mass of H2C2O4 = 90.0g/mol

Moles of oxalic acid = mass/Molar mass = 0.20/90.0 = 0.0022 moles

Volume of oxalic acid solution = 50.0ml

Molarity of oxalic acid solution = moles/Litre = 0.0022/0.050 = 0.044M

Step 2: Calculating the molarity of KMnO4 solution:-

   According to the stochiometric equation:

5 x (MxV)KMnO4 = 2 x (MxV)H2C2O4

(i) Trial 1:

5M1V1 = 2MV

5 x M1 x 14.80 = 2 x 0.044 x 10.00

74 x M1 = 0.88

M1 = 0.01189M

(ii) Trial 2:-

5 x M2 x 14.92 = 0.88

74.6 x M2 = 0.88

M2 = 0.01179

(iii) Trial 3:-

5 x M2 x 13.00 = 0.88

65 x M3 = 0.88

M3 = 0.01354

S.No. Trial #1 #2 #3 1. Volume of H2C2O4 10.00 10.00 10.00 2. Molarity of H2C2O4 0.044M 0.044M 0.044M 3. Volume of KMnO4 14.80 14.92 13.00 4. Molarity of KMnO4 0.01189M 0.01179 0.01354 5. Average Molarity of KMnO4 0.0124M

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