Experiment Vll:Oxidation Reduction Lab Report I. Standardization of KMnO4 Soluti

Question

Experiment Vll:Oxidation Reduction Lab Report I. Standardization of KMnO4 Solution 1. Mass of Na C,04 samples (include units) Sample 1:0 Volume of KMno4 solution used (include units) Sample 1: 1.9 Sample 2 Smlammpa l Sample34 10 Sample 2:0 2. Sample 3: b 3. Calculate the molarity of the KMno, solution for Sample 1 4. Calculate the molarity of the KMnO, solution for Sample 2. 5. Calculate the molarity of the KMnO4 solution for Sample 3. 6. Calculate the average molarity of the KMnO, solution. 7. Calculate the standard deviation of the molarity of the KMn04 solution.

Explanation / Answer

Q3

Molarity of KMnO4

5 mol of Na2C2O4 = 2 mol of KMnO4

then,

mol of Na2C2O4 = mass/MW = 0.11/133.9985 = 0.000820 mol of Na2C2O4

then

mol of KMnO4 = 2/5*0.000820 = 0.000328 mol of KMnO4

V = 1.8 mL

[KMnO4] = mol/V = (0.000328 )/(1.8*10^-3) =  0.1822 M

Q4.

mol of Na2C2O4 = mass/MW = 0.12/133.9985 = 0.000895 mol of Na2C2O4

then

mol of KMnO4 = 2/5*0.000895= 0.000358 mol of KMnO4

V = 1.8 mL

[KMnO4] = mol/V = (0.000358)/(1.8*10^-3) =  0.1988 M

Q5

mol of Na2C2O4 = mass/MW = 0.11/133.9985 = 0.000820 mol of Na2C2O4

then

mol of KMnO4 = 2/5*0.000820 = 0.000328 mol of KMnO4

V = 1.6 mL

[KMnO4] = mol/V = (0.000328 )/(1.6*10^-3) =  0.205 M

Q6.

avg. molarity = (0.1822 + 0.1988 +0.205 )/3 = 0.19533 M

Q7

std.dev = sqrt(((0.1822 -0.19533 )^2 + (0.1988 - 0.19533 )^2 + (0.205 -0.19533)^2)/3) = 0.00962

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