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What volume of 0.1 N oxalic acid solution can be reduced by 250 gram of an 8% by weight KMnO4 solution?
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Calculation of Volume of Oxalic Acid Solution Reduced by KMnO4


Given Information



  • Concentration of oxalic acid solution = 0.1 N

  • Weight of KMnO4 solution = 250 g

  • Percent weight of KMnO4 solution = 8%



Calculations


Step 1: Calculation of moles of KMnO4


Molar mass of KMnO4 = 158.03 g/mol

Percentage weight of KMnO4 solution = 8%

Therefore, weight of KMnO4 in 250 g solution = 0.08 x 250 g = 20 g

Number of moles of KMnO4 = 20 g / 158.03 g/mol = 0.1265 moles

Step 2: Calculation of moles of oxalic acid


Balanced chemical equation for the reaction between KMnO4 and oxalic acid:

5C2H2O4 + 2KMnO4 + 3H2SO4 → 10CO2 + 2MnSO4 + K2SO4 + 8H2O

From the equation, we can see that 2 moles of KMnO4 react with 5 moles of oxalic acid.

Therefore, number of moles of oxalic acid = 0.1265 moles x (5/2) = 0.31625 moles

Step 3: Calculation of volume of oxalic acid solution


Molarity of oxalic acid solution = 0.1 N

Number of moles of oxalic acid = 0.31625 moles

Therefore, volume of oxalic acid solution = 0.31625 moles / 0.1 N = 3.1625 liters or 3162.5 ml

Answer


The volume of 0.1 N oxalic acid solution that can be reduced by 250 g of an 8% by weight KMnO4 solution is 3162.5 ml.
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What volume of 0.1 N oxalic acid solution can be reduced by 250 gram of an 8% by weight KMnO4 solution?
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