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1.0 g of H_{2}O_{2} solution (30 wt%) was titrated with 29.29 mL of KMnO_{4} solution. What is the molarity of the KMnO_{4} solution

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1.0 g of 30 wt% H₂O₂ is used to titrate 29.29 mL of 0.20 M KMnO₄ solution.

Let's consider the balanced equation for the reaction between H₂O₂ and KMnO₄.

2 KMnO₄ + 3 H₂O₂ → 3 O₂ + 2 MnO₂ + 2 KOH + 2 H₂O

1.0 g of H₂O₂ solution react. We can calculate the reacting moles of KMnO₄ considering that:

  • The concentration of H₂O₂ is 30 wt%.
  • The molar mass of  H₂O₂ is 34.01 g/mol.
  • The molar ratio of  H₂O₂ to KMnO₄ is 3:2.

[tex]1.0gH_2O_2(Solution) \times \frac{30gH_2O_2}{100gH_2O_2(Solution)} \times \frac{1molH_2O_2}{34.01gH_2O_2} \times \frac{2molKMnO_4}{3molH_2O_2} = 0.0059molKMnO_4[/tex]

0.0059 moles of KMnO₄ are in 29.29 mL of solution. The molarity of KMnO₄ is:

[tex][KMnO_4] = \frac{0.0059mol}{0.02929 L} = 0.20 M[/tex]

1.0 g of 30 wt% H₂O₂ is used to titrate 29.29 mL of 0.20 M KMnO₄ solution.

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