Question

19.3 mL of a solution of KMnO₄ is required to completely oxidize 1.89 g Mo₂O₃ to MoO₄^-2 . What is the molarity of the KMnO₄ solution?

3MnO^-₄ + 5Mo⁺³ + 8H₂O ---------> 3 Mn⁺² + 5MoO₄^-2 + 16H⁺

Answer 1

From the equation it is found that 3 moles of KMnO₄ will react with 5 moles of Mo₂O₃.

Molar mass of Mo₂O₃ = 95.94 x 2 + 3 x 16 = 239.88 g/mol

so moles of Mo₂O₃ in 1.89g is = 1.89 / 239.88 = 0.00788 mol

the Mo³⁺ is = 0.00788 x 2 = 0.01576

This will react with 0.01576 x 3 / 5 = 0.009456 moles KMnO₄

19.3 mL of KMnO₄ solution contains 0.09456 moles of KMnO₄

1000 mL solution contains 1000 / 19.3 x 0.09456 = 0.4899 moles

That means 0.4899 moles of KMnO₄ present in 1.0 L solution

So Molarity of KMnO₄ solution = 0.4899 M