0.1MKMn O _ 4 is used for following titration. What volume of the solution in mL will be required to react with 0.158 g of Na _ 2 S _ 2 O _ 3 ? Not balanced: S _ 2 O _ 3 ^ 2 - + Mn O _ 4 ^ - + H _ 2 O → Mn O _ 2 ( s ) +

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Correct answer: A. 26.7 mL

+ 2 S _ 2 O _ 3 ^ 2 - + Mn O _ 4 ^ - + H _ 2 O → Mn O _ 2 ( s ) + + 6 S O _ 4 ^ 2 - + OH^ - 'n' factor of N a _ 2 S _ 2 O _ 3 = 8 'n' factor of KMn O _ 4 = 3 m.eq of N a _ 2 S _ 2 O _ 3 = m.eq of KMn O _ 4 (0.158)/(158) × 8 × 1000 = 0.1 × V × 3 V = 26.7mL

0.1MKMn O _ 4 is used for following titration. What volume of the...

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