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0 .1 M K M n O 4 is used for following titration. What volume of the solution in mL will be required to react with 0 .158 g of N a 2 S 2 O 3 ? Not balanced: S 2 O 3 2 - + M n O 4 - + H 2 O → M n O 2 ( s ⁡ ) + S O 4 2 - + O H -

Options

  1. A26.7 mL
  2. B50 mL
  3. C65 mL
  4. D75 mL

Correct answer

A. 26.7 mL

Step-by-step solution

S + 2 2 O 3 2 - + M n O 4 - + H 2 O → M n O 2 ( s ⁡ ) + S + 6 O 4 2 - + O H - 'n' factor of N a 2 S 2 O 3 = 8 'n' factor of K M n O 4 = 3 m.eq of N a 2 S 2 O 3 = m.eq of K M n O 4 0.158 158 × 8 × 1000 = 0.1 × V × 3 V = 26.7 m L

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