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What will be the volume of O 2 Liberated at NTP by passing 5 A current For 193 Sec. through acidified water.

Options

  1. A56 mL
  2. B112 mL
  3. C158 mL
  4. D965 mL

Correct answer

A. 56 mL

Step-by-step solution

W = i × t × E F = 5 × 1 9 3 × 8 9 6 5 0 0 ∵ At NTP, volume of 32 g O 2 = 22400 mL ∴ Volume of 0.08 g O 2 = 2 2 4 0 0 × 0 . 0 8 3 2 = 56 mL

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