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2.56 × 10 - 3 equivalent of KOH is required to neutralise 0 .12544 g H 2 XO 4 . The atomic mass of X (in g/mol ) is: [Given: H 2 XO 4 is a dibasic acid]

Options

  1. A16
  2. B8
  3. C7
  4. D32

Correct answer

D. 32

Step-by-step solution

Let the atomic weight of X = y No. of equivalents of KOH = No. of equivalents of H 2 XO 4 i.e. 0 .12544 66 + y 2 = 0 .12544 ∴ y = 32

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