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The freezing point depression of 0 . 001   M of A x B y Fe CN 6 is 5 . 58 × 10 - 3   K . If the oxidation state of Fe is + 2 and K f = 1 . 86   K   kgmol - 1 , then the total number of possibilities for different types of A and B cations are

Options

  1. A1
  2. B2
  3. C3
  4. D4

Correct answer

B. 2

Step-by-step solution

According   to   depression   in   freezing   point : - ∆ T f = iK f M A x B y Fe CN 6   → XA +   +   YB +   +   Fe CN 6 2 - i = ∆ T f K f × M ∆ T f   =   5 . 58 × 10 - 3 K f   =   1 . 86   and   M =   0 . 001 = 1 × 10 - 3 So ,   i =   5 . 58 × 10 - 3 1 . 86 × 1 × 10 - 3 = 3 Value   of   i   for   anion = 1 Then   the   value   of &#1

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