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Excess of NaOH ( aq ) was added to 100 mL of FeCl 3 aq resulting into 2 . 14 g of Fe OH 3 . The molarity of FeCl 3 aq is: (Given the molar mass of Fe = 56 g mol - 1 and molar mass of Cl = 35 .5 g mol - 1 )

Options

  1. A0 . 3   M
  2. B0 . 2   M
  3. C0 . 6   M
  4. D1 . 8   M

Correct answer

B. 0 . 2   M

Step-by-step solution

Moles of Fe OH 3 = 2 .14 107 = 2 × 10 - 2 mol Moles FeCl 3 = 100   ml   ×   10 − 3   ×   M According to stoichiometry, mole of FeCl 3 = Mole   of   Fe ( OH ) 3 100 × 10 − 3   × M = 2   ×   10 − 2 M = 2 × 10 - 2 100 × 1000 = 0 .2   M

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