Mass of KHC2O4 (potassium acid oxalate) required to reduce 100 ml of 0.02 M KMnO4 in acidic medium to Mn2+ is x g and to neutralize 100 ml of 0.05 M CaOH2 is y g, then

Mass of KHC2O4 (potassium acid oxalate) required to reduce 100 ml of 0.02 M KMnO4 in acidic medium to Mn2+ is x g and to neutralize 100 ml of 0.05 M CaOH2 is y g, then

  1. A

    x = y

  2. B

    2x = y

  3. C

    x = 2y

  4. D

    none is correct

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    Solution:

    Eq. of KMnO4 = Eq. of KHC2O4 Mn+7O4-+C+32O4-2Mn+2+2C+4O2     100 × 10-3 × 0.02 × 5 = x1282     x = 0.64 g Eq. of KHC2O4 = Eq. of CaOH2   y1281 = 100  × 10-3 × 0.05 × 2 (KHC2O4 contains only one replaceable Hydrogen, Bascity=1)   y = 1.28 g        2 x = y

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