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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 neutralise 100 ml of 0.05 M Ca(OH)2 is y g, then -

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a

x = y

b

2x = y

c

x = 2y

d

none is correct

answer is A.

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Detailed Solution

Eq. of KMnO4= Eq. of KHC2O4 ∴ 100×10−3×0.02×5=x1281 Eq. of KHC2O4= Eq. of Ca(OH)2⇒ y1281=100×10−3×0.05×2⇒y=1.28g ∴ x=y

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