Question

Mass of H2C2O4 required to reduce 100 ml of 0.008M KMnO4 in acidic medium is x g and to
neutralize 100 ml of 0.01 M Mg(OH)2 is y g, then the relation between x and y is
(A) X= y
(B) x = 2y
(C) 2x = y
(D) x = 3y​

Answer 1

Answer: A

Explanation: A

Solution :

Eq. of KMnO4= Eq. of KHC2O4

∴100×10−3×0.02×5=x1281

⇒x=1.28g

Eq. of KHC2O4= Eq. of Ca(OH)2

=y1281=100×10−3×0.05×2

⇒y=1.28g

∴X=y