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Question

If x g is the mass of NaHC2O4 is required to neutralise 100 mL of 0.2 (M) NaOH and y g is required to reduce 100 mL of 0.02 (M) KMnO4 in acidic medium, then:

A
x=y
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B
x=2y
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C
x=4y
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D
x=3y
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Solution

The correct option is C x=4y
Let the molar wt. of NaHC2O4=M

NaHC2O4+NaOHNa2C2O4+H2O

only one H exchange is taking place so, n factor of both NaHC2O4 and NaOH is 1.

xM=100×0.2×103=20×103 ...(I)


Case 2:

5HC2O4+MnO4+11H+Mn2++10CO2+8H2O

from here,
Calculating n-factor of NaHC2O4

In reactant side O.S. of carbon=3
In product side O.S. of carbon=4

n factor of NaHC2O4=1×2=2

n factor of KMnO4=5, as medium is acidic.


2yM=100×0.02×5×103=10×103

yM=100×0.02×5×103=5×103 ...(ii)

Dividing equation (i) by equation (ii)
xy=205
or x = 4y

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