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 Cx=4y Let the molar wt. of NaHC2O4=M
NaHC2O4+NaOH→Na2C2O4+H2O
only one H exchange is taking place so, n factor of both NaHC2O4 and NaOH is 1.
∴xM=100×0.2×10−3=20×10−3 ...(I)
Case 2:
5HC2O−4+MnO−4+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=100×0.02×5×10−3=10×10−3
∴yM=100×0.02×5×10−3=5×10−3 ...(ii)
Dividing equation (i) by equation (ii) xy=205
or x = 4y