Mass of `KHC_(2)O_(4)` (potassium acid oxalate) required to reduce `100 ml` of `0.02 M KMnO_(4)` in acidic medium (to `Mn^(2+)`) is `x g` and to neutralise `100 ml` of `0.05 M Ca(OH)_(2)` is `y g`, then

Options

A. `a=b`
B. `2a=b`
C. `a=2b`
D. None of these

Answer

2

Explanation

Equivalent of `KMnO_(4)=` Equivalent of `KHC_(2)O_(4)`
`0.02xx(100)/(1000)xx5=(a)/(m_(KHC_(2)O_(4))xx2)" ".........(i)`
Equivalent of `Ca(OH)_(2)=` Equivalent of `KHC_(2)O_(4)`
`0.005xx(100)/(1000)xx2=(b)/(m_(KHC_(2)O_(4))xx1)" ".......(ii)`
Equation `(i) +` Equation `(ii)`
`2a=b`