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Mass of KHC(2)O(4) (potassium acid oxala...

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

A

x= y

B

2x= y

C

x = 2y

D

None is Correct

Text Solution

Verified by Experts

The correct Answer is:
B
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Moles of KHC_(2)O_(4) (potassium acid oxalate) required to reduce 100ml of 0.02M KMnO_(4) in acidic medium (to Mn^(2+) ) is :

The weight of oxalic acid required to reduce 80 ml of 0.4 M KMnO_(4) in acidic medium is

'a' g KHC_(2)O_(4) are required to reduce 100mL of 0.02M KMnO_(4) in acid medium and 'b' g KHC_(2)O_(4) neutralises 100mL of 0.05M Ca(OH)_(2) then :

'a' gm KHC_(2)O_(4) is used to neutralize 100 mL of 0.02 M KMnO_(4) in acid medium, where as 'b' gm KHC_(2)O_(4) is used to neutralize 100 mL of 0.02 M Ca(OH)_(2) then,

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