What volume of 0.1 M `KMnO_(4)` is needed to oxidise 5 " mg of " ferrous oxalate in acidic medium (Mw of ferrous oxalate is 144.)

Options

A. 0.20 mL
B. 0.1 mL
C. 0.4 mL
D. 2.08 mL

Answer

A

Explanation

`m" Eq of "KMnO_(4)=Vxx0.1xx5`
`m" Eq of "Fe(C_(2)O_(4))_(2)=(Weight I n g)/(Ew)xx1000`
`{:([Fe^(+2)toFe^(+3)+e^(-)),(C_(2)O_(4)^(2-)to2CO_(2)+2e^(-))]:}` `(n=3)`
Ew of `Fe(C_(2)O_(4))=(Mw)/(3)=(144)/(4)`
`therefore m" Eq of "Fe(C_(2)O_(4))=(5xx10^(-3))/((144)/(3))xx10^(3)=(5xx3)/(144)`
`therefore m" Eq of "KMnO_(4)=m" Eq of "Fe(C_(2)O_(4))`
`thereforeVxx0.1xx5=(5xx3)/(144)`
`thereforeV=0.20mL`