Question

`KMnO_(4)` reacts with oxalic acid according to the reaction :
`2KMnO_(4) +5C_(2)O_(4)^(2-)+16 H^(+) to 2Mn^(2+) + 10 CO_(2) + 7H_(2)O`
Then, 20 mL of 0.1M `KMnO_(4)` is equivalent to :
A. 30 mL of 0.5 M `C_(2)H_(2)O_(4)` (oxalic acid)
B. 50 mL of 0.1 M` C_(2)H_(2)O_(4)` (oxalic acid)
C. 20 mL of `0.5 M C_(2)H_(2)O_(4)` (oxalic acid)
D. 10 mL of `0.1 M C_(2)H_(2)O_(4)` (oxalic acid)

Answer 1

Correct Answer - B
`(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2))(H_(2)C_(2)O_(4))`
`(0.1xx20)/(2)=(M_(2)V_(2))/(5)`
`M_(2)V_(2)=5`
(It is possible in case of b.)