25 ml of `H_(2)O_(2)` solution were added to excess of acidified KI solution. The iodine so liberated required 20 ml of 0.1 N `Na_(2)S_(2)O_(3)` solution. Calculate strength in terms of normality and percentage.

Options

A. 0.04 N, `0.136%`
B. 0.08 N, `0.136%`
C. 0.08 N, `0.163%`
D. 0.02 N, `0.163%`

Answer

C

Explanation

`O_(2)^(-)+2e^(-) rarr 2O^(-2)," " 2I^(-) rarr I_(2)+2e^(-)`
`2(S^(+2))_(2) rarr (S^(+5//2))_(4)+2e^(-), I_(2)+2e^(-) rarr 2I^(-)`
Meq. Of `H_(2)O_(2)=` Meq. of `I_(2)=` Meq. of `Na_(2)S_(2)O_(3)`
`w/(34//2) times 1000=20 times 0.1`
`therefore W_(H_(2)O_(2))=0.034 gr//25ml`
`therefore N_(H_(2)O_(2))=0.034/(34//2) times 1000/25=0.08`
Vol. strength `=5.6 times 0.08=0.448`
% strength `=17/56 times 5.6 times 0.08=0.136%`