50 mL of an aqueous solution of H_(2)O_(2) was reacted with excess of KI solution and dilute H_(2)SO_(4) The liberated iodine requried 20 mL 0.1 N Na_(2)S_(2)O_(3) solution for complete interaction calculate the concentration of H_(2)O_(2) in gL^(-1)

50 mL of an aqueous solution of H_(2)O_(2) was reacted with excess of

1.	50 mL of an aqueous solution of H_(2)O_(2) was reacted with excess of KI solution and dilute H_(2)SO_(4) The liberated iodine requried 20 mL 0.1 N Na_(2)S_(2)O_(3) solution for complete interaction calculate the concentration of H_(2)O_(2) in gL^(-1)
Answer» SOLUTION :`H_(2)O_(2)^(-1)KI+H_(2)SO_(4)rarrK_(2)SO_(4)+I_(2)+2H_(2)O^(-2)` `Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2` Nal Total change in O.N of O=2(-1)-2(-2)=2 `therefore` Eq of `H_(2)O_(2)=(2+2xx16)/(2)=17` Let `N_(1)` be the normality of `I_(2)` solution Since one equivalent of `H_(2)O` produces 1 equilvent of `I_(2)` `therefore` 50mL of `N_(1) I_(2)` solution =50mL of `N_(1)H_(2)O_(2)` solution `therefore N_(1)=(20xx0.1)/(50)=0.04` N or STREANGTH of `H_(2)O_(2) "solution" =0.04xx17=0.68 gL^(-1)`