10 mL of H_(2)O_(2) solution is treated with KI and titration of liberated I_(2) required 10 mL of 1 N hypo . Thus H_(2)O_(2) is

10 mL of H_(2)O_(2) solution is treated with KI and titration of liber

1.	10 mL of H_(2)O_(2) solution is treated with KI and titration of liberated I_(2) required 10 mL of 1 N hypo . Thus H_(2)O_(2) is
Answer» 1N 5.6 VOLUME `17 gL^(-1)` all are correct Solution :Let the normality of `H_(2)O_(2)=N_(1)` `therefore N_(1)xx10(H_(2)O_(2))=1xx10(Na_(2)S_(2)O_(3))` `therefore ` Normality of `H_(2)O_(2)` solution =1N `1NH_(2)O_(2)=5.6 vol. H_(2)O_(2)=17 gL^(-1)` Therefore , all statement are correct.