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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)` isA. 1NB. 5.6 volumeC. `17 gL^(-1)`D. all are correct |
Answer» Correct Answer - D 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. |
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