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» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1n-282830" style="font-weight:bold;" target="_blank" title="Click to know more about 1N">1N</a> <br/>5.6 <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a><br/>`17 gL^(-1)`<br/>all are correct</p>Solution :Let the normality of `H_(2)O_(2)=N_(1)` <br/>`therefore N_(1)xx10(H_(2)O_(2))=1xx10(Na_(2)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))` <br/>`therefore ` Normality of `H_(2)O_(2)` solution =1N <br/> `1NH_(2)O_(2)=5.6 vol. H_(2)O_(2)=17 gL^(-1)` <br/>Therefore , all statement are correct.</body></html>