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50 mL sample of ozonised oxygen at NTP was passed through a solution of potassium iodide. The liberated iodine required 15 mL of 0.08N Na_(2)S_(2)O_(3) solution for complete titration. Calculate the volume of ozone at NTP in the given sample.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/reactions-20919" style="font-weight:bold;" target="_blank" title="Click to know more about REACTIONS">REACTIONS</a> involved may be given as : <br/> `2KI+H_(2)O+O_(3) to 2KOH+I_(2)+O_(2) uarr`<br/> `I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)O_(6)`<br/> 1 mole `O_(3)=2 mol e Na_(2)S_(2)O_(3)` <br/>No. of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of hypo `=("Mass")/("Molecular mass "(158))` <br/> `(ExxNxxV)/(1000xx158)` <br/> where, `E_(Na_(2)S_(2)O_(3))=158, <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = 0.08, V=15` <br/> `therefore`No. of moles of hypo `=(158xx0.08xx15)/(1000xx158)=1.2xx10^(-3)` <br/> No. of moles of `O_(3)=(1)/(2)` mole of hypo <br/> `=(1)/(2)xx1.2xx10^(-3)` <br/> `=6xx10^(-4)` mole<br/> Volume of `O_(3)` at NTP =No. of moles`xx 22400` <br/> `=6xx10^(-4)xx22400` <br/> =13.44 mL at NTP</body></html>


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