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A 200 mL solution of I_(2) divided into two unequal parts. Part I reacts with pypo solution in acidic medium and required 8 " mL of " 2 M hypo solution for complete neutralisation. Part II was added with 300 " mL of " 0.1 M NaOH solution residual base required 30 " mL of " 0.1 M H_(2)SO_(4) solution for complete neutralisation. Calculate the value of 20 times the initiall concentration I_(2)?

Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Solution :First part: mmoles of `Na_(2)S_(2)O_(3)` used <br/> `=8xx2xx1 ("n-factor")=16` <br/> `I_(2)+2Na_(2)S_(2)O_(3)to2NaI+Na_(2)S_(4)O_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)` <br/> m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "`Na_(2)S_(2)O_(3)=(16)/(2)=8`..(i) <br/> Second Part: <br/> `3I_(2)+6NaOHto5NaI+NaIO_(3)+3H_(2)O` <br/> mmoles of `H_(2)SO_(4)=` excess `NaOH=30xx0.1xx2` <br/> `xx("n-factor")=6` <br/> m" mol of "total `NaOH=300xx0.1xx1("n-factor")=<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>` <br/> m" mol of "NaOH `used=30-6=24` <br/> m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "NaOH used <br/> `=(24)/(2)=12 m" mol of "I_(2) used` <br/> Total m" mol of "`I_(2)` used `=` part I`+` part II <br/> `=8+12=20mmol` <br/> `M of I_(2)=(mmol)/(V_(mL))=(20)/(200)=0.1M` <br/> 20 <a href="https://interviewquestions.tuteehub.com/tag/times-1420471" style="font-weight:bold;" target="_blank" title="Click to know more about TIMES">TIMES</a> the initial `M_(I_(2))=0.1xx20=2`</body></html>


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