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)?

A 200 mL solution of I_(2) divided into two unequal parts. Part I reac

1.	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» 1 2 3 4 Solution :First part: mmoles of `Na_(2)S_(2)O_(3)` used `=8xx2xx1 ("n-factor")=16` `I_(2)+2Na_(2)S_(2)O_(3)to2NaI+Na_(2)S_(4)O_(6)` m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "`Na_(2)S_(2)O_(3)=(16)/(2)=8`..(i) Second Part: `3I_(2)+6NaOHto5NaI+NaIO_(3)+3H_(2)O` mmoles of `H_(2)SO_(4)=` excess `NaOH=30xx0.1xx2` `xx("n-factor")=6` m" mol of "total `NaOH=300xx0.1xx1("n-factor")=30` m" mol of "NaOH `used=30-6=24` m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "NaOH used `=(24)/(2)=12 m" mol of "I_(2) used` Total m" mol of "`I_(2)` used `=` part I`+` part II `=8+12=20mmol` `M of I_(2)=(mmol)/(V_(mL))=(20)/(200)=0.1M` 20 TIMES the initial `M_(I_(2))=0.1xx20=2`