A 150ml of solution of `I_2` is divided into two unequal parts. One part reacts with hypo solution in acidic medium . 15ml of `0.4 M` hypo was consumed. The second part was added to 100ml of `0.3M` hot NaOH sodium to produce `IO_(3)^(-)`. Residual base required 10ml of `0.3M H_(2) SO_(4)` solution for complete neutralization.What was the initial concentration of `I_(2)` ?

Options

A. `0.08 M`
B. `0.1 M`
C. `0.2 M`
D. `0.03 M`

Answer

B

Explanation

`I_(2)+2Na_(2)S_(2)O_(3) to 2NaI +Na_(2)S_(4)O_(6)"…."(i)`
m moles of `Na_(2)S_(2)O_(3)` consumed `=15xx0.4=6` m-mole
m-moles of `I_2` consumed =3m-mole
`3I_(2)+6NaOH to 5NaI+NaIO_(3)+3H_(2)O"…."(ii)`
m- moles of `I_2` reacted with `NaOH` are `=(30-(2xx3))/(2)=12 m` mole
Total m mole of `I_2` consumed in reaction (i)and (ii) `=3+12=15 m` mole
Molanty of `I_(2)=(15)/(150)=0.1M`.