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. `0.2M`
D. `0.03 M`

Answer

B

Explanation

`I _(2) + 2 N a _(2) S _(2) O _(3) to 2 Na I + Na _(2) S _(4) O _(6) ...(i)`
m-moles of `N a _(2) S _(2) O _(3)` consumed `= 15 xx 0.4 = 6 ` m-mole
m-moles of `I _(2)` consumed = 3 m -mole
`3 I _(2) + 6 Na OH to 5 Na I + Na I O _(3) + 3 H _(2) O ....(ii)`
m-moles of `I _(2)` reacted with NaOH are `= ( 30 - (2 xx 3))/(2) =12 ` m mole
Total m-moles of `I _(2)` consumed in reaction (i) and (ii) `= 3 + 12 = 15 m` mole
Molarity of `I _(2) = ( 15)/(150) = 0.1 M`