1.5 g of a mixture containing `As_2O_5`, `Na_2HAsO_3`, and some inert impurities is dissolved in water, and the solution is kept neutral by adding excess of `NaHCO_3`. The solution when titrated with 0.1 M `I_2` required 35.0 nL of it for complete titration. The solution is then acidified and excess of KI is added to it. The `I_2` liberated required `35.0mL` of `0.3 M Na_2S_2O_3` solution for complete titration. Find the percentage composition of the mixture (Atomic mass of As is 75, Na is 23 and O is 16.)

Explanation

In First experiment:
m" Eq of "`Na_2HAsO_3=mEq` of ` I_2`
`(g)/(E)xx1000=2xx0.1xx35`
`g=(2xx0.1xx35xxE)/(1000)(E=(170)/(2))`
`=0.595%Na_2HAsO_3=39.67%`