A mixture containing `As_(2)O_(3) and As_(2)O_(5)` required 20 mL of `0.05` N iodine for titration .the resulting solution is then acidified and excess of KI was added . The liberated iodine required 1.0 g of `Na_(2)S_(2).5H_(2)O ` fr complete reaction .Calculate the weight of the mixture .(As = 75 , O = 16 , S = 32 , Na =23)

Explanation

M.e of `A_(2)O_(3)` = m.e of iodine
` = 20 xx 0.05 = 1.0`
Wt. of `As_(2)O_(3) = (1.0)/1000 xx198/2 g = 0.0495 g `
`{:((As_(2)O_(3) + I_(2)to,As_(2)O_(5)+I^(-),E_(As_(2)O_(3))=198/4),(+6,+10,):}) `
`{:(" m.e of "As_(2)O_(5)+,"m.e of "As_(2)O_(5)=,"m.e of "I_(2)),(("produced by "As_(2)O_(3)),(" in the mixture"),),(,,= "m.e of"Na_(2)S_(2)O_(3)):}`
or m.e of `As_(2)O_(3) + " m.e of "As_(2)O_(5) =" m.e of " Na_(2)S_(2)O_(3)`
`1. + " m.e of " As_(2)O_(5) = 1/(248//1) xx 1000`
m.e of `As_(2)O_(5) = 4.032 `
Wt. of `As_(2)O_(5)` in the mixture ` = (4.032)/1000 xx 230/4 g `
`= 0.232 g `
`({:(AS_(2)O_(5)+I^(-) to, As_(2)O_(3)+I_(2),E_(As_(2)O_(5) =230/4)),(+10,+6,):})`
` :. " wt. of " (As_(2)O_(3) + As_(2)O_(5)) = 0.0495 + 0.232 g = 0.2815 g `