A mixture containing `As_(2)O_(3)` and `As_(2)O_(3)` requried 20 " mL of " 0.05 N `I_(2)` for titration. The resulting solution is then acidified and excess KI was added. The liberated `I_(2)` required 1.24 g hypo `(Na_(2)S_(2)O_(3).H_(2)O)` for complete reaction. Calculate the mass of the mixture.
The reactions are
`As_(2)O_(3)+2I_(2)+2H_(2)OtoAs_(2)O_(3)+4H^(o+)+4I^(ɵ)`
`As_(2)O_(5)+4H^(o+)+4I^(ɵ)toAs_(2)O_(3)+2I_(2)+2H_(2)O`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the equivalents of I2 used in the titration.
The amount of I2 used is given as 20 mL of 0.05 N.

\[ \text{Equivalents of } I_2 = \text{Volume (L)} \times \text{Normality} = 0.020 \, \text{L} \times 0.05 \, \text{N} = 0.001 \, \text{equivalents} \]
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