10.0 g of `CaOCl_2` is dissolved in water to make 200 mL solution 20 " mL of " it is acidified with acetic acid and treated with KI solution the `I_2` liberated required 40 " mL of " `(M)/(20)Na_2S_2O_3` solution. Find the percentage of available chlorine.

Explanation

`CaOCl_2-=Cl_2-=I^(ɵ)-=S_2O_3^(2-)`
`mEq-=mEq-=mEq-=mEq`
`mEq-=mEq-=mEq-=mEq`
`--=--=--=40 mLxx(M)/(20)xx1` (n factor)
`-=2mEq-=--=2mEq`
Thus m" Eq of "`Cl_2` in " mL of " solution `=2mEq`
m" Eq of "`Cl_2` in 200 " mL of " solution `=(2xx200)/(20)=20mEq`
weight of `Cl_2=mEqxx10^(-3)xxEw of Cl_2`
`=20xx10^(-3)xx(71)/(2)(Cl_2+2e^(-)to2e^(-))`
`=0.71g`
`%` of available `Cl_2=("Weight of " Cl_2xx100)/("Weight of "CaOCl_2)`
`=(0.71xx100)/(10)=7.1%`