`5.7 g` of bleaching powder was suspended in `500 mL` of water. `25 mL` of this suspension on treatment with `KI` and `HCl` liberated iodine which reacted with `24.35 mL "of" N//10Na_(2)S_(2)O_(3)`. Calculate `%` of available `Cl_(2)` in bleaching powder.

Explanation

We, know,
`"Bleaching powder"overset(KI+HCl)rarrI_(2)overset(Na_(2)S_(2)O_(3))rarrI^(-)+Na_(2)S_(4)O_(6)`
The redox changes are: `2e+I_(2)rarr2I^(-)`
`2S_(2)^(2+)rarrS_(4)^((5//2))+2e``"Meq.of bleaching powder"="Meq.of available"Cl_(2)="Meq.of"I_(2) "liberated"="Meq.of"Na_(2)S_(2)O_(3)` used
`therefore` Meq.of available `Cl_(2)` in `25 mL` bleaching powder solution`="Meq. oo" Na_(2)S_(2)O_(3)` used
`=24.35xx(1)/(10)`
`therefore` Meq.of available `Cl_(2)` in `500 mL` bleaching powder solution `=24.35xx(1)/(10)xx(500)/(25)=48.7`
`therefore (w)/(71//2)xx1000=48.7`
`therefore w_(Cl_(2))=1.729g`
`therefore %` of available `Cl_(2)` in bleaching powder
`=(1.729)/(5.7)xx100=30.33%`