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InterviewSolution
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0.71 g of a sample of bleaching powder `(CaOCl_2)` is dissolved in 100 " mL of " water. 50 " mL of " this solution is titrated with KI solution. The `I_2` so liberated required 10 mL 0.1 M `Na_2S_2O_3` (hypo) solution in acidic medium for complete neutralisation. Calculate the percentage of available `Cl_2` from the sample of bleaching power. |
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Answer» `CaOCl_2-=Cl_2-=I^(ɵ)-=S_2O_3^(2-)` `mEq-=mEq-=mEq-=mEq` `mEq-=mEq-=mEq-=mEq` `--=--=--=10mLxx0.1xx1` (n-factor) `--=1mEq-=1mEq` Thus m" Eq of "`Cl_2` in `50 mL` of solution `=1mEq` m" Eq of "`Cl_2` in 100 " mL of " solution `=1xx2=2mEq` Weight of `Cl_2=mEqxx10^(-3)xxEq(Cl_2)` `=2xx10^(-3)xx(71)/(2)` `{:((n-fact or fo r Cl_2=2),(Cl_2+2e^(-)to2Cl^(ɵ)))` `=0.071 g` `%` of available `Cl_2=("Weight of" Cl_2)/("Weight of" CaOCl_2)xx100` `=(0.071xx100)/(0.71)=10%` |
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