Question

Calculate the percentage of available chlorine in a sample of 3.55g of bleaching powder which was dissolved in 100mL of water. 25mL of this solution, on treatment with KI and dilute acid, required 20 mL of 0.125 N sodium thiosulphate solution.

Answer 1

`CaOCl_(2)+H_(2)OrarrCa(OH)_(2)+Cl_(2)`
`Cl_(2)+2KIrarr2KCl+I_(2)`
`I_(2)+2Na_(2)S_(2)O_(3)rarrNa_(2)S_(2)O_(6)+2NaI`
In 25 mL solution, moles of `Na_(2)S_(2)O_(3)=(20)/(1000)xx(0.125)/(1)=25xx10^(-4)`
So, moles of `I_(2)=(1)/(2)xx"moles of " Na_(2)S_(2)O_(3)`
`=(1)/(2)xx25xx10^(-4)=12.5xx10^(-4)`
So, in 100 mL solution, moles of `Cl_(2)=4xx12.5xx10^(-4)=50xx10^(-4)`
So, weight of `Cl_(2)=50xx10^(-4)xx71g`
% of available `Cl_(2)=(50xx10^(-4)xx71)/(3.55)xx100=10%`