A 5.0 cm^(3) solution of H_(2)O_(2) liberates 0.508 g of iodine from an acidified KI solution. Calculate the strength of H_(2)O_(2) solution in terms of volume strength of STP.

A 5.0 cm^(3) solution of H_(2)O_(2) liberates 0.508 g of iodine from a

1.	A 5.0 cm^(3) solution of H_(2)O_(2) liberates 0.508 g of iodine from an acidified KI solution. Calculate the strength of H_(2)O_(2) solution in terms of volume strength of STP.
Answer» Solution :`2Kl + H_(2)SO_(4)+ H_(2)O_(2) to K_(2)SO_(4)+ 2H_(2)O + I_(2)` From the above equation `H_(2)O_(2) -=I_(2)` or 34G of `I_(2)` `therefore 0.508 g` of `I_(2)` will be liberated from `H_(2)O_(2)` `=(34)/(254)xx0.508` `=0.068` g The decomposition of `H_(2)O_(2)` occurs as `underset(2xx34=68 g )(2H_(2)O_(2))to underset(22400 cm^(3) " at " NTP)(2H_(2)O) + O_(2)` `therefore 0.068 of `H_(2)O_(2)` upon decomposition will give `O_(2)=(22400)/(68)xx0.068` `=22.4 cm^(3)` Now `5.0cm^(3)` of `H_(2)O_(2)` solution GIVES `O_(2)` `=22.4 cm^(3)` at STP `therefore 1.0 cm^(3) ` of `H_(2)O_(2)` solution will give `O_(2)` `=(22.4)/(5)=4.48 cm^(8)` at STP