# A 5.0 cm3 solution of H2O liberates 0.508 g of iodine from an acidified KI solution. Calculate the strength of H2O2 solution in terms of volume strength at STP.

Jitender Pal
8 years ago
 H2O2 + H2SO4 + 2KI → K2SO4 + I2 +2H2O 34 gm Acidified 254 gm 3 KI sol 0.508 gm 5 cm3 or ml
i.e. 254 gm of I2 is released by 34 gm H­2O2
∴ 0.508 gm of I2 will be released by
= 34/254 * 0.508 = 0.608
5 ml of H2O2 sol. Contains 0.068 gm of H2O2
∴ 1 ml of H2O2 contains 0.068/5 gm H2O2
NOTE : The strength of H2O2 is generally calculate in terms of volume strength. According to which 10 volume of H2O2 means that 1 ml of H2O2 sol gives 10 ml of O2 at STP.
2H2O2 → 2H2O + O2
2*34 gm 32 gm or
22,400 ml at STP
i.e., 68 gm of H2O2 gives 22, 400 ml of O2 at STP or 1 ml of H2O2 sol
or 0.068/5 gm of H2O2 sol gives 4.48 ml of O2 i.e. strength of H2O2 sol is 4.48 volumes
Jitender Pal
8 years ago
Hello Student,
 H2O2 + H2SO4 + 2KI → K2SO4 + I2 +2H2O 34 gm Acidified 254 gm 3 KI sol 0.508 gm 5 cm3 or ml
i.e. 254 gm of I2 is released by 34 gm H­2O2
∴ 0.508 gm of I2 will be released by
= 34/254 * 0.508 = 0.608
5 ml of H2O2 sol. Contains 0.068 gm of H2O2
∴ 1 ml of H2O2 contains 0.068/5 gm H2O2
NOTE : The strength of H2O2 is generally calculate in terms of volume strength. According to which 10 volume of H2O2 means that 1 ml of H2O2 sol gives 10 ml of O2 at STP.
2H2O2 → 2H2O + O2
2*34 gm 32 gm or
22,400 ml at STP
i.e., 68 gm of H2O2 gives 22, 400 ml of O2 at STP or 1 ml of H2O2 sol
or 0.068/5 gm of H2O2 sol gives 4.48 ml of O2 i.e. strength of H2O2 sol is 4.48 volumes

Thanks
jitender