2 moles of a mixture of `O_(2)` and `O_(3)` is reacted with excess of acidified solution of KI. The iodine liberated require 1L of 2M hypo solution for complete reaction. The weight % of `O_(3)` in the initial sample is x. Find `(x)/(10)`

Answer

6

Explanation

`{:(KI,+,O_(3),rarr,I_(2)+H_(2)O+O_(2)),(,,v.f = 2,,v.f = 2),(,,"moles",,),(,,=x.,,):}`
`{:(I_(2),+,Na_(2)S_(2)O_(3),rarr,I^(-)+Na_(2)S_(4)O_(6)),(v.f = 2,,M = 2,,),(,,V = 1L,,),(,,v.f = 1,,):}`
eq. of `O_(3)` = eq of `I_(2)` = eq of hypo
`rArr 2 xx x = 2`
x = 1 mole
`rArr` weight % of `O_(3) = (1 xx 48)/(1 xx 48 + 1xx32) xx 100`
`= (3)/(5) xx 100 60%` Ans. `= (60)/(10) = 6`.