Correct Answer - A
Let the volume of `O_3` be `x` ml
Volume of `O_2` present `=(600-x)ml`
22400ml of `O_3` and `O_2` at STP will weigh 48 and 32 gm respectively.
The weight of `xml` of `O_3=(x xx48)/(22400)gm`
The weight of `(600-x)ml` of `O_2=((600-x))/(22400)xx32`
Total weight of ozonised `O_2(600ml)` is `(48x)/(22400)+((600-x)xx32)/(22400)=1.0`
`x=200ml`