1.

The weight of 600 ml of a mixture of ozone and oxygen is 1 gm at STP. The volume of oxygen in the mixture is

Answer»

200 ML
500 ml
400 ml
300 ml

Solution :`M_(mix)=M`. Then `(1)/(M)=N`
`"volume"_("STP")=(1)/(M)xx22400=600impliesM=(112)/(3)`
If moles of `O_(2)=X % O_(3)=y`
then `M=(32x+48y)/(x+y)=(112)/(3)`
`IMPLIES 96x+144y=112x+112y`
`(x)/(y)=(1)/(2)`
`therefore V_(O_(2))=(1)/(3)xx600=200mL`


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