1.

20.0 mL of a mixture of oxygen (O_(2)) and ozone (O_(3)) was heated till ozone was completely decomposed. The mixture of cooling was found to have a volume of 21 mL. Calculate the percentage of ozone by volume in the mixture.

Answer»

Solution :Decomposition of OZONE TAKES place as follows :
`underset("2 volume")(2O_(3))rarr underset("3 volume")(3O_(2))`
SUPPOSE ozone in the mixture = x mL.
Then `O_(2)=(20-x)mL`
2 mL of ozone on decomposition give `O_(2)=3mL`
`THEREFORE"x mL of ozone on decomposition will give "O_(2)=(3)/(2)x="1.5 x mL"`
`therefore" Total volume of mixture after decomposition "=1.5x+(20-x)mL=20+0.5x`
`therefore""20+0.5x=21"(Given)or"0.5x=1"or"x=2mL`
`therefore"Percentage of ozone in the mixture"=(2)/(20)xx100=10%.`


Discussion

No Comment Found

Related InterviewSolutions