A 672 ml of a mixture of oxygen - ozone at N.T.P. were found to be weigh 1 gm. Calculate the volume of ozone in the mixture.

Explanation

Let V ml of ozone are there in the mixture
`rArr ( 672 - V ) m, = `vol. of oxygen
Mass of ozone at N.T.P. `= ( V )/( 22400 )xx 48`
Mass of oxygen at N.T.P. `= ( 672 - V )/( 22400 ) xx 32 `
`rArr ( V )/( 22400 ) xx 48 + ( 672 -V)/( 22400 ) xx 32 = 1 `
`rArr` On solving we get `:` V = 56 ml