Solution:
Solution
The molecular mass of oxygen= 32
The molecular mass of ozone=48
Therefore,
Let volume of $${ O }_{ 2 }=x\quad ml$$
volume of $${ O }_{ 3 }=600- x\ ml$$
At STP, wt of x ml of $${ O }_{ 2 }=\dfrac { 32 \times x }{ 22400 } gm$$
wt of $$600-x\ ml\ { O }_{ 3 }=\dfrac { 48\times (600-x) }{ 22400 } gm$$
Hence,$$\dfrac { 32\ \times x }{ 22400 } + \dfrac { 48\times (600-x) }{ 22400 } =1\\$$
Therefore, volume of ozone in the mixture is $$400\ ml$$