Vd of O2 = 16
Vd of O3 = 16
Vd of Mixture = 20
Let no. of moles of O2 : O3 = n : 1
\(20 = \frac{16 \times n + 24 \times 1}{n + 1}\)
\(20n + 20 = 16n + 24\)
\(4n = 4\)
\(n = 1\)
\(\% \text{ of }O_2 = \frac{16 \times n}{16n +24} \times 100\)
\(= \frac{16 \times 1}{16\times 1 + 24} \times 100\)
\(= \frac{16}{40} \times 100\)
\(= 40\%\)