The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$

The given reaction is:
$3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$
The expression for the equilibrium constant is:
$K_{c} = \frac{[O_{3}]^{2}}{[O_{2}]^{3}}$
Given values are:
$K_{c} = 2.0 \times 10^{-50}$
$[O_{2}] = 1.6 \times 10^{-2} \, M$
Substituting the values into the expression:
$2.0 \times 10^{-50} = \frac{[O_{3}]^{2}}{(1.6 \times 10^{-2})^{3}}$
$[O_{3}]^{2} = 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^{3}$
$[O_{3}]^{2} = 2.0 \times 10^{-50} \times 4.096 \times 10^{-6}$
$[O_{3}]^{2} = 8.192 \times 10^{-56}$
Taking the square root:
$[O_{3}] = \sqrt{8.192 \times 10^{-56}}$
$[O_{3}] \approx 2.86 \times 10^{-28} \, M$
Thus,the concentration of $O_{3}$ is $2.86 \times 10^{-28} \, M$.