Answer
$ 4.11\times 10^{19}$ molecules
Work Step by Step
Gas equation,
$$PV=nRT$$
$P = 1.4 \times 10^{-7} \text{ atm} $
$V = 1.0m^{3}=1000\text{ L}$
$R$ (Universal gas constant) = $0.08206 L\cdot \text{atm}K^{-1}\text{mol}^{-1}$
Temperature in Celsius can be converted to Kelvin by addition of 273
$T = -23^{\circ} C + 273 K = 250 K$
Substitute above for values in gas equation to find n(number of moles of ozone),
$n = \frac{PV}{RT} = \frac{1.4 \times 10^{-7} \times 1000}{0.08206 \times 250} \approx 6.82 \times 10^{-5} \text{ mol}$
One mole is equal to $6.022 \times 10^{23}$ molecules of ozone hence the number of molecules of ozone in $1m^{3}$ can be found as
$6.82 \times 10^{-5} \times 6.022 × 10^{23}\approx 4.11\times 10^{19}$
