Question

$Onelitre$ of a mixture of $O_2$ and $O_3$ at NTP was allowed to react with an excess of an acidified solution of $KI$. The iodine liberated required $40mL$ of $M/10$ sodium thiosulphate solution for titration. What is the weight percent of the ozone in the mixture? Ultraviolet radiation of wavelength $300nm$ can decompose ozone. Assuming that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture?

Answer 1

Let the total moles of $O_2$ and $O_3$ in the mixture be $n$.
Applying $PV=nRT$
$1\times 1=n\times 0.0821\times 273$
$n=0.044mole$
Moles of $O_3=$ moles of $I_2=1/2$ moles of $N{a}_2{S}_2{O}_3$
$=\frac{1}{2}\times \frac{1}{10}\times \frac{40}{1000}=0.002moles$
Moles of $O_2$ in the mixture $=0.044-0.002=0.042moles$
Mass of $O_2=0.042\times 32=1.344g$
Mass of $O_3=0.002\times 48=0.096g$
% $O_3=\frac{0.096}{1.44}\times 100=6.67$
No. of photons required to decompose $0.002$ moles of ozone
$=0.002\times 6.02\times {10}^{23}$
$=1.204\times {10}^{21}$.