Question

One litre of a mixture of $O_2$ and $O_3$ at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required $40ml$ of $M/10$ sodium thiosulphate solution for titration. What Is the weight percent of ozone in the mixture? Ultraviolet radiations of wavelength 300 nm 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

$O_3\to O_2+O$
$\begin{array}{cc}2KI+H_2{SO}_4+\left(O\right)& ⟶K_2{SO}_4+H_{2}O+I_2\\ 2{Na}_2{SO}_3+I_2& {Na}_2{S}_4{O}_6+2NaI\\ 2{Na}_2{S}_2{O}_3+2KI+O_3+H_2{SO}_4⟶& \\ O_2+K_2{SO}_4+H_{2}O+{Na}_2{S}_4{O}_6+2NaI& \end{array}$
2 moles of ${Na}_2{S}_2{O}_3$ react with one mole of $O_3$
No. of moles of ${Na}_2{S}_2{O}_3$ in $40ml$ of $\frac{M}{10}{Na}_2{S}_2{O}_3$
$=\frac{1}{10}\times \frac{1}{1000}\times 40=4\times {10}^{-3}mole$
$\therefore O_3$ reacted (present) $=2\times {10}^{-3}mole$
$(\because 2mole{Na}_2{S}_2{O}_3\equiv 1mole{O}_3)$
Volume of $O_3=2\times {10}^{-3}\times 22\cdot 4L$
$=0\cdot 0448L$
$\therefore$ Volume of $O_2=1-0\cdot 0448=0\cdot 9552L$
Weight of $O_3=2\times {10}^{-3}\times 48=0\cdot 096g$
Weight of $O_2=\frac{0\cdot 9552}{22\cdot 4}\times 32=1\cdot 3646g$
$\therefore$ Weight of mixture
$=0\cdot 096+1\cdot 3646=1\cdot 4606g$
$\therefore \%$ of $O_3$ by weight
$=\frac{0\cdot 096}{1\cdot 4606}\times 100=6\cdot 573\%$
No. of molecules in $2\times {10}^{-3}mole$ of $O_3$
$\begin{array}{c}=2\times {10}^{-3}\times 6\cdot 022\times {10}^{23}\\ =1\cdot 2044\times {10}^{21}\end{array}$
$\therefore$ No. of photons required $=1\cdot 2044\times {10}^{21}$