Question

One litre 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 $40$ mL of $\frac{M}{10}$ sodium thiosulphate solution for titration. Ultraviolet radiation of wavelength $300$ nm can decompose ozone. Assuming that one photon can decompose one ozone molecule.

What is the total amount of $O_2$ and $O_3$ present in $1\ell$ of the mixture at STP?

• A. $0.044$
• B. $0.022$
• C. $0.22$
• D. $0.44$

Answer 1

The correct option is A $0.044$

The reaction of $O_3$ with $I^{-}$ in acidic medium is,

$O_3+2I^{-}+2H^{+}⟶I_2+O_2+H_{2}O$

Hence, $1mol{O}_3=1mol{I}_2$

The reaction of $I_2$ with $S_2{O}_3^{2-}$ is,

$2S_2{O}_3^{2-}+I_2⟶S_4{O}_6^{2-}+2I^{-}$

Hence, $2mol{S}_2{O}_3^{2-}=1mol{I}_2$

Amount of $S_2{O}_3^{2-}$ consumed$=(40\times {10}^{-3})\ell \left(\frac{1}{10}㏖/\ell \right)$

$=40\times {10}^{-4}mol$

Thus, $40\times {10}^{-4}㏖S_2{O}_3^{2-}=20\times {10}^{-4}mol{I}_2$

$=20\times {10}^{-4}mol{O}_3$

Mass of $O_3$ present in $1\ell$ of mixture$=(20\times {10}^{-4}mol)\left(48ℊ/mol\right)$

$=9.6\times {10}^{-2}ℊ$

Total amount of $O_2$ and $O_3$ present in $1\ell$ of mixture at STP is,

$n_{total}=\frac{PV}{RT}=\frac{\left(1atm\right)\left(1\ell \right)}{\left(0.082atm\ell /kmol\right)\left(273K\right)}$

$=4.462\times {10}^{-2}mol$

$=0.04462mol$

Hence, amount of $O_2$ present in $1\ell$ of mixture,

$=(4.6262\times {10}^{-2}-20\times {10}^{-4})mol=4.6262\times {10}^{-2}mol$

Mass of $O_2$ present in $1\ell$ of mixture

$=(4.262\times {10}^{-2}mol)\left(32ℊ/mol\right)=1.364ℊ$

Mass percent of $O_3$ in the mixture$=\frac{9.6\times {10}^{-2}\times 100}{9.6\times {10}^{-2}+1.364}$

$=6.575$