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 $40$ 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.

How many photons would have been required for the complete decomposition of ozone in the original mixture?

• A. $1.2\times {10}^{21}$
• B. $1.2\times {10}^{23}$
• C. $3.6\times {10}^{23}$
• D. $3.6\times {10}^{22}$

Answer 1

Answer: (A)

$1.2\times {10}^{21}$

The redox reactions are as follows:

$O_3+2KI+H_{2}O⟶2KOH+I_2+O_2$

$I_2+2N{a}_2{S}_2{O}_3⟶N{a}_2{S}_4{O}_6+2NaI$

$\therefore$ Millimoles of $O_3=$ MIllimoles of $I_2$ $=\frac{1}{2}\times$ millimoles of $N{a}_2{S}_2{O}_3$

$=\frac{1}{2}\times 40\times \frac{1}{10}=2$ millimoles $=0.002$ moles

Total millimoles of $O_2$ and $O_3$ in the mixture are calculated from $PV=nRT$.

$1\times 1=n\times 0.0821\times 273$

$\therefore n=0.044$ mol

$\therefore$ Moles of $O_2=0.044-0.002=0.042$

Now, mass of $O_2=0.042\times 32g=1.344$ g

Mass of $O_3=0.002\times 48g=0.096$ g

$\%$ of $O_3=\frac{0.096}{\left(1.344+0.096\right)}\times 100=6.6\%$

Number of photons or molecules of ozone $=0.002\times 6.023\times {10}^{23}$ $=1.2\times {10}^{21}$

Hence, the correct option is $\text{A}$