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50 mL sample of ozonised oxygen at NTP w...

50 mL sample of ozonised oxygen at NTP was passed through a solution of potassium iodide. The liberated iodine required 15 mL of `0.08 N Na_(2)S_(2)O_(3)` solution for complete titration. Calculate the volume of ozone at NTP in the given sample.

Text Solution

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Reactions involved may be given as :
`2KI+H_(2)O+O_(3) to 2KOH+I_(2)+O_(2) uarr`
`I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)`
1 mole `O_(3)=2 mol e Na_(2)S_(2)O_(3)`
No. of moles of hypo `=("Mass")/("Molecular mass "(158))`
`(ExxNxxV)/(1000xx158)`
where, `E_(Na_(2)S_(2)O_(3))=158, N = 0.08, V=15`
`therefore` No. of moles of hypo `=(158xx0.08xx15)/(1000xx158)=1.2xx10^(-3)`
No. of moles of `O_(3)=(1)/(2)` mole of hypo
`=(1)/(2)xx1.2xx10^(-3)`
`=6xx10^(-4)` mole
Volume of `O_(3)` at NTP =No. of moles`xx 22400`
`=6xx10^(-4)xx22400`
=13.44 mL at NTP
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