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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. |
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Answer» 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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