One litre of a sample of ozonized oxygen at 0°C and 1 atm on passing through a Kl solution, liberated iodine which
required 9 ml of a thiosulphate solution. A volume of 12 ml of a '5.675 volume' hydrogen peroxide solution liberated
iodine from another iodide solution, which required 24 ml of the same thiosulphate solution. The volume percent of
ozone in the ozonized oxygen sample is approximately
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Answer:
Reactions involved may be given as:
2KI+H
2
O+O
3
→2KOH+I
2
+O
2
↑
I
2
+2Na
2
S
2
O
3
→2NaI+Na
2
S
4
O
6
1 mole O
3
=2 mole Na
2
S
2
O
3
....(i)
No. of moles of hypo =
Molecular mass(158)
Mass
=
1000×158
E×N×V
where, E
Na
2
S
2
O
3
=158,N=0.08,V=15
∴ No. of moles of hypo =
1000×158
158×0.08×15
=1.2×10
−3
No. of moles of O
3
=
2
1
mole of hypo [from eq. (i)]
=
2
1
×1.2×10
−3
=6×10
−4
mole
Volume of O
3
at NTP= No. of moles ×22400
=6×10
−4
×22400
=13.44 mL at NTP.
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