10 ml of H2O2 solution on treatment with KI with titartiin liberated I2 required 10 ml of 1N hypo. Thus H2O2 is
Answers
Answered by
1
Explanation:
We know that
H
2
O
2
+2H
+
+2I
−
→I
2
+2H
2
O
∴ 1mole H
2
O gives I mole I
2
we know, 1 mole H
2
O
2
=34g
1 mole I
2
=254g
∴254g of I
2
are liberated 34 g
H
2
O
2
∴0.5g will be liberated =
254
34
×0.5
=0.0669grm
% strength of H
2
O
2
=
vol
weightofH
2
O
2
=
10
0.0669
×100=0.669%
Answered by
1
Answer:
27/422=(x2^3+0^5) 3^2 is the answer
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