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Test Index
Redox Reactions
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Section:
Chemistry
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© examsnet.com
Question : 11 of 73
Marks:
+1
,
-0
0.1 M solution of KI reacts with excess of
H
2
S
O
4
\mathrm{H}_2\mathrm{SO}_4
H
2
SO
4
and
K
I
O
3
\mathrm{KIO}_3
KIO
3
solutions. According to equation
5
I
−
+
I
O
3
−
+
6
H
+
⟶
3
I
2
+
3
H
2
O
5\mathrm{I}^{-}+\mathrm{IO}_3^{-}+6\mathrm{H}^{+}\longrightarrow 3\mathrm{I}_2+3\mathrm{H}_2\mathrm{O}
5
I
−
+
IO
3
−
+
6
H
+
⟶
3
I
2
+
3
H
2
O
Identify the correct statements :
(A) 200 mL of KI solution reacts with 0.004 mol of
K
I
O
3
\mathrm{KIO}_3
KIO
3
(B) 200 mL of KI solution reacts with 0.006 mol of
H
2
S
O
4
\mathrm{H}_2\mathrm{SO}_4
H
2
SO
4
(C) 0.5 L of KI solution produced 0.005 mol of
I
2
\mathrm{I}_2
I
2
(D) Equivalent weight of
K
I
O
3
\mathrm{KIO}_3
KIO
3
is equal to
(
Molecular weight
5
)
\left(\frac{\text{Molecular weight}}{5}\right)
(
5
Molecular weight
)
Choose the correct answer from the options given below :
[29 Jan 2025 Shift 2]
(A) and (D) only
(C) and (D) only
(A) and (B) only
(B) and (C) only
Validate
Solution:
E
K
I
O
3
=
Molecular weight
n
f
\;E_{\mathrm{KIO}_3}=\frac{\text{Molecular weight}}{n_f}
E
KIO
3
=
n
f
Molecular weight
n
f
=
5
\;n_f=5
n
f
=
5
E
K
I
O
3
=
Molecular weight
5
\;E_{\mathrm{KIO}_3}=\frac{\text{Molecular weight}}{5}
E
KIO
3
=
5
Molecular weight
(D) is correct
meq of
K
I
=
0.1
×
200
=
20
\mathrm{KI}=0.1\times200=20
KI
=
0.1
×
200
=
20
meq of
K
I
O
3
=
4
×
5
=
20
\mathrm{KIO}_3=4\times5=20
KIO
3
=
4
×
5
=
20
(A) is correct
© examsnet.com
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