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AIEEE 2004 Chemistry Solved Paper
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© examsnet.com
Question : 2
Total: 75
To neutralize completely
20
mL
of
0.1
M
aqueous solution of phosphorous acid
(
H
3
PO
3
)
, the volume of
0.1
M
aqueous
KOH
solution required is
40
mL
20
mL
10
mL
60
mL
Validate
Solution:
H
3
PO
3
+
2
KOH
→
K
2
HPO
3
+
2
H
2
O
Let the volume of
KOH
solution is
=
V
mL
=
V
1000
litre
Volume of
H
3
PO
3
=
20
mL
=
20
1000
litre
No of moles of
H
3
PO
3
present
=
20
1000
×
0.1
No of moles of KOH present
=
V
1000
×
0.1
Stoichiometric coefficient of
H
3
PO
3
=
1
Stoichiometric coefficient of
KOH
=
2
∴
20
10001
×
0.1
1
=
V
1000
×
0.1
2
⇒
V
=
40
mL
© examsnet.com
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