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JEE Main 17 March 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 59
Total: 90
The oxygen dissolved in water exerts a partial pressure of
20
k
P
a
in the vapour above water. The molar solubility of oxygen in water is .....
×
10
−
5
m
o
l
d
m
−
3
. (Round off to the nearest integer).
[Given, Henry's law constant
(
K
H
)
=
8.0
×
10
4
k
P
a
for
O
2
, density of water with dissolved oxygen
=
1.0
k
g
d
m
−
3
].
[17 Mar 2021 Shift 1]
Your Answer:
Validate
Solution:
Given, partial pressure of
O
2
=
20
k
P
a
K
H
(
Henry's constant
)
=
8
×
10
4
k
P
a
From Henry's law,
p
(
g
)
=
[
K
H
]
χ
O
2
where,
χ
O
2
=
solubility of oxygen
20
×
10
3
=
(
8
×
10
4
×
10
3
)
χ
o
2
⇒
χ
O
2
=
20
8
×
10
4
Solubility
=
2.5
×
10
−
4
=
25
×
10
−
5
© examsnet.com
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