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JEE Mains 11-Apr-2023 Shift 1 Solved Paper
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© examsnet.com
Question : 83
Total: 90
0.004
M
K
2
SO
4
solution is isotonic with
0.01
M
glucose solution. Percentage dissociation of
K
2
SO
4
is (Nearest integer)
[11-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
For isotonic solution
(
ic
)
glucose
=
(
ic
)
K
2
SO
4
0.01
=
i
(
0.004
)
i
=
0.01
0.004
=
10
4
=
5
2
1
+
(
n
−
1
)
α
=
5
2
1
+
(
3
−
1
)
α
=
5
2
(
∵
n
=
3
for
K
2
SO
4
)
2
α
=
3
2
α
=
3
4
→
75
%
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