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JEE Main Chemistry Class 11 Some Basic Concepts of Chemistry Part 1 Questions
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© examsnet.com
Question : 66
Total: 100
3 g of activated charcoal was added to
50
m
L
of acetic acid solution
(
0.06
N
)
in a flask. After an hour it was filtered and the strength of the filtrate was found to be
0.042
N
. The amount of acetic acid adsorbed (per gram of charcoal) is:
[2015]
42
m
g
54
m
g
18
m
g
36
m
g
Validate
Solution:
Let the weight of acetic acid initially be
w
1
in
50
m
L
of
0.060
N
solution.
N
=
w
1
×
1000
M
.
w
t
⋅
×
50
(
Normality
=
0.06
N
)
0.06
=
w
1
×
1000
60
×
50
⇒
w
1
=
0.06
×
60
×
50
1000
=
0.18
g
=
180
m
g
After an hour, the strength of acetic acid
=
0.042
N
so, let the weight of acetic acid be
w
2
N
=
w
2
×
1000
60
×
50
0.042
=
W
2
×
1000
3000
⇒
w
2
=
0.126
g
=
126
m
g
So amount of acetic acid adsorbed per
3
g
=
180
−
126
m
g
=
54
m
g
∴
amount of acetic acid absorbed per
g
=
54
∕
3
=
18
m
g
© examsnet.com
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