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JEE Main 16 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 53
Total: 90
A
and
B
decompose via first order kinetics with half-lives
54.0
min and
18.0
m
i
n
respectively. Starting from an equimolar non-reactive mixture of
A
and
B
, the time taken for the concentration of
A
to become 16 times that of
B
is ......... min. (Round off to the nearest integer).
[16 Mar 2021 Shift 2]
Your Answer:
Validate
Solution:
Given,
(
t
1
∕
2
)
1
=
54
m
i
n
(
t
1
/
2
)
2
=
18
m
i
n
A
B
t
=
0
,
′
x
′
M
t
=
0
,
′
x
′
M
To calculate,
[
A
t
]
=
16
×
[
B
t
]
...(i)
time
=
?
For Ist order reaction,
[
A
t
]
=
A
0
(
2
)
n
n
=
number of half-lives
From Eq. (i),
[
A
t
]
=
16
×
[
B
t
]
x
(
2
)
n
1
=
x
(
2
)
n
2
×
16
(
2
)
n
2
=
(
2
)
n
1
×
(
2
)
4
⇒
n
2
=
n
1
+
4
t
(
t
1
∕
2
)
2
=
t
(
t
1
∕
2
)
1
+
4
⇒
t
(
1
18
−
1
54
)
=
4
⇒
t
=
4
×
18
×
54
36
=
108
m
i
n
© examsnet.com
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