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JEE Mains 13-Apr-2023 Shift 1 Solved Paper
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© examsnet.com
Question : 86
Total: 90
A metal surface of
100
cm
2
area has to be coated with nickel layer of thickness
0.001
mm
. A current of
2
A
was passed through a solution of
Ni
(
NO
3
)
2
for '
x
' seconds to coat the desired layer. The value of
x
is (Nearest integer)
(
ρ
Ni
(
.
.
density of Nickel) is
10
g
mL
−
1
, Molar mass of Nickel is ________
60
g
mol
−
1
F
=
96500
C
mol
−
1
)
[13-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
Volume of nickel required
=
100
×
0.001
×
10
−
3
×
100
=
0.01
cm
3
Mass of Nickel required
=
0.01
×
10
=
0.1
gm
Moles
=
0.1
60
=
1
600
mol
Ni
2
+
+
2
e
−
→
Ni
(
s
)
for coating of
1
mol
Ni
, charge required
=
2
×
96500
C
for coating of
1
600
mol, charge required
=
2
×
96500
×
1
600
C
=
965
3
C
I
=
q
t
t
=
965
∕
3
2
=
160.83
s
e
c
≈
161
© examsnet.com
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