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Tardigrade
Question
Chemistry
5 textN textH2 textS textO4 was diluted from 1 textl texti textt textr texte to 10 textl texti textt textr texte texts . Normality of the solution obtained is
Q.
5
N
H
2
S
O
4
was diluted from
1
l
i
t
r
e
to
10
l
i
t
r
e
s
. Normality of the solution obtained is
53
166
NTA Abhyas
NTA Abhyas 2022
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A
10
N
B
5
N
C
1
N
D
0.5
N
Solution:
N
1
V
1
=
N
2
V
2
5
N
×
1
l
i
t
=
x
×
10
l
i
t
.
∴
x
=
0.5
N