Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
4 g caustic soda is dissolved in 100 cc of solution. The normality of solution is:
Q.
4
g
caustic soda is dissolved in
100
cc
of solution. The normality of solution is:
1757
203
AFMC
AFMC 2001
Report Error
A
0
B
0.5
C
1
D
1.5
Solution:
Normality is defined as number of equivalents of a substance present in
1
L
of solution
N
=
volume
(
L
)
gram equivalents of solute
=
equivalent mass of solute
×
vol
.
(
L
)
mass of solute
(
g
)
N
=
40
4
×
0.1
1
=
1
N