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JEE Main Chemistry Class 12 Analytical Chemistry Questions
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© examsnet.com
Question : 18
Total: 106
If
50
mL
of
0.5
M
oxalic acid is required to neutralise
25
mL
of
NaOH
solution, the amount of
NaOH
in
50
mL
of given
NaOH
solution is ____g.
[29-Jan-2024 Shift 2]
Your Answer:
Validate
Solution:
Equivalent of Oxalic acid = Equivalents of
NaOH
50
×
0.5
×
2
=
25
×
M
×
1
M
NaOH
=
2
M
W
NaOH
in
50
ml
=
2
×
50
×
40
×
10
−
3
g
=
4
g
© examsnet.com
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