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Tardigrade
Question
Chemistry
The amount of oxalic acid (mol. wt. 63 ) required to prepare 500 mL of its 0.10 N solution is
Q. The amount of oxalic acid (mol. wt.
63
) required to prepare
500
m
L
of its
0.10
N
solution is
2274
300
Solutions
Report Error
A
0.315 g
B
3.150 g
C
6.300 g
D
63.00 g
Solution:
1000
m
L
of
1
N
oxalic acid sol
=
63
g
500
m
L
of
0.1
N
oxalic acid sol
=
1000
63
×
500
×
0.1
=
3.15
g