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Byju's Answer
Standard XII
Chemistry
Molarity
Match the rea...
Question
Match the reactions given in column I with molarities given in column II.
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Solution
(A) millimoles of
Z
n
=
196.2
m
g
65.4
m
g
=
3
m
m
o
l
3 mmol of Zn
≡
2 mmol of
l
F
e
(
C
N
)
6
]
4
−
≡
2
m
m
o
l
40
m
L
=
0.05
M
0.05 M
K
4
[
F
e
(
C
N
)
6
]
should be used so that 40 mL of a solution titrates 196.2 mg
Z
n
(dissolved) by forming
K
2
Z
n
3
[
F
e
(
C
N
)
6
]
2
.
(Atomic weight
Z
n
=
65.4
g
m
o
l
−
1
)
(B) Moles of
B
a
S
O
4
=
2.33
233
=
0.01
m
o
l
=
0.01
m
o
l
N
a
2
S
O
4
M
N
a
2
S
O
4
=
m
m
o
l
m
L
=
0.01
×
10
3
40
=
0.25
M
The molarity of the
N
a
2
S
O
4
solution is 0.25 M.
(C) mmol of
T
h
4
+
=
116
m
g
232
m
g
=
0.5
m
m
o
l
=
0.5
×
2
×
mmol of
H
2
C
2
O
4
M
H
2
C
3
O
4
=
m
m
o
l
m
L
=
0.5
×
2
50
=
0.02
M
The molarity of
H
2
C
2
O
4
is 0.02 M.
(D) mmol of
A
g
N
O
3
=
10
×
0.5
=
5
m
m
o
l
=
2
×
5
m
m
o
l
of
N
a
C
N
M
N
a
C
N
=
m
m
o
l
m
L
=
2
×
5
100
=
0.1
M
The molarity of
N
a
C
N
is
0.1
M
.
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