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Standard XII
Chemistry
Concentration Terms - An Introduction (w/w etc)
Neutralisatio...
Question
Neutralisation of
0.3504
g
m
of acid
(
A
)
requires
27.24
m
l
of
0.15
M
N
a
O
H
, and
M
w
is found from the mass spectral data to be
172.1
g
m
m
o
l
−
1
.
(
a
)
Calculate the
N
.
E
.
of
(
A
)
(
b
)
How many ionisable
H
atoms are there in
(
A
)
?
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Solution
(a) mEq of
N
a
O
H
=
27.24
×
0.15
=
4.086
m
E
q
.
=
4.086
1000
e
q
.
=
0.004086
e
q
.
N.E (Ew)
=
0.3504
0.004086
=
85.76
g
m
.
e
q
−
1
(b) Ionisable H atoms
=
M
w
N
.
E
or
M
w
E
w
=
172.1
85.76
=
2
e
q
.
m
o
l
−
1
=
2
ionisable
H
atoms
m
o
l
−
1
=
A
is dicarboxylic acid
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