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Question
Chemistry
28 g KOH is required to completely neutralise CO 2 produced on heating 60 g of impure CaCO 3 The percentage purity of CaCO 3 is approximately (molar masses of KOH and CaCO 3 are 56 and 100 g mol -1, respectively )
Q.
28
g
K
O
H
is required to completely neutralise
C
O
2
produced on heating
60
g
of impure
C
a
C
O
3
The percentage purity of
C
a
C
O
3
is approximately (molar masses of
K
O
H
and
C
a
C
O
3
are
56
and
100
g
m
o
l
−
1
, respectively )
1678
242
TS EAMCET 2018
Report Error
A
41.6
B
40
C
20.8
D
83.3
Solution:
The given relation
C
a
C
O
3
2 KOH
Δ
C
O
2
means,
2 moles of KOH will neutralise 1 mole of
C
O
2
Given,
(i)
K
O
H
( used )
=
28
g
(ii) Molar mass of
K
O
H
(
M
)
=
56
g
(iii) Molar mass of
C
a
C
O
3
=
100
g
(iv) Impure
C
a
C
O
3
=
60
g
∵
112
g
(
56
×
2
)
of
K
O
H
will neutralise
=
100
g
or
C
a
C
O
3
∴
28
g
K
O
H
will neutralise
=
112
100
×
28
=
25
g
of
C
a
C
O
3
Also,
∵
60
g
(impure) of
C
a
C
O
3
has
25
g
pure
C
a
C
O
3
.
∴
100
g
(impure)
C
a
C
O
3
has pure
C
a
C
O
3
=
60
25
×
100
=
41.6
g