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Tardigrade
Question
Chemistry
A 0.001 molal solution of [Pt ((NH)3)4 (Cl)4] in water has a freezing point depression of 0.0054° C . If Kf of water is 1.80 , then correct formulation of above molecule is :
Q. A
0.001
molal solution of
[
Pt
(
(
N
H
)
3
)
4
(
Cl
)
4
]
in water has a freezing point depression of
0.005
4
∘
C
. If
K
f
of water is
1.80
, then correct formulation of above molecule is :
99
168
NTA Abhyas
NTA Abhyas 2020
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A
[
Pt
(
(
N
H
)
3
)
4
(
Cl
)
3
]
Cl
B
[
Pt
(
(
N
H
)
3
)
4
(
Cl
)
2
]
(
Cl
)
2
C
[
Pt
(
(
N
H
)
3
)
4
Cl
]
(
Cl
)
3
D
[
Pt
(
(
N
H
)
3
)
4
(
Cl
)
4
]
Solution:
Δ
T
f
=
i
K
f
⋅
m
0.0054
=
i
×
1.80
×
0.001
i
=
3
if
α
=
100%
then,
i
=
n
n
=
3