Tardigrade
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Tardigrade
Question
Chemistry
The freezing point of the solution M is
Q. The freezing point of the solution
M
is
1997
218
IIT JEE
IIT JEE 2008
Solutions
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A
268.7 K
B
268.5 K
C
234.2 K
D
150.9 K
Solution:
In the given solution '
M
',
H
2
O
is solute.
Therefore, molality of
H
2
O
=
0.9
×
46
0.1
×
1000
=
2.4
⇒
−
Δ
T
f
=
K
f
ethanol
×
2.4
=
2
×
2.4
=
4.8
⇒
T
f
=
155.7
−
4.8
=
150.9
K