Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
If 3 g of glucose (molar mass =180 g ) is dissolved in 60 g of water at 15° C, the osmotic pressure of the solution will be
Q. If
3
g
of glucose (molar mass
=
180
g
)
is dissolved in
60
g
of water at
1
5
∘
C
, the osmotic pressure of the solution will be
546
108
KCET
KCET 2022
Solutions
Report Error
A
6.57 atm
33%
B
5.57 atm
16%
C
0.34 atm
26%
D
0.65 atm
25%
Solution:
π
=
C
.
R
.
T
=
M
2
w
2
V
(
m
ℓ
)
1000
×
R
.
T
⇒
180
3
×
60
1000
×
0.0821
×
288
=
6.568
a
t
m