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JEE Main Chemistry Class 11 Solutions Part 2 Questions
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© examsnet.com
Question : 35
Total: 80
The vapour pressure of
30
%
(
w
∕
v
)
aqueous solution of glucose is _______
mm
Hg
at
25
∘
C
.
[Given : The density of
30
%
(
w
∕
v
)
, aqueous solutions of glucose is
1.2
g
cm
−
3
and vapour pressure of pure water is
24
mm
Hg
.]
(Molar mass of glucose is
180
g
mol
−
1
)
[15-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
24
−
P
s
P
s
=
m
×
18
1000
wt of solute
=
30
gm
Volume of solution
=
100
mL
wt. of solution
=
1.2
×
100
=
120
gm
wt. of solvent
=
120
−
30
=
90
gm
m
30
×
1000
180
×
90
=
185
24
−
P
s
P
s
=
1.85
×
18
1000
24
−
P
s
=
0.0333
P
s
P
s
(
1.033
)
=
24
P
s
=
23.22
© examsnet.com
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