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JEE Advanced 2023 Paper 2 Solutions
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© examsnet.com
Question : 29
Total: 52
50
mL
of 0.2 molal urea solution (density
=
1.012
g
mL
−
1
at
300
K
) is mixed with
250
mL
of a solution containing
0.06
g
of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in torr) of the resulting at
300
K
is
[Use : Molar mass of urea
=
60
g
mol
−
1
; gas constant,
R
=
62
L
torr K
−
1
mol
−
1
;
Assume,
∆
mix
H
=
0
,
∆
mix
V
=
0
]
Your Answer:
Validate
Solution:
Mole of urea
=
0.2
Weight of urea
=
0.2
×
60
=
12
g
Weight of solvent
=
1000
g
Weight of solution
=
1012
g
∴
Volume of solution
=
1012
1.012
=
1000
ml
∵
1000
ml
solution contain 0.2 mole
∴
50
ml
solution contain
=
0.2
×
50
1000
=
0.01
Mole of urea in other solution
=
0.06
60
=
0.001
∴
Concentration of solution
=
0.01
+
0.001
300
1000
=
0.0366
∴
π
=
CRT
=
0.0366
×
62
×
300
=
682
© examsnet.com
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