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Byju's Answer
Standard XII
Biology
Osmotic Pressure and Osmotic Potential
1. Pcl5=pcl3+...
Question
1. Pcl5=pcl3+cl2 Vapour density is found to be hundred when 1 mole of pcl5 is taken in 10dm3 flask at 300k. Thus, equilibrium pressure is :- 1. 1.00 atm 2. 4.92 atm 3. 2.46 atm 4. 2.57 atm
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Q.
In an evacuated closed isolated chamber at
250
o
C
.
.02
mole
P
C
l
5
and
.01
mole
C
l
2
are mixed
(
P
C
l
5
⇌
P
C
l
3
+
C
l
2
)
. At equilibrium density of mixture was
2.49
g
/
L
and pressure was
1
atm. The number of total moles at equilibrium will be approximately:
Q.
In the reaction,
P
C
l
5
⇌
P
C
l
3
+
C
l
2
, the amount of
P
C
l
5
,
P
C
l
3
and
P
C
l
2
at equilibrium are
2
moles each and the total pressure is
3
atm. The equilibrium constant
K
p
is:
Q.
'
a
' mol of
P
C
l
5
undergoes thermal dissociation as
P
C
l
5
⇌
P
C
l
3
+
C
l
2
; the mole fraction
P
C
l
3
at equilibrium is 0.25 and the total pressure is 2 atm. The partial pressure is
C
l
2
at equilibrium is:
Q.
A reaction at equilibrium involving 2 moles each of
P
C
l
5
,
P
C
l
3
,
and
C
l
2
is maintained at
250
o
C
and a total pressure of 3 atm. The value of
K
p
is :
Given:
P
C
l
5
⇌
P
C
l
3
+
C
l
2
Q.
2 mole each of
P
C
l
5
and
C
l
2
and 1 mole of
P
C
l
3
are taken in 5lt. flask at
27
o
C
. After following equilibrium is attained in gaseous phase,
P
C
l
5
⇌
P
C
l
3
+
C
l
2
,
P
C
l
5
is found to be 40% dissociated. Calculate equilibrium constant
K
c
.
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