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Byju's Answer
Standard XII
Chemistry
Molality
"PCl5(g) PCl3...
Question
"PCl5(g) PCl3(g) + Cl2(g) Vapour density is found to be 100 when 1 mole of PCl5 is taken in a 10 litre flask at 27* C." How did the initial density equals to M/2? Since volume is 10 L shouldn't ot be M/10?
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Similar questions
Q.
K
c
for
P
C
l
5
(
g
)
⇌
P
C
l
3
(
g
)
+
C
l
2
(
g
)
, is
0.04
at
250
o
C
. How many moles of
P
C
l
5
must be added to a
3
litre flask to obtain a
C
l
2
concentration of
0.15
M
?
Q.
The equilibrium constant for the equilibrium
P
C
l
5
(
g
)
⇌
P
C
l
3
(
g
)
+
C
l
2
(
g
)
at a particular temperature is
2
×
10
−
2
m
o
l
L
−
1
. The number of moles of
P
C
l
5
that must be taken in a one litre flask at the same temperature to obtain a concentration of 0.20 mole of chlorine at equilibrium is:
Q.
K
c
for
P
C
l
5
(
g
)
⇌
P
C
l
3
(
g
)
+
C
l
2
(
g
)
is 0.45 at
250
∘
C
. How many mole of
P
C
l
5
must be added to a 3 - litre flask to obtain a
C
l
2
concentration of 0.15 M?
Q.
One mole of
P
C
l
5
(
g
)
dissociates and attains equilibrium after t seconds at
27
∘
C
:
P
C
l
5
(
g
)
⇋
P
C
l
3
(
g
)
+
C
l
2
(
g
)
At equillibrium, the vapour density is found to be 100 and the volume of the system is 10.425 L at
27
∘
C
.
The total pressure of the system at equilibrium will be
atm.
R
=
0.082
L
a
t
m
m
o
l
−
1
K
−
1
Q.
How much
P
C
l
5
must be added to a one litre vessel kept at
250
o
C in order to obtain
0.1
mole of
C
l
2
gas?
[
K
C
for
P
C
l
5
(
g
)
⇌
P
C
l
3
(
g
)
+
C
l
2
(
g
)
is
0.0414
mol/L]
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