40dm3 of oxygen gas was collected over water at 834 torr pressure at 25°C find the volume of dry oxygen gas at standard "Temperature" "pressure"
( The vopour pressure of water at 25°C is 24 torr).
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P
H
2
O
=
35
torr
Explanation:
At constant number of mole
n
and Temperature
T
, using the Ideal Gas Law
P
V
=
n
R
T
we can write:
P
1
V
1
=
P
2
V
2
We should calculate the pressure of pure
O
2
in a
2.01
L
volume:
P
1
=
P
2
V
2
V
1
=
785
torr
×
1.92
L
2.01
L
=
750
torr
In the original mixture, the total pressure
P
t
is the sum of partial pressures of
H
2
O
and
O
2
:
P
t
=
P
H
2
O
+
P
O
2
⇒
P
H
2
O
=
P
t
−
P
O
2
=
785
torr
−
750
torr
=
35
torr
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