Calculate the no of atoms in 5.6 litre diatomic molecule
Answers
Answer:
2.80
×
10
23
atoms
Explanation:
We're asked to calculate the number of atoms in
5.6
L
of a diatomic gas at normal temperature and pressure (NTP).
To do this, we can use the ideal-gas equation:
P
V
=
n
R
T
where
P
is the pressure exerted by the gas, in units of
atm
(at NTP, the pressure is defined as
1
atm
V
is the volume occupied by the gas, in units of
L
(given as
5.6
L
)
n
is the number of moles of gas present (we'll need to find this)
R
is the universal gas constant, equal to
0.082057
L
⋅
atm
mol
⋅
K
T
is the absolute temperature of the gas, in units of
K
(the temperature at NTP is defined as
20
o
C
, which is
20
o
C
+
273.15
=
293.15
K
)
Plugging in known values, and solving for the quantity,
n
, we have
n
=
P
V
R
T
=
(
1
atm
)
(
5.6
L
)
(
0.082057
L
⋅
atm
mol
⋅
K
)
(
293.15
K
)
=
0.233
mol gas
Using Avogadro's number, we can convert from moles of gas to molecules:
0.233
mol gas
(
6.022
×
10
23
l
molecules gas
1
mol gas
)
=
1.40
×
10
23
molecules gas
We're given that the gas is diatomic, meaning there are two atoms per molecule, so the total number of atoms is
1.40
×
10
23
molecules gas
(
2
l
atoms gas
1
molecule gas
)
=
2.80
×
10
23
atoms
Answer:
Therefore, 5.6 L is occupied by 0.25 moles of a gas, which will therefore contain 0.25 x 6.023 x 10^23 particles.1.50575 * 10^23 molecules of Oxygen gas in 5.6 L. One molecule contains 2 atoms of oxygen, so the number of atoms is double the number of molecules. 3.0115 * 10^23 atoms of oxygen in 5.6 L oxygen gas.