What is the molecular weight of a gas if 27.0 g sample has a pressure of 836 mm hg at 25 c in a 2.00 L flask (R=.0821 L atm/ mol k)
Answers
Answer:
The molecular weight of the gas is
167 u
.
The molar mass is
167 g/mol
.
Explanation:
Use the equation for the ideal gas law to determine moles of the gas. Then divide the mass of gas by the moles to get molar mass.
P
V
=
n
R
T
,
where:
P
is pressure,
V
is volume,
n
is moles,
R
is the gas constant, and
T
is the temperature in Kelvins.
Since the given value for
R
is in
L atm
mol K
, you need to convert the pressure in
mmHg
to
atm
.
1 atm
=
760.0 mmHg
836
mmHg
×
1 atm
760.0
mmHg
=
1.10 atm
The given temperature is
25
∘
C
, so you need to convert it to Kelvins.
25
∘
C + 273.15
=
298 K
Organize the data:
Known/Given
P
=
1.10 atm
V
=
2.00 L
R
=
0.0821 L atm K
−
1
mol
−
1
T
=
298 K
Unknown
n
Solution
Rearrange the equation to isolate
n
. Plug in the known values and solve.
n
=
P
V
R
T
n
=
(
1.10
atm
)
×
(
2.00
L
)
(
0.0821
L
atm
K
−
1
mol
−
1
)
×
(
298
K
)
=
0.0899 mol
Molar mass of gas
M
=
15.0 g
0.0899 mol
=
167 g/mol
(rounded to three significant figures)
Molecular weight is numerically equal to the molar mass.
M
r
=
167 u
Explanation: