Question

The density of a certain gas is 1.19g litre at 50°C and 730 mm pressure​

Answer 1

26.3 g mol
−
1

Explanation:
Your starting point here will be the ideal gas law equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
P
V
=
n
R
T
a
a
∣
∣
−−−−−−−−−−−−−−−

, where
P
- the pressure of the gas
V
- the volume it occupies
n
- the number of moles of gas
R
- the universal gas constant, usually given as
0.0821
atm
⋅
L
mol
⋅
K

T
- the absolute temperature of the gas
Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,
ρ
, under those conditions for pressure and temperature, and its molar mass,
M
M
.
You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass
m
of this gas, you can express its molar mass as the ratio between
m
and
n
, the number of moles it contains
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
M
M
=
m
n
a
a
∣
∣
−−−−−−−−−−−−−


(
1
)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.
This means that for the mass
m
of this gas, you can express its density as the ratio between
m
and the volume it occupies
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
ρ
=
m
V
a
a
∣
∣
∣
−−−−−−−−−−−


(
2
)

Plug equation
(
1
)
into the ideal gas law equation to get
P
V
=
m
M
M
⋅
R
T

Rearrange to get
P
V
⋅
M
M
=
m
⋅
R
T

P
⋅
M
M
=
m
V
⋅
R
T

M
M
=
m
V
⋅
R
T
P

Finally, use equation
(
2
)
to write
M
M
=
ρ
⋅
R
T
P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find
M
M
=
1.02
g
L
⋅
0.0821
atm
⋅
L
mol
⋅
K
⋅
(
273.15
+
37
)
K
0.990
atm
M
M
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
26.3 g mol
−
1
a
a
∣
∣
−−−−−−−−−−−−−−−−


Hope this helps you