Calculate the volume accupied by 6.52×10^22 molecules of ammonia gas at STP
Answers
The volume is 2.271 L.
Explanation:
Method 1. Using the Ideal Gas Law
We can use the Ideal Gas Law to solve this problem.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
a
a
p
V
=
n
R
T
a
a
∣
∣
−−−−−−−−−−−−−−−
where
p
is the pressure
V
is the volume
n
is the number of moles
R
is the gas constant
T
is the temperature
We can rearrange the Ideal Gas Law to get
V
=
n
R
T
p
Step 1. Calculate the moles of ammonia
n
=
6.023
×
10
22
molecules NH
3
×
1 mol NH
3
6.022
×
10
23
molecules NH
3
=
0.100 02 mol NH
3
Step 2. Calculate the volume at STP
Remember that STP is defined as 0 °C and 1 bar.
n
=
0.100 02 mol
R
=
0.083 14 bar⋅L⋅K
-1
mol
-1
T
=
(0 + 273.15) K
=
273.15 K
p
=
1 bar
V
=
n
R
T
p
=
0.100 02
g
×
0.083 14
bar
⋅
L
⋅
K
-1
mol
-1
×
273.15
K
1
bar
=
2.271 L
Method 2. Using the molar volume
We know that there are 0.100 02 mol of
NH
3
.
We also know that the molar volume of a gas is 22.71 L at STP.
∴
V
=
0.100 02
mol NH
3
×
22.71 L NH
3
1
mol NH
3
=
2.271 L NH
3
1 molecule of NH3 occupies 22.4 l
6.52×10^22 molecule of NH3 occupies 22.4* 6.52×10^22
Multiply and get the answer....