Answer :
Answer:
To solve this problem, we can use the ideal gas law to relate the pressure, volume, temperature, and number of moles of the gas. The ideal gas law is given by:
�
�
=
�
�
�
PV=nRT
Where:
�
P is the pressure of the gas (in bar)
�
V is the volume of the gas (in liters)
�
n is the number of moles of the gas
�
R is the ideal gas constant (
0.0831
 
L
â‹…
bar
â‹…
K
−
1
â‹…
mol
−
1
0.0831Lâ‹…barâ‹…K
−1
â‹…mol
−1
)
�
T is the temperature of the gas (in Kelvin)
First, let's calculate the number of moles of
�
�
2
CO
2
​
 dissolved in the carbonated water using the given pressure and volume:
�
=
�
�
�
�
n=
RT
PV
​
�
=
2.4
 
bar
â‹…
1.00
 
L
0.0831
 
L
â‹…
bar
â‹…
K
−
1
â‹…
mol
−
1
â‹…
298
 
K
n=
0.0831Lâ‹…barâ‹…K
−1
â‹…mol
−1
â‹…298K
2.4barâ‹…1.00L
​
�
=
0.0996
 
moles
n=0.0996moles
Now, we can convert the moles of
�
�
2
CO
2
​
 to grams using the molar mass of
�
�
2
CO
2
​
, which is approximately
44.01
 
g/mol
44.01g/mol:
�
=
�
×
molar mass
m=n×molar mass
�
=
0.0996
 
moles
×
44.01
 
g/mol
m=0.0996moles×44.01g/mol
�
=
4.37
 
grams
m=4.37grams
So, there are approximately 4.37 grams of
�
�
2
CO
2
​
 dissolved in the 1.00 L bottle of carbonated water.
Next, to find the volume the dissolved
�
�
2
CO
2
​
 would occupy as a gas at 298 K and 1.00 bar, we can use the ideal gas law again. We'll use the number of moles we calculated earlier and the new pressure and temperature:
�
=
�
�
�
�
V=
P
nRT
​
�
=
0.0996
 
moles
×
0.0831
 
L
â‹…
bar
â‹…
K
−
1
â‹…
mol
−
1
â‹…
298
 
K
1.00
 
bar
V=
1.00bar
0.0996moles×0.0831L⋅bar⋅K
−1
â‹…mol
−1
â‹…298K
​
�
≈
2.46
 
L
V≈2.46L
So, the dissolved
�
�
2
CO
2
​
 would occupy approximately 2.46 liters as a gas at 298 K and 1.00 bar.
Explanation:
