How many grams of CO 2 are dissolved in a 1.00 L bottle of carbonated water at 298 K if the pressure used in the carbonation process was 2.4 bar? What volume would the dissolved CO 2 occupy as a gas at 298 K and 1.00 bar?



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

bar

K

1

mol

1

0.0831L⋅bar⋅K

−1

⋅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

1.00

L

0.0831

L

bar

K

1

mol

1

298

K

n=

0.0831L⋅bar⋅K

−1

⋅mol

−1

⋅298K

2.4bar⋅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

bar

K

1

mol

1

298

K

1.00

bar

V=

1.00bar

0.0996moles×0.0831L⋅bar⋅K

−1

⋅mol

−1

⋅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: