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?

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: