Question

The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO2 dissolved in water is 1.65 x 103 atm, and the density of water is 1.0 g/cm3.

Answer 1

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

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The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.