An atmospheric concentration of 380 ppm CO2 corresponds to a partial pressure of 0.00038 atm. What percentage of the CO2 originally dissolved in the solution in Part A remains in solution after the soft drink reaches equilibrium with the ambient atmosphere

8.44 * 10^-3 %

Explanation:

The solubility of CO2 gas in water is 0.15g/100 ml at a CO2 pressure of 760 mmHg.

Determine the percentage of the CO2 that dissolved originally in the solution that remains

Amount remaining ( Sg = kpg )

Sg = solubility of gas = 0.15 / 100 g/ml

pg. = partial pressure of gas = 760 mmHg = 1 atm

therefore ; K = 1.5 * 10^-3 gm^-1 atm^-1

solubility of gas inside container

Sg = 0.675 * 10^-2 g/ml

solubility of gas during/after dissolution ( amount that remains )

sg = 1.5 * 10^-3 * 0.00038

    =  5.7 * 10^-7  g/ml

therefore the percentage remaining

= (5.7 * 10^-7  / 0.675 * 10^-2 ) * 100

= 8.44 * 10^-3 %

a is the answer to this question

explanation: