Question

How many grams of CO2 are dissolved in a 1.00 L bottle of carbonated water at 298 K if the pressure used in the carbonation process was 1.8 bar? The density of water at this temperature is 998 kg⋅m−3. The Henry's law constant for aqueous solution of CO2 at this temperature is 1.65×103bar.

Answer 1

Using Henry's Law, the solubility of CO2 at 1.8 bar and 298 K is found to be 2970 g/m3. Since there is 1 L of water, this equates to 2.97 g of CO2 dissolved in the 1.00 L bottle of carbonated water.

Step-by-step explanation:

The question asks how many grams of CO2 are dissolved in a 1.00 L bottle of carbonated water at a pressure of 1.8 bar and 298 K using the Henry's Law constant for CO2. To find the concentration of CO2 in the water, we use Henry's Law which states that the solubility of a gas in a liquid is proportional to the pressure of that gas above the liquid. The Henry's Law equation is Sg = kH * Pg, where Sg is the solubility of the gas, kH is the Henry's Law constant and Pg is the partial pressure of the gas.

The density of water is given as 998 kg/m3, which we use to convert the volume of water to mass. Since 1 L of water has a mass of approximately 998 g and the Henry's Law constant for CO2 at 298 K is 1.65×103 bar, we can plug in our values: Sg = (1.65×103 bar)(1.8 bar) = 2970 g/m3 of CO2. Since we have 1 L of water, or 0.998 kg, we can convert to grams: 0.998 kg × 1000 g/kg = 998 g.

To find the grams of CO2 dissolved, we use the solubility in g/m3: 2970 g/m3 × 0.001 m3 = 2.97 g of CO2.