The Henry's law constant for carbon dioxide dissolved in water at 298 K is 0.163* 10⁹ Pa. Determine the mole fraction of carbon dioxide in water in equilibrium with air at a pressure of 1 atm and temperature of 298 K, assuming the proportion of carbon dioxide in dry air is 0.04%.

Question

Quizplus · Accepted Answer

First, we need to convert the percent of carbon dioxide in dry air to a partial pressure. 
0.04% of 1 atm = 0.0004 atm

Next, we can use Henry's law equation to calculate the mole fraction:

P = K_H*x

where P is the partial pressure of the gas above the solution, K_H is the Henry's law constant, and x is the mole fraction of the gas in the solution.

We know P and K_H, so we can solve for x:

x = P/K_H = 0.0004 atm / (0.163 x 10^3 Pa) = 2.45 x 10^-6

To convert this to ppm (parts per million), we can multiply by 10^6:

2.45 x 10^-6 * 10^6 = 2.45 ppm

Therefore, the answer is B) 25 ppm.

The Henry's Law Constant for Carbon Dioxide Dissolved in Water