Question

Kh for CO2 in water is 1.67X10^8 pa at 298K. Calculate the mass of CO2 that can be dissolved in 500 ml of water at a pressure of 2.5 atmosphere at 298 K. (1atm = 1.013X10^5Pa)

Answer 1

Henry's law:

$.\ \ \ partial\ pressure\ p=k_H\ c,\\. \ \ \ \ \ \ where\ c=concentration\ of\ gas\ in\ solution,\\.\ \ \ \ \ \ k_H=Henry's\ constant$

$k_H=1.67*10^8\ Pa\ at\ 298\ {}^0 K\\p=2.5\ atm=2.5*1.013*10^5\ Pa\\\\c=\frac{p}{k_H}=\frac{2.5*1.013*10^5}{1.67*10^8}=1.516*10^{-3}\ \frac{moles}{Litre}\\\\mass=c*volume\ of\ solution=1.516*10^{-3}\ \frac{mole}{Litre}*\frac{500}{1000}litre$

Molecular mass of carbon dioxide = 12+2*16=44

Mass of CO₂ dissolving in 500 ml of water at a pressure of 2.5 atm at 298 °K =
            44 * 0.758 * 10⁻³ = 33.352 mg.

Answer 2

Henry's law:





Molecular mass of carbon dioxide = 12+2*16=44

Mass of CO₂ dissolving in 500 ml of water at a pressure of 2.5 atm at 298 °K =
            44 * 0.758 * 10⁻³ = 33.352 mg.