Question

3rd attempt
calculate the solubility of argon in water at an atmospheric pressure of 0.440 atm (a typical value at high altitude).
2nd attempt
1st attempt

Explanation:

Step1: Calculate the partial pressure of argon

According to Dalton's law of partial pressures, \(P_{Ar}=\chi_{Ar}\times P_{total}\). Here, \(\chi_{Ar} = 9.34\times 10^{-3}\) (mole fraction of \(Ar\)) and \(P_{total}=0.440\ atm\).
So, \(P_{Ar}=9.34\times 10^{-3}\times0.440\)

$$P_{Ar}=9.34\times0.440\times 10^{-3}=4.1096\times 10^{-3}\ atm$$

Step2: Use Henry's law to find the solubility of argon

Henry's law is \(S = k_{H}\times P\). For argon, \(k_{H}=1.40\times 10^{-3}\ mol/(L\cdot atm)\) and \(P = P_{Ar}=4.1096\times 10^{-3}\ atm\)

$$S=(1.40\times 10^{-3}\ mol/(L\cdot atm))\times(4.1096\times 10^{-3}\ atm)$$

$$S = 1.40\times4.1096\times10^{-3}\times10^{-3}\ mol/L$$

$$S=5.75344\times 10^{-6}\ mol/L\approx5.75\times 10^{-6}\ mol/L$$

Answer:

\(5.75\times 10^{-6}\ mol/L\)