Question

Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constant for Ar under these conditions is 1.5 × 10−3 mol/L·atm.

Answer 1

Answer : The solubility is, $1.4\times 10^{-5}mol/L$

Explanation :  Given,

$k_H$ = Henry's law constant  of argon = $1.5\times 10^{-3}mol/L.atm$

First we have to calculate the pressure of argon.

Pressure of argon = $1.0atm\times \frac{0.93}{100}$

Pressure of argon = 0.0093 atm

Now we have to calculate the solubility.

As, the solubility of argon in 1 atm pressure = $1.5\times 10^{-3}mol/L$

So, the solubility of argon in 0.0093 atm pressure = $\frac{0.0093atm}{1atm}\times 1.5\times 10^{-3}mol/L$

= $1.4\times 10^{-5}mol/L$

Thus, the solubility is, $1.4\times 10^{-5}mol/L$