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 : The solubility is,

Explanation :  Given,

= Henry's law constant  of argon =

First we have to calculate the pressure of argon.

Pressure of argon =

Pressure of argon = 0.0093 atm

Now we have to calculate the solubility.

As, the solubility of argon in 1 atm pressure =

So, the solubility of argon in 0.0093 atm pressure =

=

Thus, the solubility is,

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