Question

The Henry’s law constant for Ar(g) at 25 o C is 1.4 × 10 - 3 M/atm. What mass (i n mg) of Ar is dissolved in a typical 8.0 - fl. oz. glass of water at 25 o C and at 1.00 atm total pressure? Argon constitutes about 1.0% of the atmosphere.

Answer 1

partial Pressure of Ar = Total Pressure of atmosphere x Argon % / 100 = 1 x ( 1/100) = 0.01 atm

Now Henery law is   Kh = C / P   where Kh is henry constant = 1.4 x 10^ -3 M/atm

C = solubility,   P = partial Pressure = 0.01 atm

C = Kh x P = 1.4x10^ -3 x 0.01 = 1.4 x 10^ -5 M

now water volume = 8 fl.oz   = 8 x 29.5735 ml   = 236.6 ml = 0.2366 L ( 1fl.oz = 29.575 ml)

Molarity = moles / volume in L

1.4 x10^ -5 = ( moles of Ar / 0.2366)

Moles ofAR dissolved = 3.3 x 10^ -6

Ar mass = moles x atomic mass of AR = 3.3 x 10^ -6 x 39.948 = 0.0001323 g = 0.1323 mg