Question

Helium has a Henry’s law constant of 3.8 *10^-4 mol/kg *bar at 25 °C when dissolving in water. If the total pressure of gas (He gas plus water vapor) over water is 1.00 bar, what is the concentration of He in the water in grams per milliliter? (The vapor pressure of water at 25 °C is 23.8 torr.)

Answer 1

Sol:- From Henry's law , C_He = K_H P_He      ...............(1)

here C_He = concentration of He gas = ?

K_H = Henry's constant = 3.8 x 10^-4 mol/kg x bar (given)

Total pressure i.e P_T = P_He + P_{water vapours} = 1.00 bar (given)

P_{water vapour} = 23.8 torr (given) = 0.03173 bar   , ( beacuse 1 torr = 133.32 bar)

therefore partial pressure of He i.e P_He = P_T - P_{water vapours}

P_He = 1.00 bar - 0.03173 bar = 0.9683 bar

Now convert unit of K_H i.e mol/kg bar into g / mL bar

We know number of moles = given mass in g / gram molar mass

and gram molar mass of He gas = 4g/mol

therefore 3.8 x 10^-4 mol of He = 4 g/mol x 3.8 x 10^-4 mol = 1.52 x 10^-3 g of He gas .

also mass of water = 1 kg ( given)

and density of water at 25 ⁰C = 1 kg/L = 10^-3 kg/mL

volume of water = mass of water / density of water

volume of water = 1 kg / 10^-3 Kg mL^-1 = 10³ mL

K_H= 1.52 x 10^-3g / 10³ mL bar = 1.52 x 10^-6 g/mL bar

Now from equation (1) , we have

C_He = 1.52 x 10^-6 g /mL bar x 0.9683 bar

C_He = 1.47 x 10^-6 g / mL

Hence concentration of He in water = 1.47 x 10^-6 g/mL