Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 ∘ C and 721 mmHg?

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT ... Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass ... Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass) RT

⇒ Mass/molar mass = PV/RT ... Equation 3

But mass = Density * Volume

⇒ M = D * V ... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT ... Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT ... Equation 6

Cross multiplying equation 6

D * RT = P * molar mass

∴ Molar mass = (D * RT) / P ... Equation 7

Where D = 0.518 g/L, R = 0.0821 atm dm³/K. mol,

T = 25°C = 25 + 273 = 298 K,

P = 721 mmHg = (721/760) atm = 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 * 0.0821 * 298) / 0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is = 13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4 (y) + 32 (1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y = 0.666

y = 0.666 * 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%