Answer:
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%
