Question

In a cylinder their are 60g ne and 64g o2. if pressure of mixture of gases in cylinder is 30 bar then in this cylinder partial pressure of o2 is

Answer 1

Given Conditions ⇒

Mass of the Neon = 60 g.
Molar Mass of the Neon = 20 g/mole.

∵ No. of moles = Mass/Molar Mass
∴ No. of moles of Neon = 60/20
∴ No. of moles of Neon = 3 moles.

Mass of the Oxygen given = 64 g.
Molar Mass of the Oxygen(O₂) = 32 g/mole.
[2 × 16 = 32]

∴ No. of moles = Mass/Molar Mass.
∴ No. of moles of O₂ = Mass/Molar Mass.
∴ No. of moles of O₂ = 64/32
No. of moles of Oxygen = 2 moles.

Now, Total number of moles in the cylinder = No. of moles of Ne + No. of moles of O₂
= 3 + 2
= 5 moles.

We know,
Partial Pressure of the Oxygen in the Cylinder = Mole Fraction of the Oxygen in the Cylinder.
= Moles of Oxygen/Total Number of moles in the Cylinder.
= 2/5
= 0.4.

Hence, the Partial Pressure of the Oxygen in the cylinder is 0.4.


Hope it helps.

Answer 2

Hi.

For Oxygen, 
Mass of the Oxygen = 64 g.
Molar Mass of the Oxygen = 32 g/mole.

∵ No. of moles of Oxygen = Mass/Molar mass
 ∴ No. of moles of Oxygen = 64/32
 ⇒ No. of moles of Oxygen = 2 moles.

For Neon,
Mass of the Neon = 60 g.
Molar Mass of the Neon = 20 g/mole.

∵ No. of moles = Mass/Molar mass
∴ No. of moles = 60/20
∴ No. of moles = 3 moles.

Now, Total Number of Moles in the Cylinder = Moles of Oxygen + Moles of Neon.
= 2 + 3= 5 moles.

Now,
We know, Partial Pressure of the Oxygen = Mole Fraction of the Oxygen.

∵ Mole Fraction of the Oxygen in the Cylinder = No. of moles of the Oxygen/Total number of moles ion the solutions.

∴ Mole Fraction of the Oxygen = 2/5

∴ Mole Fraction of the Oxygen = 0.4

 Hence, Partial Pressure of the Oxygen in the Cylinder is 2.5.


Hope it helps.