[Shing]

The partial pressure of CH₄ is 0.175atm and partial pressure of O₂ is 0.250 atm in a mixture of gases.. what is the mole fraction of each gas in the mixture?

What I did was that i divided the partial pressure of each gas in the mixture by the total pressure.

Mole fraction of CH₄ : 0.175atm/ 0.425atm
Mole fraction of O₂: 0.250 atm /0.425atm

Is this correct??
Help

[Borek]

OK

[Shing]

OK

Does that mean its right?

[Borek]

It means its OK

Yes, looks right to me.

As usual, could mean I am missing something.

[Shing]

It means its OK

Yes, looks right to me.

As usual, could mean I am missing something.

The Whole Question is...

The partial pressure of CH4 is 0.175atm and partial pressure of O2 is 0.250 atm in a mixture of gases..
a) what is the mole fraction of each gas in the mixture?
b) If the volume of teh mixture is 10.5 L At 65 deg C, calculate total number of moles of gas in the mixture
c) Calculate grams of each gas in the mixture.

What I did:

partial presure / total pressure
a) mole fraction of CH4 := 17.7kPa/43.02kPa
                      of O2 := 25.3 kPa/ 43.02 KPa

b) V- 10.5 L
    T = 338 K
    P = 43.02 kPa (total)
so using PV=nRt
n= PV/RT
n= 0.160 mol

c) using the mole fraction
moles of CH4 present := 17.7/43.02 * 0.160 = 0.065 mol
moles of CH4 present := 25.3/43.02 * 0.160 = 0.094 mol

mass of CH4 = 0.065 mol * 16g/mol (molar mass) = 1.04 g
mass of O2 = 0.094mol * 32g/mol = 3.0 g

[Borek]

I got differences on last digits - for example total seems to be 0.161, not 0.160.

Generally OK.

I mean - right.

[Shing]

I got differences on last digits - for example total seems to be 0.161, not 0.160.

Generally OK.

I mean - right.

Thank you.