[guilherme]

Hey guys, I've been struggling with a question my teacher assigned me. This is it:
A balloon containing air is at the pressure of one atmosphere and at a temperature of 0 degrees Celsius. Assuming that the molar fraction of oxygen contained in the balloon is 0,20 and that the number of oxygen molecules in the baloon is 2,5 · 10^24, calculate the volume of air present in the baloon.
The answer must be given in dm^3.

Can you help me figure it out? I'd appreciate that so much!

[guilherme]

Edit: 2,5 · 10^4 to 2,5 · 10^24

[Borek]

You have to show your attempts at solving the problem to receive help, this is a forum policy.

You definitely should know some formulas/laws related to the problem, can you list them?

[guilherme]

You have to show your attempts at solving the problem to receive help, this is a forum policy.

You definitely should know some formulas/laws related to the problem, can you list them?

Oh, I'm so sorry. I'm a newbie. Of course I attempted to solve it.
First, 0 degrees Celsius and 1 atmosphere equals Vm=22.4dm^3·mol^-1
Next, I thought that if N(0₂)=2,5·10^24, then N(O)=5·10^24 (the molar fraction is of oxigen atoms).
Then, I discovered the moles of O (N=n·Na <=> n=N/Na <=> n=0,83·10^-2<=>n=0,0083 mol
I know it is supposed to use Vm=V(air)/n(air), but I don't know how to find the n(air).

[hypervalent_iodine]

You have to show your attempts at solving the problem to receive help, this is a forum policy.

You definitely should know some formulas/laws related to the problem, can you list them?

Oh, I'm so sorry. I'm a newbie. Of course I attempted to solve it.
First, 0 degrees Celsius and 1 atmosphere equals Vm=22.4dm^3·mol^-1
Next, I thought that if N(0₂)=2,5·10^24, then N(O)=5·10^24 (the molar fraction is of oxigen atoms).
Then, I discovered the moles of O (N=n·Na <=> n=N/Na <=> n=0,83·10^-2<=>n=0,0083 mol
I know it is supposed to use Vm=V(air)/n(air), but I don't know how to find the n(air).

I'm a little confused as to why you are calculating the number of oxygen atoms present. I would recommend going back and re-looking at the definition of molar fractions.

[MNIO]

how's your dimensional analysis these days?  Can you follow along?

moles air

    2.5x10^24 O2 molecules      100 molecules air             1 mol air molecules
   ------------------------------ x --------------------- x -------------------------------- = 20.8mol
                      1                       20 molecules O2       6.022x10^23 molecules air

then
   PV = nRT
   V = nRT/P
   V = 20.8 mol air * 0.08206 dm³atm/molK * 273.15K / 1 atm = 470dm³ air... (to 2 sig figs)

[guilherme]

Yeah, I think I understood. I found a different, more laborious way of doing it:
1 atm and 0 ºC -> NTP -> V_m = 22,4 dm³mol^-1

N=n·Na <=> n=2,5·10²⁴/6.022·10²³ <=> n=4,151 mol

1 mol of air    0,2 mol of O₂
n_air  4,15 mol of O₂
n_air=4,15/0,2=20,76 mol

V_m= V/n <=> V=22,4 x 20,76 <=> V=465,02 dm³