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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: INeedSerotonin on July 28, 2019, 04:29:24 PM

Title: Gaseous solution
Post by: INeedSerotonin on July 28, 2019, 04:29:24 PM

There is a gaseous solution of CO2 and O2. The temperature is 60,3 °C, and the pressure is 2,0 atm. If the density of the solution is 3,0 g/L, what is the percentage of O2 (in volume)?

Could you guys give me a hint on how to solve this exercise? I know that d = (P.M) / (R.T), and I found that the total mass is 41g, but I'm really totally lost here.

Thanks

Title: Re: Gaseous solution
Post by: AWK on July 28, 2019, 05:24:58 PM

Calculate M. Then the rest of your calculation will be quite easy.

Title: Re: Gaseous solution
Post by: INeedSerotonin on July 28, 2019, 05:49:54 PM

Quote from: AWK on July 28, 2019, 05:24:58 PM

Calculate M. Then the rest of your calculation will be quite easy.

I found that M = 41 g, so I tried to use d = M / V. So 3 = 41 / V .:. V = 13,6 L.

P.V = N.R.T

2 . 13,6 = n . 0,082 . 333,33 (60,3 °C in K)
n = 1

But I don't know how to continue. Could you please give me one more hint?

Title: Re: Gaseous solution
Post by: AWK on July 28, 2019, 06:03:35 PM

1 mole of mixture od two known gases weights 41 g (and shows mean molecular mass 41 g).

???

Title: Re: Gaseous solution
Post by: INeedSerotonin on July 28, 2019, 06:21:29 PM

Quote from: AWK on July 28, 2019, 06:03:35 PM

1 mole of mixture od two known gases weights 41 g (and shows mean molecular mass 41 g).

???

Thank you! I didn't know that the total mass of a gaseous solution is the mean molecular mass. I found this to be a hard exercise.

Here is how I solved it:

(32.x + 44.y)/100 = 41, where x is the percentage of O2 and y is the percentage of CO2.

32.x + 44.y = 4100

32.x + 44.(100-x) = 4100

x = 25

 :)