# Average molecular weight?

by tag16
Tags: average, molecular, weight
 P: 97 I just assumed what I originally thought the volume was must be wrong or you wouldn't asked about the volume so I just came up with something random because I didn't know what else it could be, hence the 250 cm^3. So I have to do the calculations for each of the three bulbs seperately as well as for each temperature, correct?
P: 23,540
To quote the document you have read:

 Treat each bulb separately, do not average the values from the two bulbs.
 P: 97 So the volume of the bulbs using V= mass (w/water)/ water density bulb 1: 397.122 bulb 2: 362.841 bulb 3: 398.928 Which brings me back to my problem of finding the average molecular weight of the gas mixture at each temperature. I originally thought I was suppose to use this equation: M(mix)=x(NO2)M(NO2)+x(N2O4)M(N2O4), is that right?
 Admin P: 23,540 These are not volumes, volumes should be much closer to 250. You forgot to subtract mass of empty bulb. Calculate volumes, calculate masses of gas mixtures. Knowing mass, volume, temperature and pressure you should be able to calculate average molar mass, later you can use this number to calculate molar fractions of both gases.
 P: 97 opps thats what I did originally but I also averaged them all together which was wrong. This is minus the mass of the bulb: bulb1:254.882 bulb2:263.277 bulb3:284.049
 P: 97 I'm not really sure what to do next, I think I'm suppose to use this equation: M(mix)=x(NO2)M(NO2)+x(N2O4)M(N2O4), if so I'm not sure how to get the mole fractions from the data I have or the degree of dissociation.
 Admin P: 23,540 Calculate volumes, calculate masses of gas mixtures. Knowing mass, volume, temperature and pressure you should be able to calculate average molar mass, later you can use this number to calculate molar fractions of both gases.
 P: 97 so I used M=DRT/P to find the average molecular mass and got this: For bulb 1: P=717.41 Torr= 0.94396 atm V= 254.882 cm^3 T=303.15 K mass= 141.9780 g M=(141.9780 g/254.882 cm^3)(0.0821)(303.15 K)/(0.94396 atm) ave. molecular mass of bulb 1: 14.686 Am I going about this the right way or am I completely off?
 Admin P: 23,540 You are off. Perhaps not completely, but seriously. First - 141.8780 g can't be mass of the gas. I guess you forgot to subtract bulb mass again. Second - your volume units are inconsistent with R units.
 P: 97 For bulb 1: P=717.41 Torr= 0.94396 atm V= 254.882 cm^3 T=303.15 K mass= 141.9780 g(bulb w/gas)-141.229 g(bulb empty)= 0.749 g M=(0.749g/254.882 cm^3)(8.314)(303.15 K)/(0.94396 atm) ave. molecular mass of bulb 1: 7.846g
 P: 97 opps I think I used the wrong gas constant again: M=(0.749g/254.882 cm^3)(82.057)(303.15 K)/(0.94396 atm) ave. molecular mass of bulb 1: 77.4395 g
P: 23,540
 Quote by tag16 ave. molecular mass of bulb 1: 77.4395 g
That looks OK.

Now, write the equation for average molar mass using x & y for molar fractions of both gases. You know that x+y=1, that means you have two equations and two unknowns. Solve and you will know how much of each gas is present.
 P: 97 x + y = 1 92.011 x + 46.0055 y = 77.4395 x= -y + 1 92.011 (-y +1) + 46.0055 y = 77.4395 -92.011 y + 92.011 + 46.0055 y = 77.4395 -46.0055 y + 92.011= 77.4395 -46.0055 y = -14.5715 y = 0.316734 1- y = x x = 0.683266 To find the degree of dissociation I did this: n(N2O4)=77.4395/92=0.841 mol degree dissociation: x(N2O4)/n(N2O4)= 0.683266/0.841= 0.812445= 81.24% To find Kp I did: n(N2O4)=77.4395/92=0.841 mol n(NO2)= 77.4395/46=1.683 mol Kp= n(N2O4)/(n(NO2))^2= 0.841/(1.683)^2= 0.296912 I think to find delta G standard I'd want to use this equation: delta G= -RTlnKp and for delta S: delta S= delta G/T Did I go about any of that right or is that not what I'm suppose to do?
P: 23,540
 Quote by tag16 y = 0.316734 1- y = x x = 0.683266
Looks OK.

 To find the degree of dissociation I did this: n(N2O4)=77.4395/92=0.841 mol degree dissociation: x(N2O4)/n(N2O4)= 0.683266/0.841= 0.812445= 81.24%
So you have calculated composition of the mixture only to ignore it now and not use the information in following calculations?
 P: 97 Is this right? If not I don't know how to find it. PV = nRT --> n =PV/RT--> n = (1 − α)n + 2αn = PV/RT α = (PV/nRT)− 1