Question

A gas mixture contains 1.19 g `N_2` and 0.79 g `O_2` in a 1.60-L container at 25 deg C. How do you calculate the mole fraction of `N_2`?

Answer 1

`chi_"dinitrogen"` `=` `0.64`

Explanation:

`chi_i=(n_i)/(n_"total")`.

Now `n_"total"=(1.19*g)/(28.01*g*mol^-1)+(0.79*g)/(32.00*g*mol^-1)`

`=6.67xx10^-2*mol`

And `chi_"dinitrogen"` `=` `((1.19*g)/(28.01*g*mol^-1))/(6.67xx10^-2*mol)=0.64`.

Given that it is a binary system, what is `chi_"dioxygen"?`

Note that the information given with respect to `"volume"` and `"temperature"` was included to distract you.