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 $N_2$ ?

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Answer 1 · 2016-12-10T19:20:14

$chi_"dinitrogen"$ $=$ $0.64$

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

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.

A gas mixture contains 1.19 g N_2N_2 and 0.79 g O_2O_2 in a 1.60-L container at 25 deg C. How do you calculate the mole fraction of N_2N_2?