# 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?

Dec 10, 2016

${\chi}_{\text{dinitrogen}}$ $=$ $0.64$

#### Explanation:

${\chi}_{i} = \frac{{n}_{i}}{{n}_{\text{total}}}$.

Now ${n}_{\text{total}} = \frac{1.19 \cdot g}{28.01 \cdot g \cdot m o {l}^{-} 1} + \frac{0.79 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1}$

$= 6.67 \times {10}^{-} 2 \cdot m o l$

And ${\chi}_{\text{dinitrogen}}$ $=$ $\frac{\frac{1.19 \cdot g}{28.01 \cdot g \cdot m o {l}^{-} 1}}{6.67 \times {10}^{-} 2 \cdot m o l} = 0.64$.

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

Note that the information given with respect to $\text{volume}$ and $\text{temperature}$ was included to distract you.