We are asked to** calculate the mole fractions of each of the gases. **

*Mole Fraction (X)** relates the moles of components and the total moles:*

$\overline{){\mathit{m}}{\mathit{o}}{\mathit{l}}{\mathit{e}}{\mathbf{}}{\mathit{f}}{\mathit{r}}{\mathit{a}}{\mathit{c}}{\mathit{t}}{\mathit{i}}{\mathit{o}}{\mathit{n}}{\mathbf{}}{\mathbf{\left(}}{\mathit{X}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{component}}{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{}\mathit{t}\mathit{o}\mathit{t}\mathit{a}\mathit{l}}}$

Now to get the **total moles** we have:

$\overline{){{\mathit{n}}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{n}}}_{\mathbf{N}\mathbf{e}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{{\mathit{n}}}_{\mathbf{A}\mathbf{r}}}$

So for the **mole fraction of Ne and Ar** we have:

$\overline{){{\mathit{X}}}_{\mathbf{N}\mathbf{e}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{{\mathbf{n}}_{\mathbf{N}\mathbf{e}}}{{\mathbf{n}}_{\mathbf{N}\mathbf{e}}\mathbf{+}{\mathbf{n}}_{\mathbf{A}\mathbf{r}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathit{X}}}_{{\mathbf{Ar}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{{\mathbf{n}}_{\mathbf{Ar}}}{{\mathbf{n}}_{\mathbf{Ne}}\mathbf{+}{\mathbf{n}}_{\mathbf{Ar}}}}\phantom{\rule{0ex}{0ex}}$

In the flask diagrammed below, the left bulb contains 3.0L of neon at a pressure of 2.0 atm. The right bulb contains 2.0L of argon at a pressure of 1.0 atm. What is the mole fraction of neon and argon in the final mixture after the valve is opened and a uniform pressure is reached at constant temperature (25°C).

(a) Ne 0.33, Ar 0.67

(b) Ne 0.25, Ar 0.75

(c) Ne 0.67, Ar 0.33

(d) Ne 0.92, Ar 0.82

(e) Ne 0.75, Ar 0.25

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