Dec 15, 2015

$\frac{1}{9}$

#### Explanation:

You can approach this problem in an intuitive manner, i.e. without doing any actual calculations, or in a practical manner, i.e. getting the answer via a series of calculations.

I'll show you both approaches so that you'll know how to solve similar problems in the future.

For a mixture of gases, the partial pressure exerted by each gas will be proportional to its mole fraction.

As you know, mole fraction is determined by dividing the number of moles of a given compound by the total number of moles present in that mixture.

You know that you're mixing equal masses of methane, ${\text{CH}}_{4}$, and hydrogen gas, ${\text{H}}_{2}$, so all you have to do now is figure out how many moles of each would such a mixture contain.

Now, the molar mass of a substance tells you what the mass of one mole of that substance is. For these two gases, you have

• $\text{For CH"_4: " 16.04 g/mol" ~~ "16 g/mol}$
• $\text{For H"_2: " 2.016 g/mol" ~~ "2 g/mol}$

What these values tell you is that you get eight times more moles of hydrogen gas than of methane per gram.

This implies that mixing equal masses of these two gases will be equivalent to mixing eight times more moles of ${\text{H}}_{2}$ than of ${\text{CH}}_{4}$.

This means that the mole fraction of methane will always be equal to $\textcolor{g r e e n}{\frac{1}{9}}$, since you get one part ${\text{CH}}_{4}$ for every nine parts of mixture.

You can prove this by using $m$ as the mass of each gas added to the mixture. The number of moles of each gas will be

${\text{For CH"_4: m color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16color(red)(cancel(color(black)("g")))) = m/16" moles CH}}_{4}$

${\text{For H"_2: m color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2color(red)(cancel(color(black)("g")))) = m/2" moles H}}_{2}$

The total number of moles will be

${n}_{\text{total}} = \frac{m}{16} + \frac{m}{2} = \frac{m \cdot \left(16 + 2\right)}{16 \cdot 2}$

The mole fraction of methane will once again be

${\chi}_{C {H}_{4}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{m}}}}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{16}}}} \cdot \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{16}}} \cdot 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \cdot \left(16 + 2\right)} = \frac{2}{18} = \textcolor{g r e e n}{\frac{1}{9}}$