# Question b4dc0

Oct 11, 2016

The answer is (B) $\frac{8}{9}$

#### Explanation:

The idea here is that the partial pressures of the two gases will be proportional to the number of moles each gas contributes to the total number of moles present in the container.

You can take the molar masses of the two gases to be

M_ ("M CH"_ 4) = "16 g mol"^(-1)

M_( "H"_ 2) = "2 g mol"^(-1)

This means that for equal masses of methane and hydrogen gas, let's say $m$ grams, you will have

m color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16color(red)(cancel(color(black)("g")))) = (m/16)" moles CH"_4

m color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2color(red)(cancel(color(black)("g")))) = (m/2)" moles H"_2

Now, the partial pressure of a gas that's part of a gaseous mixture, as given by Dalton's Law of Partial Pressures, is equal to

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{P}_{\text{gas i" = "moles of i"/"total moles of gas" xx P_"total}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The ratio between the number of moles of gas $i$ and the total number of moles of gas present in the mixture is the mole fraction of $i$ in the mixture, ${\chi}_{i}$.

In this case, the total number of moles of gas present in the mixture is

$\left(\frac{m}{16}\right) \text{ moles" + (m/2)" moles" = ((9m)/16)" moles}$

This means that the mole fraction of hydrogen gas will be

chi_( "H"_ 2) = ( color(red)(cancel(color(black)(m)))/2 color(red)(cancel(color(black)("moles"))))/((9color(red)(cancel(color(black)(m))))/16 color(red)(cancel(color(black)("moles")))) = 1/2 * 16/9 = color(green)(bar(ul(|color(white)(a/a)color(black)(8/9)color(white)(a/a)|)))

This means that for a total pressure ${P}_{\text{total}}$, the partial pressure of hydrogen gas is

P_( "H"_ 2) = 8/9 xx P_"total"#