# Question #ad366

##### 1 Answer

#### 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,

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

#"For CH"_4: " 16.04 g/mol" ~~ "16 g/mol"# #"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

This means that the mole fraction of methane will **always** be equal to **one part** **nine parts** of mixture.

You can prove this by using

#"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#

#"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_"total" = m/16 + m/2 = (m * (16 + 2))/(16 * 2)#

The mole fraction of methane will once again be

#chi_(CH_4) = color(red)(cancel(color(black)(m)))/color(blue)(cancel(color(black)(16))) * (color(blue)(cancel(color(black)(16))) * 2)/(color(red)(cancel(color(black)(m))) * (16 + 2)) = 2/18 = color(green)(1/9)#