# Question c8a2e

Mar 21, 2016

${\chi}_{{N}_{2}} = 0.574$

#### Explanation:

The idea here is that you need to use the ideal gas law equation to determine how many moles of gas are present in that sample.

Once you know the total number of moles present in the sample, use the molar masses of the two gases to find a relationship between the mass of each gas and the total mass of the gaseous mixture.

So, the ideal gas law equation looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Rearrange to solve for $n$, the total number of moles of gas present in the mixture

$P V = n R T \implies n = \frac{P V}{R T}$

Now, do not forget that the units must match those used in the expression of the ideal gas constant, $R$. This means that you must convert the pressure from mmHg to atm and the temperature from degrees Celsius to Kelvin

Use the conversion factors

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 atm " = " 760 mmHg}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$ $\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

$n = \left(\frac{870.2}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 15.1color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 31.2)color(red)(cancel(color(black)("K}}}}\right)$

$n = \text{0.6919 moles}$

So, you know that the mixture contains a total of $0.69129$ moles of gas and that it has a total mass of $\text{24.1 g}$.

Now, if you take $x$ to be the number of moles of nitrogen, ${\text{N}}_{2}$, and $y$ to be the number of moles of carbon dioxide, ${\text{CO}}_{2}$, you can use the molar masses of the two gases to write the mass each gas contributes to the total

m_(N_2) = x color(red)(cancel(color(black)("moles N"_2))) * "28.01 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = 28.01 * x color(white)(a)"g"

m_(CO_2) = y color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = 44.01 * ycolor(white)(a)"g"#

You can now write two equations with two unknowns

$x + y = 0.6919$

$28.01 \cdot x + 44.01 \cdot y = 24.1$

This will get you

$x = 0.6919 - y$

$28.01 \cdot \left(0.6919 - y\right) + 44.01 y = 24.1$

$19.38 - 28.01 y + 44.01 y = 24.1$

$16 y = 4.72 \implies y = \frac{4.72}{16} = 0.295$

The value of $x$ will be

$x = 0.6919 - 0.295 = 0.3969$

So, the mixture contains $0.3969$ moles of nitrogen gas and $0.295$ moles of carbon dioxide.

To get the mole fraction of nitrogen gas, divide the number of moles of nitrogen gas by the *total number of moles8 present in the mixture

${\chi}_{{N}_{2}} = \left(0.3969 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(0.6919color(red)(cancel(color(black)("moles}}}}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 0.574 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.