# Question ecc4f

Jan 22, 2016

Here's what I got.

#### Explanation:

The most important thing to remember about gaseous mixtures is that the total pressure of the mixture will be equal to the sum of the partial pressures of the components of the mixture - this is known as Dalton's Law of Partial Pressure.

color(blue)(P_"total" = sum_i P_i)" ", where

${P}_{\text{total}}$ - the total pressure of the mixture
${P}_{i}$ - the partial pressure of gas $i$

The idea here is that isolating each individual gas in the same volume in which you'd keep the gaseous mixture would return the partial pressure of the gas in the mixture.

Now, an important consequence of Dalton's Law is that you can express the partial pressure of a gas that's part of a gaseous mixture by using

• the mole fraction of that gas in the mixture
• the total pressure of the mixture

Mathematically, you can say that

color(blue)(P_i = chi_i xx P_"total")" ", where

${\chi}_{i}$ - the mole fraction of gas $i$

The mole fraction of a gas that's part of a gaseous mixture is simply the ratio between the number of moles of that gas and the total number of moles present in the mixture.

Your mixture is said to contain three gases, nitrogen gas, ${\text{N}}_{2}$, oxygen gas, ${\text{O}}_{2}$, and argon, $\text{Ar}$. The total pressure of the mixture will thus be equal to

${P}_{\text{total}} = {P}_{{N}_{2}} + {P}_{{O}_{2}} + {P}_{A r}$

Plug in your values to get

${P}_{\text{total" = "320 torr" + "275 torr" + "285 torr}}$

P_"total" = color(green)("880 torr")#

To get the mole fraction of oxygen, you would use

${P}_{{O}_{2}} = {\chi}_{{O}_{2}} \cdot {P}_{\text{total" implies chi_(O_2) = P_(O_2)/P_"total}}$

Plug in your values to get

${\chi}_{{O}_{2}} = \left(275 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{torr")))) /(880color(red)(cancel(color(black)("torr}}}}\right) = 0.3125$

Rounded to two sig figs, the answer will be

${\chi}_{{O}_{2}} = \textcolor{g r e e n}{0.31}$