Question 1cee1

Jun 6, 2016

$\text{7.07 mmHg}$

Explanation:

The idea here is that you can determine the partial pressure of argon by using Dalton's Law of Partial Pressures, which states that the partial pressure of a gas $i$, ${P}_{i}$, that's part of a gaseous mixture is equal to

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{i} = {\chi}_{i} \times {P}_{\text{total}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${\chi}_{i}$ - the mole fraction of gas $i$ in the mixture
${P}_{\text{total}}$ - the total pressure of the mixture

Now, you know that argon makes up 0.93% by volume of the atmosphere, which is equivalent to saying that for every $\text{100 L}$ of air, you have $\text{0.93 L}$ of argon.

Since volume is directly proportional to number of moles at constant temperature and pressure -- think Avogadro's Law here -- you can say that you have $0.93$ moles of argon for every $100$ moles of air.

This means that the mole fraction of argon, which is defined as the ratio between the number of moles of argon and the *total number of moles8 of gas present in a given sample of air, will be

chi_"argon" = (0.93 color(red)(cancel(color(black)("moles"))))/(100color(red)(cancel(color(black)("moles")))) = 0.0093

The partial pressure of argon in air will be

P_"argon" = 0.0093 * "760 mmHg" = color(green)(|bar(ul(color(white)(a/a)color(black)("7.07 mmHg")color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.