Question

Question #1cee1

Answer 1

`"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

`color(blue)(|bar(ul(color(white)(a/a)P_i = chi_i xx P_"total"color(white)(a/a)|)))`

Here

`chi_i` - the mole fraction of gas `i` in the mixture
`P_"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 `"100 L"` of air, you have `"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.