# The atmospheric pressure at the summit of Mt. Everest is 255 torr. The percentage of #O_2# and #N_2# is 20% and 79%, respectively. How many #N_2# and #O_2# molecules are there in a volume of 10.0L at 12°C?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Your strategy here will be to use the ideal gas law equation to determine the **total number of moles** of air you have in that

Once you know how many moles of air you have, you can use the given percentages to find how many *moles* of nitrogen gas and oxygen gas you have.

Finally, **Avogadro's number** will allow you to convert the number of moles to number of *molecules*.

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "# , where

*universal gas constant*, usually given as

So, plug in your values and solve for *torr* to *atm* and the temperature from *degrees Celsius* to *Kelvin*

#PV = nRT implies n = (PV)/(RT)#

#n = (255/760 color(red)(cancel(color(black)("atm"))) * 10.0 color(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 + 12)color(red)(cancel(color(black)("K")))) = "0.1433 moles"#

Now, out of these moles of air, you know that

#0.1433 color(red)(cancel(color(black)("moles air"))) * "20 moles O"_2/(100color(red)(cancel(color(black)("moles air")))) = "0.02866 moles O"_2#

#0.1433 color(red)(cancel(color(black)("moles air"))) * "79 moles N"_2/(100color(red)(cancel(color(black)("moles air")))) = "0.1132 moles N"_2#

Now, as you know, in order to have **one mole** of a substance, you need exactly

In your case, the sample will contain

#0.02866 color(red)(cancel(color(black)("moles O"_2))) * (6.022 * 10^(23)"molecules of O"_2)/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(1.7 * 10^(22)"molecules of O"_2#

and

#0.1132 color(red)(cancel(color(black)("moles N"_2))) * (6.022 * 10^(23)"molecules of N"_2)/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(6.8 * 10^(22)"molecules of N"_2#

The answers are rounded to two sig figs.