# Question #1499b

##### 1 Answer

#### Explanation:

The first thing to do here is figure out what the average molar mass of air is by using the molar masses of nitrogen gas,

This will then help you find a relaationship between the ideal gas law and the density of air.

So, if you know that air is

#M_"M air" = 80/100 * "28.0134 g/mol" + 20/100 * "32.0 g/mol"#

#M_"M air" = 22.411 + 6.4 = "28.81 g/mol"#

According to the ideal gas law equation, you have

#PV = n * RT" "# , where

*ideal gas constant*, equal to

Now, you should also know that the number of moles of a substance can be determined by using a sample of that substance and its molar mass.

#n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT#

Rearrange this equation to get

#PV * M_M = m * RT implies m/V = (P * M_M)/(RT)#

The ratio between the mass of a substance and the volume it occupies is actually equal to its density,

#rho = (P * M_M)/(RT)#

Plug in your values to find the value of

#rho = (1.000color(red)(cancel(color(black)("atm"))) * 28.81"g"/color(red)(cancel(color(black)("mol"))))/(0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "1.178 g/L"#

I'll leave the answer rounded to three sig figs

#rho = color(green)("1.18 g/L")#