##rho = “1.18 g/L”##

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, ##”N”_2##, and oxygen gas, ##”O”_2##.

This will then help you find a relaationship between the and the of air.

So, if you know that air is ##80%## nitrogen and ##20%## oxygen, you can say that its molar mass will depend on those proportions

##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 equation, you have

##PV = n * RT” “##, where

##P## – the pressure of the gas; ##V## – the volume of the gas; ##n## – the number of moles of gas; ##R## – the ideal gas constant, equal to ##0.082″atm L”/”mol K”## ##T## – the temperature of the gas.

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 ##m/V## on one side

##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##. Therefore,

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

Plug in your values to find the value of ##rho##

##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

##rho = color(green)(“1.18 g/L”)##