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# Calculate the density of nitrogen gas, in grams per liter, at STP?

##"1.25 g L"^(-1)##

The idea here is that you need to use the equation and the **molar mass** of nitrogen gas, ##"N"_2##, to find a relationship between the mass of a sample of nitrogen gas and the volume it occupies at **STP**.

Now, **STP conditions** will most likely be given to you as a pressure of ##"1 atm"## and a temperature of ##0^@"C"##, or ##"273.15 K"##.

The equation looks like this

##color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "##, where

##P## - the pressure of the gas
##V## - the volume it occupies
##n## - the number of moles of gas
##R## - the universal gas constant, usually given as ##0.0821("atm" * "L")/("mol" * "K")##
##T## - the **absolute temperature** of the gas

As you know, a compound's **molar mass** tells you the mass of **one mole** of that substance. IN your case, nitrogen gas has a molar mass of

##M_M = "28.0134 g mol"^(-1)##

This means that **every mole** of nitrogen gas has a mass of ##"28.0134 g"##.

You can thus replace the number of moleso f nitrogen gas, ##n##, in the ideal gas law equation by the ratio between a mass ##m## and the **molar mass** of the gas

##n = m/M_M##

Plug this into the ideal gas law equation to get

##PV = m/M_M * RT##

Rearrange to get

##PV *M_M = m * RT##

This will be equivalent to

##P * M_M = m/V * RT##

But since , ##rho##, is defined as mass per unit of volume, you can say that you have

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

and therefore

##color(purple)(|bar(ul(color(white)(a/a)color(black)(rho = (P * M_M)/(RT))color(white)(a/a)|)))##

Now all you have to do is use the STP values for pressure and temperature and the molar mass of nitrogen gas to get

##rho = (1 color(red)(cancel(color(black)("atm"))) * "28.0134 g" color(red)(cancel(color(black)("mol"^(-1)))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 273,15color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))##

##color(white)(a)##**ALTERNATIVE APPROACH**

As you know, **one mole** of any ideal gas kept at ##"1 atm"## and ##0^@"C"## occupies ##"22.4 L"## -- this is known as the .

Moreover, you know that **one mole** of nitrogen gas has a mass of ##"28.0134 g"##. Since density tells you mass **per unit of volume**, all you have to do now is determine the mass of one liter of nitrogen gas at STP

##1 color(red)(cancel(color(black)("L N"_2))) * overbrace((1 color(red)(cancel(color(black)("mole N"_2))))/(22.4color(red)(cancel(color(black)("L N"_2)))))^(color(purple)("molar volume of a gas at STP")) * overbrace("28.0134 g"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(blue)("molar mass of N"_2)) = "1.25 g"##

Since this is how many grams you get **per liter** of nitrogen gas at STP, it follows that its density will be

##rho = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))##