##”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)|)))##
