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QUESTION

# 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)|)))