Your starting point here will be the ideal gas law equation
#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#
Here you have
#P#- the pressure of the gas
#V#- the volume it occupies
#n#- the number of moles of gas
#R#- the universal gas constant
#T#- the absolute temperature of the gas
Rearrange the above equation to solve for
#PV = nRT implies R = (PV)/(nT)#
Now, notice that the problem doesn't provide you with the number of moles of nitrogen gas present in your sample. This means that you're going to have to improvise a bit.
As you know, the number of moles present in a sample of a given compound is equal to the mass of the sample divided by the molar mass of the compound.
If you take
#n = m/M_M#
Plug this into the above equation to get
#R = (PV)/(m/M_M * T) = (PV)/(m * T) * M_M#
Now, the problem tells you that the gas has a density of
You already used
#rho = m/V implies V/m = 1/(rho)#
Plug this into the equation to get
#R = P/T * 1/(rho) * M_M#
#R = "1 atm"/((273.15 + 0)"K") * 1/(1.250 color(red)(cancel(color(black)("g"))) "dm"^(-3)) * 28.0134 color(red)(cancel(color(black)("g"))) "mol"^(-1)#
#color(green)(|bar(ul(color(white)(a/a)color(black)(R = 0.082 ("atm" * "dm"^3)/("mol" * "K"))color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but don't forget that you only have one sig fig for the temperature and pressure of the sample.
The actual value of
#R ~~ 0.0820575("atm" * "dm"^3)/("mol" * "K")#
so this is an excellent result.