Calculate the density of HBr gas in g/L at 733 mmHg & 46 degrees C

PV = nRT
n = mass / mw

substitute and rearrange
PV = (mass / mw) RT
(mass / V) = mw x P/(RT)

since density = mass / volume...
density = mw x P/(RT)

solving..
density = (80.91g/mole) x (733mmHg x 1atm/760mmHg) / ((0.08206 Latm/moleK) x (319.15K))
density = 2.98g/L

There is a reason I did that. To avoid intermediate calculations and preserve sig figs & precision.

the way you enter it in your calculator is
80.91 x 733 / 760 / 0.08206 / 319.15 =

vs..
733/760 = 0.9645 with 1 extra sigfig for intermediate step
then.
80.91 x 0.9645 / 0.08206 / 319.15 =

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Start with the equation PV=nRT
n is number of moles.
number of moles is equal to mass/molar mass , or m/M

substitute this into the equation

PV=nRT

PV=(m/M)RT

PVM=mRT <------------------------------ (Memorize this equation along with PV=nRt, its very important)

now solve this for density
density is mass/volume so just divide m by V, and you have another equation

PVM=mRT=
PM=(m/V)RT=

PM=DRT <---------------------------------------… (another important equation to memorize)

Now just substitute the values in.
733 mm Hg = 0.964 atm
46 degrees C = 319 K
molar mass of HBr = 80.908 g/mol

substitute into the equation

PM=DRT
(0.964)(80.908)=D(0.08206)(319)
D= 2.98 g/L