Question

What is the density of 2.5 g of gaseous sulfur held at 130. kPa and 10.0 degrees Celsius?

Answer 1

Assuming that the gaseous sulfur will behave as an ideal gas,
(PV)/T = constant
(P₁V₁)/T₁ = (P₂V₂)/T₂
(101.3 x 22.4)/(273) = (130 x V₂)/(283)
V₂ = 18.1 L
density = mass/volume
density = 2.5/18.1
= 0.14 g/L

Answer 2

Use ideal gas equation: pV = nRT

Now pass n to mass: n = mass / MM .... [MM is the  molar mass]

pV = [mass/MM]*RT =>mass/V = [p*MM] / RT  and mass / V = density

p= 130 kPa = 130,000 Pa = 130,00 joule / m^3
T = 10.0 ° + 273.15 = 283.15 k
MM of sulfur (S) = 32 g/mol  = 32000 kg/mol

density = 130,000 Pa * 32000kg/mol / [8.31 joule / mol*k * 283.15 k] = 1.77*10^6 kg/m^3 = 1.77 g/L  ≈ 1.8 g/L

Then, I do not get any of the option choices.

Is it possbile that the pressure is 13.0 kPa instead 130. kPa? If so the answer would be 18 g/L

Note that the mass is not used. You do not need it unless you are asked for the volume, which is not the case.