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

a. 18 g/L

b. 1.5g/L

c. 45 g/L

d. 0.14 g/L

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

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.