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

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.

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