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Naddika [18.5K]
2 years ago
15

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

Chemistry
2 answers:
marta [7]2 years ago
7 0
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
worty [1.4K]2 years ago
4 0
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.



 
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A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
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Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

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∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

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∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

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Answer:

k = 23045 N/m

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