Question

At 500 °C and 93.2 kPa, the mass density of sulfur vapour is 3.710 kg m". What is the molecular formula of +. sulfur under these conditions?

Answer 1

Answer:

temperature of the sulphur vapour is T = 500o C=500 273 = 773K P = 93.2 kPa = 0.919atm mass density = mass/volume = 3.710 kgm-3 = 3.71g/L to calculate the molar mass of gas P = mass density(g/L)*RT/molar mass molar mass(g/mol)...

Answer 2

The question is:

At 500 °C and 93.2 kPa, the mass density of sulfur vapour is 3.710 kg/m³. What is the molecular formula of sulfur under these conditions?

The molecular formula of sulfur under these conditions is S₇.

Given:

Temperature= T= 500 °C= 500 °C + 273.15= 773.15K

Pressure= P= 93.2 kPa= 93.2 kPa × 1000= 93200 Pa

Mass density= ρ= 3.710 kg/m³

Universal gas constant= R= 8.314 K/mol K

To find:

The molecular formula of sulfur under these conditions.

Solution:

• To find the the molecular formula of sulfur under these conditions we have to find the emperical formula of sulfur under these conditions.
• To find emperical formula we need molar mass and molecular mass of sulfur.

Molar mass of sulfur can be determined using the formula:

PV= RT

P(m/ρ)= RT

m= RTρ/ P

m= $\frac{(8.314)(773.15)(3.710)}{93200}$

m= 0.250kg/mol

m= 0.250 × 1000 g/mol

m= 225 g/mol

Now, we have to determine the emperical formula using the formula:

Emperical formula= $\frac{Molar mass}{Molecular mass}$

Emperical formula= $\frac{225}{32.064}$

Emperical formula= 7

So, the emperical formula of sulfur under these conditions is 7.

Therefore, the molecular formula of sulfur under these conditions is S₇.

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