Question

What is the molecular weight of sulfur if 35.5 grams of sulfur dissolve in 100.0 grams of CS2 to produce a solution that has a boiling point of 49.48°C?

Answer 1

The molecular mass of sulfur is 256 u.

As you might have guessed, this is a boiling point elevation problem. The formula for boiling point elevation is

`ΔT_"b" = K_"b"m`

We need to look up the values of `T_"b"^"o"` and `K_"b"` for CS₂. They are

`T_"b"^"o"` = 46.2 °C and `K_"b"` = 2.37 °C·kg/mol

`ΔT_"b" = T_"b"^"o" – T_"b"` = (49.48 – 46.2) °C = 3.28 °C

We can use this information to calculate the molality `m` of the solution.

`m = (ΔT_"b")/K_"b" = (3.28" °C")/(2.37" °C·kg·mol⁻¹")` = 1.38 mol·kg⁻¹

We have 100.0 g or 0.1000 kg of CS₂. So we can calculate the moles of sulfur.

Molality = `"moles of [solute](http://socratic.org/chemistry/solutions-and-their-behavior/solute)"/"kilograms of [solvent](http://socratic.org/chemistry/solutions-and-their-behavior/solvent)"`

Moles of solute = molality× kilograms of solvent = 1.38 mol·kg⁻¹ × 0.1000 kg = 0.138 mol

So 35.5 g of sulfur = 0.138 mol

∴ Molar mass of sulfur = `(35.5" g")/(0.138" mol")` = 257 g/mol

∴ The experimental molecular mass of sulfur is 257 u (The actual molecular mass of S₈ is 256 u).