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?

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Answer 1 · 2014-05-17T02:26:45

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

$Δ T_{b} = K_{b} m$ $ΔT_"b" = K_"b"m$

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

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

$Δ T_{b} = T_{b}^{o} - T_{b}$ $ΔT_"b" = T_"b"^"o" – T_"b"$ = (49.48 – 46.2) °C = 3.28 °C

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

$m = \frac{Δ T_{b}}{K_{b}} = \frac{3.28 °C}{2.37 °C\cdotkg\cdotmol⁻¹}$ $m = (ΔT_"b")/K_"b" = (3.28" °C")/(2.37" °C·kg·mol⁻¹")$ = 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"/"kilograms of 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 = $\frac{35.5 g}{0.138 mol}$ $(35.5" g")/(0.138" mol")$ = 257 g/mol

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